[{"video_title": "Slope and y intercept from equation.mp3", "Sentence": "What I'd like to do in this video is a few more examples recognizing the slope and y-intercept given an equation. So let's start with something that we might already recognize. Let's say we have something in the form y is equal to five x plus three. What is the slope and the y-intercept in this example here? Well, we've already talked about that we can have something in slope-intercept form where it has the form y is equal to the slope, which people use the letter m for, the slope times x plus the y-intercept, which people use the letter b for. So if we just look at this, m is going to be the coefficient on x right over there. So m is equal to five."}, {"video_title": "Slope and y intercept from equation.mp3", "Sentence": "What is the slope and the y-intercept in this example here? Well, we've already talked about that we can have something in slope-intercept form where it has the form y is equal to the slope, which people use the letter m for, the slope times x plus the y-intercept, which people use the letter b for. So if we just look at this, m is going to be the coefficient on x right over there. So m is equal to five. That is the slope. And b is just going to be this constant term, plus three. So b is equal to three."}, {"video_title": "Slope and y intercept from equation.mp3", "Sentence": "So m is equal to five. That is the slope. And b is just going to be this constant term, plus three. So b is equal to three. So this is your y-intercept. So that's pretty straightforward, but let's see a few slightly more involved examples. Let's say if we had form y is equal to five plus three x."}, {"video_title": "Slope and y intercept from equation.mp3", "Sentence": "So b is equal to three. So this is your y-intercept. So that's pretty straightforward, but let's see a few slightly more involved examples. Let's say if we had form y is equal to five plus three x. What is the slope and the y-intercept in this situation? Well, it might have taken you a second or two to realize how this earlier equation is different than the one I just wrote. Here, it's not five x, it's just five."}, {"video_title": "Slope and y intercept from equation.mp3", "Sentence": "Let's say if we had form y is equal to five plus three x. What is the slope and the y-intercept in this situation? Well, it might have taken you a second or two to realize how this earlier equation is different than the one I just wrote. Here, it's not five x, it's just five. And this isn't three, it's three x. So if you wanna write it in the same form as we have up there, you can just swap the five and the three x, doesn't matter which one comes first, you're just adding the two. So you could rewrite it as y is equal to three x plus five."}, {"video_title": "Slope and y intercept from equation.mp3", "Sentence": "Here, it's not five x, it's just five. And this isn't three, it's three x. So if you wanna write it in the same form as we have up there, you can just swap the five and the three x, doesn't matter which one comes first, you're just adding the two. So you could rewrite it as y is equal to three x plus five. And then it becomes a little bit clearer that our slope is three, the coefficient on the x term, and our y-intercept is five, y-intercept. Let's do another example. Let's say that we have the equation y is equal to 12 minus x."}, {"video_title": "Slope and y intercept from equation.mp3", "Sentence": "So you could rewrite it as y is equal to three x plus five. And then it becomes a little bit clearer that our slope is three, the coefficient on the x term, and our y-intercept is five, y-intercept. Let's do another example. Let's say that we have the equation y is equal to 12 minus x. Pause this video and see if you can determine the slope and the y-intercept. All right, so something similar is going on here that we had over here. The standard form, slope-intercept form, we're used to seeing the x term before the constant term."}, {"video_title": "Slope and y intercept from equation.mp3", "Sentence": "Let's say that we have the equation y is equal to 12 minus x. Pause this video and see if you can determine the slope and the y-intercept. All right, so something similar is going on here that we had over here. The standard form, slope-intercept form, we're used to seeing the x term before the constant term. So we might wanna do that over here. So we could rewrite this as y is equal to negative x plus 12, negative x plus 12. And so from this, you might immediately recognize, okay, my constant term, when it's in this form, that's my b, that is my y-intercept."}, {"video_title": "Slope and y intercept from equation.mp3", "Sentence": "The standard form, slope-intercept form, we're used to seeing the x term before the constant term. So we might wanna do that over here. So we could rewrite this as y is equal to negative x plus 12, negative x plus 12. And so from this, you might immediately recognize, okay, my constant term, when it's in this form, that's my b, that is my y-intercept. So that's my y-intercept right over there. But what's my slope? Well, the slope is the coefficient on the x term."}, {"video_title": "Slope and y intercept from equation.mp3", "Sentence": "And so from this, you might immediately recognize, okay, my constant term, when it's in this form, that's my b, that is my y-intercept. So that's my y-intercept right over there. But what's my slope? Well, the slope is the coefficient on the x term. But all you see is a negative here. What's the coefficient? Well, you could view negative x as the same thing as negative one x."}, {"video_title": "Slope and y intercept from equation.mp3", "Sentence": "Well, the slope is the coefficient on the x term. But all you see is a negative here. What's the coefficient? Well, you could view negative x as the same thing as negative one x. So your slope here is going to be negative one. Let's do another example. Let's say that we had the equation y is equal to five x."}, {"video_title": "Slope and y intercept from equation.mp3", "Sentence": "Well, you could view negative x as the same thing as negative one x. So your slope here is going to be negative one. Let's do another example. Let's say that we had the equation y is equal to five x. What's the slope and y-intercept there? At first, you might say, hey, this looks nothing like what we have up here. This is only, I only have one term on the right-hand side of the equality sign."}, {"video_title": "Slope and y intercept from equation.mp3", "Sentence": "Let's say that we had the equation y is equal to five x. What's the slope and y-intercept there? At first, you might say, hey, this looks nothing like what we have up here. This is only, I only have one term on the right-hand side of the equality sign. Here, I have two. But you could just view this as five x plus zero. And then it might jump out at you that our y-intercept is zero and our slope is the coefficient on the x term."}, {"video_title": "Slope and y intercept from equation.mp3", "Sentence": "This is only, I only have one term on the right-hand side of the equality sign. Here, I have two. But you could just view this as five x plus zero. And then it might jump out at you that our y-intercept is zero and our slope is the coefficient on the x term. It is equal to five. Let's do one more example. Let's say we had y is equal to negative seven."}, {"video_title": "Slope and y intercept from equation.mp3", "Sentence": "And then it might jump out at you that our y-intercept is zero and our slope is the coefficient on the x term. It is equal to five. Let's do one more example. Let's say we had y is equal to negative seven. What's the slope and y-intercept there? Well, once again, you might say, hey, this doesn't look like what we had up here. How do we figure out the slope or the y-intercept?"}, {"video_title": "Slope and y intercept from equation.mp3", "Sentence": "Let's say we had y is equal to negative seven. What's the slope and y-intercept there? Well, once again, you might say, hey, this doesn't look like what we had up here. How do we figure out the slope or the y-intercept? Well, we could do a similar idea. We could say, hey, this is the same thing as y is equal to zero times x minus seven. And so now, it looks just like what we have over here."}, {"video_title": "Slope and y intercept from equation.mp3", "Sentence": "How do we figure out the slope or the y-intercept? Well, we could do a similar idea. We could say, hey, this is the same thing as y is equal to zero times x minus seven. And so now, it looks just like what we have over here. And you might recognize that our y-intercept is negative seven y-intercept is equal to negative seven. And our slope is a coefficient on the x term. It is equal to zero."}, {"video_title": "Converting an explicit formula of a geometric sequence to a recursive formula   Khan Academy.mp3", "Sentence": "So I have the function g of x is equal to 9 times 8 to the x minus 1 power. And it's defined for x being a positive, or if x is a positive integer. If x is a positive integer. So we could say the domain of this function, or all of the valid inputs here, are positive integers. So 1, 2, 3, 4, 5, on and on and on. So this is an explicitly defined function. What I now want to do is to write a recursive definition of this exact same function."}, {"video_title": "Converting an explicit formula of a geometric sequence to a recursive formula   Khan Academy.mp3", "Sentence": "So we could say the domain of this function, or all of the valid inputs here, are positive integers. So 1, 2, 3, 4, 5, on and on and on. So this is an explicitly defined function. What I now want to do is to write a recursive definition of this exact same function. That given an x, it will give the exact same outputs. So let's first just try to understand the inputs and outputs here. So let's make a little table."}, {"video_title": "Converting an explicit formula of a geometric sequence to a recursive formula   Khan Academy.mp3", "Sentence": "What I now want to do is to write a recursive definition of this exact same function. That given an x, it will give the exact same outputs. So let's first just try to understand the inputs and outputs here. So let's make a little table. Let's make a table here. And let's think about what happens when we put in various x's into this function definition. So the domain is positive integers."}, {"video_title": "Converting an explicit formula of a geometric sequence to a recursive formula   Khan Academy.mp3", "Sentence": "So let's make a little table. Let's make a table here. And let's think about what happens when we put in various x's into this function definition. So the domain is positive integers. So let's try a couple of them. 1, 2, 3, 4. And then see what the corresponding g of x is."}, {"video_title": "Converting an explicit formula of a geometric sequence to a recursive formula   Khan Academy.mp3", "Sentence": "So the domain is positive integers. So let's try a couple of them. 1, 2, 3, 4. And then see what the corresponding g of x is. g of x. So when x is equal to 1, g of x is 9 times 8 to the 1 minus 1 power. So 9 times 8 to the 0 power, or 9 times 1."}, {"video_title": "Converting an explicit formula of a geometric sequence to a recursive formula   Khan Academy.mp3", "Sentence": "And then see what the corresponding g of x is. g of x. So when x is equal to 1, g of x is 9 times 8 to the 1 minus 1 power. So 9 times 8 to the 0 power, or 9 times 1. So g of x is going to be just 9. When x is 2, what's going to happen? It's going to be 9 times 8 to the 2 minus 1."}, {"video_title": "Converting an explicit formula of a geometric sequence to a recursive formula   Khan Academy.mp3", "Sentence": "So 9 times 8 to the 0 power, or 9 times 1. So g of x is going to be just 9. When x is 2, what's going to happen? It's going to be 9 times 8 to the 2 minus 1. So that's the same thing as 9 times 8 to the 1st power. And that's just going to be 9 times 8. So that is 72."}, {"video_title": "Converting an explicit formula of a geometric sequence to a recursive formula   Khan Academy.mp3", "Sentence": "It's going to be 9 times 8 to the 2 minus 1. So that's the same thing as 9 times 8 to the 1st power. And that's just going to be 9 times 8. So that is 72. Actually, let me just write it that way. Let me write it as just 9 times 8. Then, when x is equal to 3, what's going on here?"}, {"video_title": "Converting an explicit formula of a geometric sequence to a recursive formula   Khan Academy.mp3", "Sentence": "So that is 72. Actually, let me just write it that way. Let me write it as just 9 times 8. Then, when x is equal to 3, what's going on here? Well, this is going to be 3 minus 1 is 2. So it's going to be 8 squared. So it's going to be 9 times 8 squared."}, {"video_title": "Converting an explicit formula of a geometric sequence to a recursive formula   Khan Academy.mp3", "Sentence": "Then, when x is equal to 3, what's going on here? Well, this is going to be 3 minus 1 is 2. So it's going to be 8 squared. So it's going to be 9 times 8 squared. So we could write that as 9 times 8 times 8. I think you see a little bit of a pattern forming. When x is 4, this is going to be 8 to the 4 minus 1 power, 8 to the 3rd power."}, {"video_title": "Converting an explicit formula of a geometric sequence to a recursive formula   Khan Academy.mp3", "Sentence": "So it's going to be 9 times 8 squared. So we could write that as 9 times 8 times 8. I think you see a little bit of a pattern forming. When x is 4, this is going to be 8 to the 4 minus 1 power, 8 to the 3rd power. So that's 9 times 8 times 8 times 8. So this gives us a good clue about how we would define this recursively. Notice, if our first term, when x equals 1, is 9, every term after that is 8 times the preceding term."}, {"video_title": "Converting an explicit formula of a geometric sequence to a recursive formula   Khan Academy.mp3", "Sentence": "When x is 4, this is going to be 8 to the 4 minus 1 power, 8 to the 3rd power. So that's 9 times 8 times 8 times 8. So this gives us a good clue about how we would define this recursively. Notice, if our first term, when x equals 1, is 9, every term after that is 8 times the preceding term. So let's define that as a recursive function. First, we'll define our base case. We could say g of x, and I'll do this in a new color because I'm overusing the red."}, {"video_title": "Converting an explicit formula of a geometric sequence to a recursive formula   Khan Academy.mp3", "Sentence": "Notice, if our first term, when x equals 1, is 9, every term after that is 8 times the preceding term. So let's define that as a recursive function. First, we'll define our base case. We could say g of x, and I'll do this in a new color because I'm overusing the red. I like the blue. g of x, well, we can define our base case. It's going to be equal to 9 if x is equal to 1. g of x equals 9 if x equals 1."}, {"video_title": "Converting an explicit formula of a geometric sequence to a recursive formula   Khan Academy.mp3", "Sentence": "We could say g of x, and I'll do this in a new color because I'm overusing the red. I like the blue. g of x, well, we can define our base case. It's going to be equal to 9 if x is equal to 1. g of x equals 9 if x equals 1. So that took care of that right over there. And then if it equals anything else, it equals the previous g of x. So if we're looking at, let's go all the way down to x minus 1 and then an x."}, {"video_title": "Converting an explicit formula of a geometric sequence to a recursive formula   Khan Academy.mp3", "Sentence": "It's going to be equal to 9 if x is equal to 1. g of x equals 9 if x equals 1. So that took care of that right over there. And then if it equals anything else, it equals the previous g of x. So if we're looking at, let's go all the way down to x minus 1 and then an x. So if this entry right over here is g of x minus 1, however many times we multiply the 8s and we have a 9 in front. So this is g of x minus 1. We know that g of x, we know that this one right over here, is going to be the previous entry, g of x minus 1, the previous entry, that's the previous entry, times 8."}, {"video_title": "Converting an explicit formula of a geometric sequence to a recursive formula   Khan Academy.mp3", "Sentence": "So if we're looking at, let's go all the way down to x minus 1 and then an x. So if this entry right over here is g of x minus 1, however many times we multiply the 8s and we have a 9 in front. So this is g of x minus 1. We know that g of x, we know that this one right over here, is going to be the previous entry, g of x minus 1, the previous entry, that's the previous entry, times 8. So we could write that right here. So times 8. So for any other x other than 1, g of x is equal to the previous entry."}, {"video_title": "Converting an explicit formula of a geometric sequence to a recursive formula   Khan Academy.mp3", "Sentence": "We know that g of x, we know that this one right over here, is going to be the previous entry, g of x minus 1, the previous entry, that's the previous entry, times 8. So we could write that right here. So times 8. So for any other x other than 1, g of x is equal to the previous entry. So it's g of, I'll do that in a blue color, g of x minus 1, g of x minus 1 times 8 if x is greater than 1, or x is integer greater than 1. Now let's verify that this actually works. So let's draw another table here."}, {"video_title": "Converting an explicit formula of a geometric sequence to a recursive formula   Khan Academy.mp3", "Sentence": "So for any other x other than 1, g of x is equal to the previous entry. So it's g of, I'll do that in a blue color, g of x minus 1, g of x minus 1 times 8 if x is greater than 1, or x is integer greater than 1. Now let's verify that this actually works. So let's draw another table here. Let's draw another table here. So once again, we're going to have x and we're going to have g of x, but this time we're going to use this recursive definition for g of x. And the reason why it's recursive is that it's referring to itself."}, {"video_title": "Converting an explicit formula of a geometric sequence to a recursive formula   Khan Academy.mp3", "Sentence": "So let's draw another table here. Let's draw another table here. So once again, we're going to have x and we're going to have g of x, but this time we're going to use this recursive definition for g of x. And the reason why it's recursive is that it's referring to itself. In its own definition, it's saying, hey, g of x, well if x doesn't equal 1, it's going to be g of x minus 1. It's using the function itself, but we'll see that it actually does work out. So let's see, when x is equal to 1, x equals 1, so g of 1, well if x equals 1, it's equal to 9."}, {"video_title": "Converting an explicit formula of a geometric sequence to a recursive formula   Khan Academy.mp3", "Sentence": "And the reason why it's recursive is that it's referring to itself. In its own definition, it's saying, hey, g of x, well if x doesn't equal 1, it's going to be g of x minus 1. It's using the function itself, but we'll see that it actually does work out. So let's see, when x is equal to 1, x equals 1, so g of 1, well if x equals 1, it's equal to 9. It's equal to 9, so that was pretty straightforward. What happens when x equals 2? Well, when x equals 2, this case doesn't apply anymore."}, {"video_title": "Converting an explicit formula of a geometric sequence to a recursive formula   Khan Academy.mp3", "Sentence": "So let's see, when x is equal to 1, x equals 1, so g of 1, well if x equals 1, it's equal to 9. It's equal to 9, so that was pretty straightforward. What happens when x equals 2? Well, when x equals 2, this case doesn't apply anymore. We go down to this case. So when x is equal to 2, it's going to be equivalent to g of 2 minus 1. Let me write this down."}, {"video_title": "Converting an explicit formula of a geometric sequence to a recursive formula   Khan Academy.mp3", "Sentence": "Well, when x equals 2, this case doesn't apply anymore. We go down to this case. So when x is equal to 2, it's going to be equivalent to g of 2 minus 1. Let me write this down. It's going to be equivalent to g of 2 minus 1 times 8, which is the same thing as g of 1 times 8. And what's g of 1? Well, g of 1 is right over here."}, {"video_title": "Converting an explicit formula of a geometric sequence to a recursive formula   Khan Academy.mp3", "Sentence": "Let me write this down. It's going to be equivalent to g of 2 minus 1 times 8, which is the same thing as g of 1 times 8. And what's g of 1? Well, g of 1 is right over here. g of 1 is 9, so this is going to be equal to 9 times 8, exactly what we got over here. And of course, this was equivalent to g of 2. So let me write this."}, {"video_title": "Converting an explicit formula of a geometric sequence to a recursive formula   Khan Academy.mp3", "Sentence": "Well, g of 1 is right over here. g of 1 is 9, so this is going to be equal to 9 times 8, exactly what we got over here. And of course, this was equivalent to g of 2. So let me write this. This is g of 2. I'm going to scroll over a little bit so I don't get all squenched up. So now let's go to 3."}, {"video_title": "Converting an explicit formula of a geometric sequence to a recursive formula   Khan Academy.mp3", "Sentence": "So let me write this. This is g of 2. I'm going to scroll over a little bit so I don't get all squenched up. So now let's go to 3. Let's go to 3. And right now I'll write g of 3 first. So g of 3 is equal to, we're going to this case, it's equal to g of 3 minus 1 times 8, so that's equal to g of 2 times 8."}, {"video_title": "Converting an explicit formula of a geometric sequence to a recursive formula   Khan Academy.mp3", "Sentence": "So now let's go to 3. Let's go to 3. And right now I'll write g of 3 first. So g of 3 is equal to, we're going to this case, it's equal to g of 3 minus 1 times 8, so that's equal to g of 2 times 8. Well, what's g of 2? Well, g of 2, we already figured out, is 9 times 8. So it's equal to 9 times 8, that's g of 2, times 8 again."}, {"video_title": "Graphing exponential growth & decay   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "This is from the graph basic exponential functions on Khan Academy. And they ask us graph the following exponential function. And they give us the function, h of x is equal to 27 times 1 3rd to the x. So our initial value is 27, and 1 3rd is our common ratio. It's written in kind of standard exponential form. And they give us this little graphing tool where we can define these two points, and we can also define, we can define a horizontal asymptote to construct our function. And these three things are enough to define, to graph an exponential if we know that it is an exponential function."}, {"video_title": "Graphing exponential growth & decay   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "So our initial value is 27, and 1 3rd is our common ratio. It's written in kind of standard exponential form. And they give us this little graphing tool where we can define these two points, and we can also define, we can define a horizontal asymptote to construct our function. And these three things are enough to define, to graph an exponential if we know that it is an exponential function. So let's think about it a little bit. So the easiest thing that I could think of is, well, let's think about its initial value. Its initial value is going to be when x equals zero."}, {"video_title": "Graphing exponential growth & decay   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "And these three things are enough to define, to graph an exponential if we know that it is an exponential function. So let's think about it a little bit. So the easiest thing that I could think of is, well, let's think about its initial value. Its initial value is going to be when x equals zero. X equals zero, it's 1 3rd to the zero power, which is just one. And so you're just left with 27 times one, or just 27. That's why we call this number here when you're written it in this form."}, {"video_title": "Graphing exponential growth & decay   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "Its initial value is going to be when x equals zero. X equals zero, it's 1 3rd to the zero power, which is just one. And so you're just left with 27 times one, or just 27. That's why we call this number here when you're written it in this form. You call this the initial value. So when x is equal to zero, h of x is equal to 27. And we're graphing y equals h of x."}, {"video_title": "Graphing exponential growth & decay   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "That's why we call this number here when you're written it in this form. You call this the initial value. So when x is equal to zero, h of x is equal to 27. And we're graphing y equals h of x. So now let's graph another point. So let's think about it a little bit. When x is equal to one, when x is equal to one, what is h of x?"}, {"video_title": "Graphing exponential growth & decay   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "And we're graphing y equals h of x. So now let's graph another point. So let's think about it a little bit. When x is equal to one, when x is equal to one, what is h of x? Well, it's going to be 1 3rd to the first power, which is just 1 3rd. And so 1 3rd times 27 is going to be nine. So when x is one, h of one is nine."}, {"video_title": "Graphing exponential growth & decay   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "When x is equal to one, when x is equal to one, what is h of x? Well, it's going to be 1 3rd to the first power, which is just 1 3rd. And so 1 3rd times 27 is going to be nine. So when x is one, h of one is nine. And we can verify, well, and now let's just think about, let's think about the asymptote. So what's going to happen here when x becomes really, really, really, really, really big? Well, if I take 1 3rd to like a really large exponent, to say to the 10th power, to the 100th power, or to the 1,000th power, this thing right over here is going to start approaching zero as x becomes much, much, much, much larger."}, {"video_title": "Graphing exponential growth & decay   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "So when x is one, h of one is nine. And we can verify, well, and now let's just think about, let's think about the asymptote. So what's going to happen here when x becomes really, really, really, really, really big? Well, if I take 1 3rd to like a really large exponent, to say to the 10th power, to the 100th power, or to the 1,000th power, this thing right over here is going to start approaching zero as x becomes much, much, much, much larger. And so something that is approaching zero times 27, well, that's going to approach zero as well. So we're going to have a horizontal asymptote at zero. And you can verify that this works for more than just the two points we thought about."}, {"video_title": "Graphing exponential growth & decay   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "Well, if I take 1 3rd to like a really large exponent, to say to the 10th power, to the 100th power, or to the 1,000th power, this thing right over here is going to start approaching zero as x becomes much, much, much, much larger. And so something that is approaching zero times 27, well, that's going to approach zero as well. So we're going to have a horizontal asymptote at zero. And you can verify that this works for more than just the two points we thought about. When x is equal to two, this is telling us that the graph y equals h of x goes through the point two comma three. So h of two should be equal to three. And you can verify that that is indeed the case."}, {"video_title": "Graphing exponential growth & decay   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "And you can verify that this works for more than just the two points we thought about. When x is equal to two, this is telling us that the graph y equals h of x goes through the point two comma three. So h of two should be equal to three. And you can verify that that is indeed the case. If x is two, 1 3rd squared is nine, oh, sorry, 1 3rd squared is 1 9th, times 27 is three. And we see that right over here. When x is two, h of two is three."}, {"video_title": "Graphing exponential growth & decay   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "And you can verify that that is indeed the case. If x is two, 1 3rd squared is nine, oh, sorry, 1 3rd squared is 1 9th, times 27 is three. And we see that right over here. When x is two, h of two is three. So I feel pretty good about that. Let's do another one of these. So graph the following exponential function."}, {"video_title": "Graphing exponential growth & decay   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "When x is two, h of two is three. So I feel pretty good about that. Let's do another one of these. So graph the following exponential function. So same logic. When x is zero, the g of zero is just going to boil down to that initial value. And so let me scroll down."}, {"video_title": "Graphing exponential growth & decay   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "So graph the following exponential function. So same logic. When x is zero, the g of zero is just going to boil down to that initial value. And so let me scroll down. The initial value is negative 30. And so let's think about when x is equal to one. When x is equal to one, two to the 1st power is just two."}, {"video_title": "Graphing exponential growth & decay   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "And so let me scroll down. The initial value is negative 30. And so let's think about when x is equal to one. When x is equal to one, two to the 1st power is just two. And so two times negative 30 is negative 60. So when x is equal to one, the value of the graph is negative 60. Now let's think about this asymptote where that should sit."}, {"video_title": "Graphing exponential growth & decay   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "When x is equal to one, two to the 1st power is just two. And so two times negative 30 is negative 60. So when x is equal to one, the value of the graph is negative 60. Now let's think about this asymptote where that should sit. So let's think about what happens when x becomes really, really, really, really, really, really negative. So when x is really negative, so two to the negative one power is 1 1st, two to the negative two is 1 4th, two to the negative three is 1 8th. As you get larger and larger negative, or higher magnitude negative values, or in other words, as x becomes more and more and more negative, two to that power is going to approach zero."}, {"video_title": "Graphing exponential growth & decay   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "Now let's think about this asymptote where that should sit. So let's think about what happens when x becomes really, really, really, really, really, really negative. So when x is really negative, so two to the negative one power is 1 1st, two to the negative two is 1 4th, two to the negative three is 1 8th. As you get larger and larger negative, or higher magnitude negative values, or in other words, as x becomes more and more and more negative, two to that power is going to approach zero. And so negative 30 times something approaching zero is going to approach zero. So this asymptote's in the right place. Our horizontal asymptote, as x approaches negative infinity, as we move further and further to the left, the value of the function is going to approach zero."}, {"video_title": "Graphing exponential growth & decay   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "As you get larger and larger negative, or higher magnitude negative values, or in other words, as x becomes more and more and more negative, two to that power is going to approach zero. And so negative 30 times something approaching zero is going to approach zero. So this asymptote's in the right place. Our horizontal asymptote, as x approaches negative infinity, as we move further and further to the left, the value of the function is going to approach zero. And we can see it kind of approaches zero from below. We can see that it approaches below because we already looked at the initial value and we used that common ratio to find one other point. Hopefully you found that interesting."}, {"video_title": "Analyzing graphs of exponential functions negative initial value   High School Math   Khan Academy.mp3", "Sentence": "And our goal in this video is to figure out at what x value, so when, when does f of x, so at what x value is f of x going to be equal to negative 1 25th? And you might be tempted to just eyeball it over here, but when f of x is negative 1 25th, that's like right below the x axis. So if I tried to, if I tried to eyeball it, it would be very difficult. It's very difficult to tell what value that is. It might be at three, it might be at four. I am not sure. So instead of, actually it looks like, no, maybe, well, I don't wanna just eyeball it, just guess it."}, {"video_title": "Analyzing graphs of exponential functions negative initial value   High School Math   Khan Academy.mp3", "Sentence": "It's very difficult to tell what value that is. It might be at three, it might be at four. I am not sure. So instead of, actually it looks like, no, maybe, well, I don't wanna just eyeball it, just guess it. Instead, I'm gonna actually find an expression that defines f of x, because they've given us some information here, and then I can just solve for x. So let's do that. Well, since we know that f of x is an exponential function, we know it's gonna take the form f of x is equal to our initial value, a, times our common ratio, r, to the xth power."}, {"video_title": "Analyzing graphs of exponential functions negative initial value   High School Math   Khan Academy.mp3", "Sentence": "So instead of, actually it looks like, no, maybe, well, I don't wanna just eyeball it, just guess it. Instead, I'm gonna actually find an expression that defines f of x, because they've given us some information here, and then I can just solve for x. So let's do that. Well, since we know that f of x is an exponential function, we know it's gonna take the form f of x is equal to our initial value, a, times our common ratio, r, to the xth power. Well, the initial value is straightforward enough. That's going to be the value that the function takes on when x is equal to zero. And you can even see it here."}, {"video_title": "Analyzing graphs of exponential functions negative initial value   High School Math   Khan Academy.mp3", "Sentence": "Well, since we know that f of x is an exponential function, we know it's gonna take the form f of x is equal to our initial value, a, times our common ratio, r, to the xth power. Well, the initial value is straightforward enough. That's going to be the value that the function takes on when x is equal to zero. And you can even see it here. If x is equal to zero, the r to the x would just be one, and so f of zero will just be equal to a. And so what is f of zero? Well, when x is equal to zero, this is essentially we're saying where does it intersect, where does it intersect the y-axis?"}, {"video_title": "Analyzing graphs of exponential functions negative initial value   High School Math   Khan Academy.mp3", "Sentence": "And you can even see it here. If x is equal to zero, the r to the x would just be one, and so f of zero will just be equal to a. And so what is f of zero? Well, when x is equal to zero, this is essentially we're saying where does it intersect, where does it intersect the y-axis? We see f of zero is negative 25. So a is going to be negative 25. Negative 25."}, {"video_title": "Analyzing graphs of exponential functions negative initial value   High School Math   Khan Academy.mp3", "Sentence": "Well, when x is equal to zero, this is essentially we're saying where does it intersect, where does it intersect the y-axis? We see f of zero is negative 25. So a is going to be negative 25. Negative 25. When x is zero, the r to the x is just one, so f of zero is going to be negative 25. We see that right over there. Now, to figure out the common ratio, there's a couple of ways you could think about it."}, {"video_title": "Analyzing graphs of exponential functions negative initial value   High School Math   Khan Academy.mp3", "Sentence": "Negative 25. When x is zero, the r to the x is just one, so f of zero is going to be negative 25. We see that right over there. Now, to figure out the common ratio, there's a couple of ways you could think about it. The common ratio is the ratio between two successive values that are separated by one. What do I mean by that? Well, you could view it as the ratio between f of one and f of zero."}, {"video_title": "Analyzing graphs of exponential functions negative initial value   High School Math   Khan Academy.mp3", "Sentence": "Now, to figure out the common ratio, there's a couple of ways you could think about it. The common ratio is the ratio between two successive values that are separated by one. What do I mean by that? Well, you could view it as the ratio between f of one and f of zero. That's going to be the common ratio, or the ratio between f of two and f of one. That is going to be the common ratio. Well, lucky for us, we know f of zero is negative 25, and we know that f of one, f of one, x is equal to one, y, or f of x, or f of one is equal to negative five."}, {"video_title": "Analyzing graphs of exponential functions negative initial value   High School Math   Khan Academy.mp3", "Sentence": "Well, you could view it as the ratio between f of one and f of zero. That's going to be the common ratio, or the ratio between f of two and f of one. That is going to be the common ratio. Well, lucky for us, we know f of zero is negative 25, and we know that f of one, f of one, x is equal to one, y, or f of x, or f of one is equal to negative five. Negative five. And so just like that, we're able to figure out that our common ratio r is negative five over negative 25, which is the same thing as 1 5th. Divide a negative by a negative, you get a positive."}, {"video_title": "Analyzing graphs of exponential functions negative initial value   High School Math   Khan Academy.mp3", "Sentence": "Well, lucky for us, we know f of zero is negative 25, and we know that f of one, f of one, x is equal to one, y, or f of x, or f of one is equal to negative five. Negative five. And so just like that, we're able to figure out that our common ratio r is negative five over negative 25, which is the same thing as 1 5th. Divide a negative by a negative, you get a positive. So you're gonna have five over 25, which is 1 5th. Which is 1 5th. So now we can write, we can write an expression that defines f of x. f of x is going to be equal to negative 25 times, times 1 5th to the x power."}, {"video_title": "Analyzing graphs of exponential functions negative initial value   High School Math   Khan Academy.mp3", "Sentence": "Divide a negative by a negative, you get a positive. So you're gonna have five over 25, which is 1 5th. Which is 1 5th. So now we can write, we can write an expression that defines f of x. f of x is going to be equal to negative 25 times, times 1 5th to the x power. And so let's go back to our question. When is this going to be equal to negative 1 25th? So when does this equal to negative 1 25th?"}, {"video_title": "Analyzing graphs of exponential functions negative initial value   High School Math   Khan Academy.mp3", "Sentence": "So now we can write, we can write an expression that defines f of x. f of x is going to be equal to negative 25 times, times 1 5th to the x power. And so let's go back to our question. When is this going to be equal to negative 1 25th? So when does this equal to negative 1 25th? Well, let's just set them equal to each other. So let, there's a siren outside, I don't know if you hear it. So negative, I'll power through."}, {"video_title": "Analyzing graphs of exponential functions negative initial value   High School Math   Khan Academy.mp3", "Sentence": "So when does this equal to negative 1 25th? Well, let's just set them equal to each other. So let, there's a siren outside, I don't know if you hear it. So negative, I'll power through. All right, negative, so let's see, at what x value does this expression equal negative 1 25th? Let's see, we can multiply, well actually, we wanna solve for x, so let's see, let's divide both sides by negative 25. And so we are going to get 1 5th to the x power is equal to, if we divide both sides by negative 25, this negative 25 is gonna go away, and on the right hand side, we're going to have, divided negative by negative is gonna be positive, it's gonna be 1 over 625."}, {"video_title": "Analyzing graphs of exponential functions negative initial value   High School Math   Khan Academy.mp3", "Sentence": "So negative, I'll power through. All right, negative, so let's see, at what x value does this expression equal negative 1 25th? Let's see, we can multiply, well actually, we wanna solve for x, so let's see, let's divide both sides by negative 25. And so we are going to get 1 5th to the x power is equal to, if we divide both sides by negative 25, this negative 25 is gonna go away, and on the right hand side, we're going to have, divided negative by negative is gonna be positive, it's gonna be 1 over 625. It's gonna be 1 over 625. And 1 5th to the x power, this is the same thing as 1 to the x power over 5 to the x power is equal to, is equal to 1 over 625. 1 over 625."}, {"video_title": "Analyzing graphs of exponential functions negative initial value   High School Math   Khan Academy.mp3", "Sentence": "And so we are going to get 1 5th to the x power is equal to, if we divide both sides by negative 25, this negative 25 is gonna go away, and on the right hand side, we're going to have, divided negative by negative is gonna be positive, it's gonna be 1 over 625. It's gonna be 1 over 625. And 1 5th to the x power, this is the same thing as 1 to the x power over 5 to the x power is equal to, is equal to 1 over 625. 1 over 625. Well, 1 to the x power is just going to be equal to, is just going to be equal to 1. So we could really, it doesn't matter that we have this to the x power over here. And so we can see, I thought I was erasing that with a black color, there you go."}, {"video_title": "Analyzing graphs of exponential functions negative initial value   High School Math   Khan Academy.mp3", "Sentence": "1 over 625. Well, 1 to the x power is just going to be equal to, is just going to be equal to 1. So we could really, it doesn't matter that we have this to the x power over here. And so we can see, I thought I was erasing that with a black color, there you go. That's a black color right over there. So we can see that 5 to the x power needs to be equal to 625. So let me write that over here."}, {"video_title": "Analyzing graphs of exponential functions negative initial value   High School Math   Khan Academy.mp3", "Sentence": "And so we can see, I thought I was erasing that with a black color, there you go. That's a black color right over there. So we can see that 5 to the x power needs to be equal to 625. So let me write that over here. 5, whoops, didn't change my color. 5 to the x power needs to be equal to 625. Now, the best way I could think of doing this is let's just think about our powers of 5."}, {"video_title": "Analyzing graphs of exponential functions negative initial value   High School Math   Khan Academy.mp3", "Sentence": "So let me write that over here. 5, whoops, didn't change my color. 5 to the x power needs to be equal to 625. Now, the best way I could think of doing this is let's just think about our powers of 5. So 5 to the 1st power is 5. 5 squared is 25. 5 to the 3rd is 125."}, {"video_title": "Analyzing graphs of exponential functions negative initial value   High School Math   Khan Academy.mp3", "Sentence": "Now, the best way I could think of doing this is let's just think about our powers of 5. So 5 to the 1st power is 5. 5 squared is 25. 5 to the 3rd is 125. 5 to the 4th, we'll multiply that by 5, you're gonna get 625. So x is going to be 4. Is going to be 4, because 5 to the 4th power is 625."}, {"video_title": "Analyzing graphs of exponential functions negative initial value   High School Math   Khan Academy.mp3", "Sentence": "5 to the 3rd is 125. 5 to the 4th, we'll multiply that by 5, you're gonna get 625. So x is going to be 4. Is going to be 4, because 5 to the 4th power is 625. So we can now say that f of 4, f of 4 is equal to, is equal to negative 1 25th, is equal to negative 1 25th. And once again, you can verify that. You can verify that right over here."}, {"video_title": "Analyzing graphs of exponential functions negative initial value   High School Math   Khan Academy.mp3", "Sentence": "Is going to be 4, because 5 to the 4th power is 625. So we can now say that f of 4, f of 4 is equal to, is equal to negative 1 25th, is equal to negative 1 25th. And once again, you can verify that. You can verify that right over here. 1 5th to, 1 5th to the 4th power is gonna be 1 over 625. 25, negative 25 over positive 625 is gonna be negative 1 25th. So hopefully that clears things up a little bit."}, {"video_title": "Ex 1 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "And let's say that Diya, Diya today, is two, two years old. And what I am curious about in this video is how many years will it take? And let me write this down. How many years will it take? Will it take, will it take for Arman, for Arman to be three times, times as old as Diya? As Diya. So that's the question right there and I encourage you to try to take a shot at this yourself."}, {"video_title": "Ex 1 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "How many years will it take? Will it take, will it take for Arman, for Arman to be three times, times as old as Diya? As Diya. So that's the question right there and I encourage you to try to take a shot at this yourself. So let's think about this a little bit. We're asking how many years will it take? That's what we don't know."}, {"video_title": "Ex 1 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So that's the question right there and I encourage you to try to take a shot at this yourself. So let's think about this a little bit. We're asking how many years will it take? That's what we don't know. That's what we're curious about. How many years will it take for Arman to be three times as old as Diya? So let's set, let's set some variable."}, {"video_title": "Ex 1 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "That's what we don't know. That's what we're curious about. How many years will it take for Arman to be three times as old as Diya? So let's set, let's set some variable. Let's say y for years. Let's say y is equal to years, years it will take. So given that, can we now set up an equation, given this information, to figure out how many years it will take for Arman to be three times as old as Diya?"}, {"video_title": "Ex 1 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So let's set, let's set some variable. Let's say y for years. Let's say y is equal to years, years it will take. So given that, can we now set up an equation, given this information, to figure out how many years it will take for Arman to be three times as old as Diya? Well, let's think about how, how old Arman will be in y years. How old will he be? Well, in y years, let me write it here."}, {"video_title": "Ex 1 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So given that, can we now set up an equation, given this information, to figure out how many years it will take for Arman to be three times as old as Diya? Well, let's think about how, how old Arman will be in y years. How old will he be? Well, in y years, let me write it here. In, in y years, Arman, Arman, Arman is going to be how old? Arman is going to be, well he's 18 right now, and in y years, he's going to be y years older. So in y years, Arman is going to be 18 plus y."}, {"video_title": "Ex 1 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Well, in y years, let me write it here. In, in y years, Arman, Arman, Arman is going to be how old? Arman is going to be, well he's 18 right now, and in y years, he's going to be y years older. So in y years, Arman is going to be 18 plus y. And what about Diya? How old will she be? How old will she be in y years?"}, {"video_title": "Ex 1 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So in y years, Arman is going to be 18 plus y. And what about Diya? How old will she be? How old will she be in y years? Well, she's two now, and in y years, she'll just be two plus y. So what we're curious about, now that we know this, is how many years will it take for this quantity, for this expression, to be three times this quantity? So we're really curious."}, {"video_title": "Ex 1 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "How old will she be in y years? Well, she's two now, and in y years, she'll just be two plus y. So what we're curious about, now that we know this, is how many years will it take for this quantity, for this expression, to be three times this quantity? So we're really curious. We want to solve for y, such that 18 plus y is going to be equal to three times, is going to be equal to three times two plus y. Three times two plus y. Notice, this is Arman in y years, this is Diya in y years, and we're saying that what Arman's going to be in y years is three times what Diya is going to be in y years."}, {"video_title": "Ex 1 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So we're really curious. We want to solve for y, such that 18 plus y is going to be equal to three times, is going to be equal to three times two plus y. Three times two plus y. Notice, this is Arman in y years, this is Diya in y years, and we're saying that what Arman's going to be in y years is three times what Diya is going to be in y years. So we've set up our equation, now we can just solve it. So let's take this step by step. So on the left-hand side, and maybe I'll do this in a new color, just so I don't have to keep switching."}, {"video_title": "Ex 1 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Notice, this is Arman in y years, this is Diya in y years, and we're saying that what Arman's going to be in y years is three times what Diya is going to be in y years. So we've set up our equation, now we can just solve it. So let's take this step by step. So on the left-hand side, and maybe I'll do this in a new color, just so I don't have to keep switching. So on the left-hand side, I still have 18 plus y, and on the right-hand side, I can distribute this three. So three times two is six, three times y is three y. Six plus three y."}, {"video_title": "Ex 1 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So on the left-hand side, and maybe I'll do this in a new color, just so I don't have to keep switching. So on the left-hand side, I still have 18 plus y, and on the right-hand side, I can distribute this three. So three times two is six, three times y is three y. Six plus three y. And then it's always nice to get all of our constants on one side of the equation, and all of our variables on the other side of the equation. So we have a three y over here. We have more y's on the right-hand side than the left-hand side."}, {"video_title": "Ex 1 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Six plus three y. And then it's always nice to get all of our constants on one side of the equation, and all of our variables on the other side of the equation. So we have a three y over here. We have more y's on the right-hand side than the left-hand side. So let's get rid of the y's on the left-hand side. You could do it either way, but you'd end up with negative numbers. So let's subtract a y from each side."}, {"video_title": "Ex 1 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "We have more y's on the right-hand side than the left-hand side. So let's get rid of the y's on the left-hand side. You could do it either way, but you'd end up with negative numbers. So let's subtract a y from each side. And we are left with, on the left-hand side, 18. And on the right-hand side, you have six plus three y's. Take away one of those y's, you're going to be left with two y's."}, {"video_title": "Ex 1 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So let's subtract a y from each side. And we are left with, on the left-hand side, 18. And on the right-hand side, you have six plus three y's. Take away one of those y's, you're going to be left with two y's. Now we can get rid of the constant term here. So we will subtract six from both sides. Subtract six from both sides."}, {"video_title": "Ex 1 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Take away one of those y's, you're going to be left with two y's. Now we can get rid of the constant term here. So we will subtract six from both sides. Subtract six from both sides. 18 minus six is 12. The whole reason why we subtracted six from the right was to get rid of this. Six minus six is zero, so you have 12 is equal to two y."}, {"video_title": "Ex 1 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Subtract six from both sides. 18 minus six is 12. The whole reason why we subtracted six from the right was to get rid of this. Six minus six is zero, so you have 12 is equal to two y. Two times the number of years it will take is 12. And you could probably solve this in your head, but if we just want a one coefficient here, we would divide by two on the right. Whatever we do to one side of the equation, we have to do it on the other side, or the equation will not still be an equation."}, {"video_title": "Ex 1 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Six minus six is zero, so you have 12 is equal to two y. Two times the number of years it will take is 12. And you could probably solve this in your head, but if we just want a one coefficient here, we would divide by two on the right. Whatever we do to one side of the equation, we have to do it on the other side, or the equation will not still be an equation. So we are left with y is equal to six. Or y is equal to six. So, going back to the question, how many years will it take for Armand to be three times as old as Dia?"}, {"video_title": "Ex 1 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Whatever we do to one side of the equation, we have to do it on the other side, or the equation will not still be an equation. So we are left with y is equal to six. Or y is equal to six. So, going back to the question, how many years will it take for Armand to be three times as old as Dia? Well, it's going to take six years. Now, I want you to verify this. Think about it."}, {"video_title": "Ex 1 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So, going back to the question, how many years will it take for Armand to be three times as old as Dia? Well, it's going to take six years. Now, I want you to verify this. Think about it. Is this actually true? Well, in six years, how old is Armand going to be? He's going to be 18 plus six."}, {"video_title": "Ex 1 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Think about it. Is this actually true? Well, in six years, how old is Armand going to be? He's going to be 18 plus six. We now know that this thing is six, so in six years, Armand is going to be 18 plus six, which is 24 years old. How old is Dia going to be? Well, she's going to be two plus six, which is eight years old."}, {"video_title": "Ex 1 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "He's going to be 18 plus six. We now know that this thing is six, so in six years, Armand is going to be 18 plus six, which is 24 years old. How old is Dia going to be? Well, she's going to be two plus six, which is eight years old. And lo and behold, 24 is indeed three times as old as eight. In six years, Armand is 24, Dia is eight, Armand is three times as old as Dia. And we are done."}, {"video_title": "Setting up proportions to solve word problems   7th grade   Khan Academy.mp3", "Sentence": "And essentially, we're going to be setting up proportions in either case. So in this first problem, we have nine markers cost $11.50. And then they ask us, how much would seven markers cost? Now let's just set x to be equal to our answer. So x is equal to the cost of seven markers. So the way to solve a problem like this is to set up two ratios and then set them equal to each other. So you could say that the ratio of nine markers to the cost of nine markers, so the ratio of the number of markers, so 9 to the cost of the nine markers to $11.50, this should be equal to the ratio of our new number of markers, 7, equal to the new number of markers, so that's 7, to whatever the new cost, or whatever the cost of seven markers are, to x."}, {"video_title": "Setting up proportions to solve word problems   7th grade   Khan Academy.mp3", "Sentence": "Now let's just set x to be equal to our answer. So x is equal to the cost of seven markers. So the way to solve a problem like this is to set up two ratios and then set them equal to each other. So you could say that the ratio of nine markers to the cost of nine markers, so the ratio of the number of markers, so 9 to the cost of the nine markers to $11.50, this should be equal to the ratio of our new number of markers, 7, equal to the new number of markers, so that's 7, to whatever the new cost, or whatever the cost of seven markers are, to x. So let me do x in green. 2 to x. So this is a completely valid proportion here."}, {"video_title": "Setting up proportions to solve word problems   7th grade   Khan Academy.mp3", "Sentence": "So you could say that the ratio of nine markers to the cost of nine markers, so the ratio of the number of markers, so 9 to the cost of the nine markers to $11.50, this should be equal to the ratio of our new number of markers, 7, equal to the new number of markers, so that's 7, to whatever the new cost, or whatever the cost of seven markers are, to x. So let me do x in green. 2 to x. So this is a completely valid proportion here. The ratio of nine markers to the cost of nine markers is equal to seven markers to the cost of seven markers. So then you could solve this to figure out how much those seven markers would cost. And you could flip both sides of this and it would still be a completely valid ratio."}, {"video_title": "Setting up proportions to solve word problems   7th grade   Khan Academy.mp3", "Sentence": "So this is a completely valid proportion here. The ratio of nine markers to the cost of nine markers is equal to seven markers to the cost of seven markers. So then you could solve this to figure out how much those seven markers would cost. And you could flip both sides of this and it would still be a completely valid ratio. You could have $11.50 to 9. So the ratio between the cost of the markers to the number of markers you're buying is equal to, so $11.50 to 9 is equal to the cost of seven markers is equal to the ratio of the cost of seven markers to the number of markers, which is obviously 7. So all I did is flip both sides of this equation right here to get this one over here."}, {"video_title": "Setting up proportions to solve word problems   7th grade   Khan Academy.mp3", "Sentence": "And you could flip both sides of this and it would still be a completely valid ratio. You could have $11.50 to 9. So the ratio between the cost of the markers to the number of markers you're buying is equal to, so $11.50 to 9 is equal to the cost of seven markers is equal to the ratio of the cost of seven markers to the number of markers, which is obviously 7. So all I did is flip both sides of this equation right here to get this one over here. You could also think about the ratios in other ways. You could say that the ratio of nine markers to seven markers is going to be the same as the ratio of their cost, is going to be equal to the ratio of the cost of nine markers to the cost of seven markers. And then obviously, you could flip both of these sides."}, {"video_title": "Setting up proportions to solve word problems   7th grade   Khan Academy.mp3", "Sentence": "So all I did is flip both sides of this equation right here to get this one over here. You could also think about the ratios in other ways. You could say that the ratio of nine markers to seven markers is going to be the same as the ratio of their cost, is going to be equal to the ratio of the cost of nine markers to the cost of seven markers. And then obviously, you could flip both of these sides. So you could say that the ratio of seven markers, same magenta color, the ratio of seven markers to nine markers is the same thing as the ratio of the cost of seven markers to the cost of nine markers. So that is $11.50. So all of these would be valid proportions, valid equations that describe what's going on here."}, {"video_title": "Setting up proportions to solve word problems   7th grade   Khan Academy.mp3", "Sentence": "And then obviously, you could flip both of these sides. So you could say that the ratio of seven markers, same magenta color, the ratio of seven markers to nine markers is the same thing as the ratio of the cost of seven markers to the cost of nine markers. So that is $11.50. So all of these would be valid proportions, valid equations that describe what's going on here. And then you would just have to essentially solve for x. So let's do this one right over here. Seven apples cost $5."}, {"video_title": "Setting up proportions to solve word problems   7th grade   Khan Academy.mp3", "Sentence": "So all of these would be valid proportions, valid equations that describe what's going on here. And then you would just have to essentially solve for x. So let's do this one right over here. Seven apples cost $5. How many apples can I buy with $8? So once again, we can say, so we're going to assume that what they're asking is how many apples, let's call that x. x is what we want to solve for. So seven apples cost $5."}, {"video_title": "Setting up proportions to solve word problems   7th grade   Khan Academy.mp3", "Sentence": "Seven apples cost $5. How many apples can I buy with $8? So once again, we can say, so we're going to assume that what they're asking is how many apples, let's call that x. x is what we want to solve for. So seven apples cost $5. So we have the ratio between the number of apples, 7, and the cost of the apples, 5, is going to be equal to the ratio between another number of apples, which is now x, and the cost of that other number of apples. And it's going to be $8. And so notice here, in this first situation, what was unknown was the cost."}, {"video_title": "Setting up proportions to solve word problems   7th grade   Khan Academy.mp3", "Sentence": "So seven apples cost $5. So we have the ratio between the number of apples, 7, and the cost of the apples, 5, is going to be equal to the ratio between another number of apples, which is now x, and the cost of that other number of apples. And it's going to be $8. And so notice here, in this first situation, what was unknown was the cost. So we kind of had the number of apples to cost, number of apples to cost. Now in this example, the unknown is the number of apples. So number of apples to cost, number of apples to cost."}, {"video_title": "Setting up proportions to solve word problems   7th grade   Khan Academy.mp3", "Sentence": "And so notice here, in this first situation, what was unknown was the cost. So we kind of had the number of apples to cost, number of apples to cost. Now in this example, the unknown is the number of apples. So number of apples to cost, number of apples to cost. Now we could do all of the different scenarios like this. You could also say the ratio between 7 apples and x apples is going to be the same as the ratio between the cost of 7 apples and the cost of 8 apples. And obviously, you could flip both sides of either of these equations to get two more equations."}, {"video_title": "Setting up proportions to solve word problems   7th grade   Khan Academy.mp3", "Sentence": "So number of apples to cost, number of apples to cost. Now we could do all of the different scenarios like this. You could also say the ratio between 7 apples and x apples is going to be the same as the ratio between the cost of 7 apples and the cost of 8 apples. And obviously, you could flip both sides of either of these equations to get two more equations. And any of these would be valid equations. Now let's do this last one. So we have a cake recipe for five people."}, {"video_title": "Setting up proportions to solve word problems   7th grade   Khan Academy.mp3", "Sentence": "And obviously, you could flip both sides of either of these equations to get two more equations. And any of these would be valid equations. Now let's do this last one. So we have a cake recipe for five people. I'll use new colors here. A cake recipe for five people requires two eggs. So we want to know how many eggs."}, {"video_title": "Setting up proportions to solve word problems   7th grade   Khan Academy.mp3", "Sentence": "So we have a cake recipe for five people. I'll use new colors here. A cake recipe for five people requires two eggs. So we want to know how many eggs. So this we'll call x. And you don't always have to call it x. You could call it e for x."}, {"video_title": "Setting up proportions to solve word problems   7th grade   Khan Academy.mp3", "Sentence": "So we want to know how many eggs. So this we'll call x. And you don't always have to call it x. You could call it e for x. Well, e isn't a good idea, because e represents another number once you get to higher mathematics. But you could call them y or z or any variable, a, b, or c, anything. How many eggs do we need for a 15-person cake?"}, {"video_title": "Setting up proportions to solve word problems   7th grade   Khan Academy.mp3", "Sentence": "You could call it e for x. Well, e isn't a good idea, because e represents another number once you get to higher mathematics. But you could call them y or z or any variable, a, b, or c, anything. How many eggs do we need for a 15-person cake? So you could say the ratio of people to eggs is constant. So if we have five people for two eggs, then for 15 people, we are going to need x eggs. This ratio is going to be constant."}, {"video_title": "Setting up proportions to solve word problems   7th grade   Khan Academy.mp3", "Sentence": "How many eggs do we need for a 15-person cake? So you could say the ratio of people to eggs is constant. So if we have five people for two eggs, then for 15 people, we are going to need x eggs. This ratio is going to be constant. 5 over 2 is equal to 15 over x. Or you could flip both sides of this. Or you could say the ratio between 5 and 15 is going to be equal to the ratio between the number of eggs for five people."}, {"video_title": "Setting up proportions to solve word problems   7th grade   Khan Academy.mp3", "Sentence": "This ratio is going to be constant. 5 over 2 is equal to 15 over x. Or you could flip both sides of this. Or you could say the ratio between 5 and 15 is going to be equal to the ratio between the number of eggs for five people. Let me do that in that blue color. The ratio between the number of eggs for five people and the number of eggs for 15 people. And obviously, you could flip both sides of this equation."}, {"video_title": "Speed of light and distance from sun (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "The speed of light is 3 times 10 to the 8th meters per second. So as you can tell, light is very fast. 3 times 10 to the 8th meters per second. If it takes 5 times 10 to the 2nd power seconds for light to travel from the sun to the earth. So just let's think about that a little bit. So 5 times 10 to the 2nd, that's 500. 500 seconds, you have 60 seconds in a minute."}, {"video_title": "Speed of light and distance from sun (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "If it takes 5 times 10 to the 2nd power seconds for light to travel from the sun to the earth. So just let's think about that a little bit. So 5 times 10 to the 2nd, that's 500. 500 seconds, you have 60 seconds in a minute. So 8 minutes would be 480 seconds. So 500 seconds would be about 8 minutes 20 seconds. So it takes 8 minutes 20 seconds for light to travel from the sun to the earth."}, {"video_title": "Speed of light and distance from sun (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "500 seconds, you have 60 seconds in a minute. So 8 minutes would be 480 seconds. So 500 seconds would be about 8 minutes 20 seconds. So it takes 8 minutes 20 seconds for light to travel from the sun to the earth. What is the distance in meters between the sun and the earth? So they're giving us a rate. They're giving us a speed."}, {"video_title": "Speed of light and distance from sun (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "So it takes 8 minutes 20 seconds for light to travel from the sun to the earth. What is the distance in meters between the sun and the earth? So they're giving us a rate. They're giving us a speed. They're giving us a time. And they want to find a distance. So this goes straight back to the standard."}, {"video_title": "Speed of light and distance from sun (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "They're giving us a speed. They're giving us a time. And they want to find a distance. So this goes straight back to the standard. Distance is equal to rate times time. So they give us the rate. The rate is 3 times 10 to the 8th meters per second."}, {"video_title": "Speed of light and distance from sun (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "So this goes straight back to the standard. Distance is equal to rate times time. So they give us the rate. The rate is 3 times 10 to the 8th meters per second. So it's 3 times 10 to the 8th meters per second. That right there is the rate. They give us the time."}, {"video_title": "Speed of light and distance from sun (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "The rate is 3 times 10 to the 8th meters per second. So it's 3 times 10 to the 8th meters per second. That right there is the rate. They give us the time. The time is 5 times 10 to the 2nd seconds. So times 5 times 10 to the 2nd seconds. I'll just use that with an s. So how many meters?"}, {"video_title": "Speed of light and distance from sun (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "They give us the time. The time is 5 times 10 to the 2nd seconds. So times 5 times 10 to the 2nd seconds. I'll just use that with an s. So how many meters? So what is the distance? And so we can just reassociate these, or actually move these around from the commutative and the associative properties of multiplication. So this right here is the same thing."}, {"video_title": "Speed of light and distance from sun (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "I'll just use that with an s. So how many meters? So what is the distance? And so we can just reassociate these, or actually move these around from the commutative and the associative properties of multiplication. So this right here is the same thing. And actually, you can multiply the units. That's called dimensional analysis. And when you multiply the units, you kind of treat them like variables."}, {"video_title": "Speed of light and distance from sun (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "So this right here is the same thing. And actually, you can multiply the units. That's called dimensional analysis. And when you multiply the units, you kind of treat them like variables. You should get the right dimensions for distance. So let's just rearrange these numbers. This is equal to 3 times 5."}, {"video_title": "Speed of light and distance from sun (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "And when you multiply the units, you kind of treat them like variables. You should get the right dimensions for distance. So let's just rearrange these numbers. This is equal to 3 times 5. I'm just recommuting and reassociating these numbers. So 3 times in this product, because we're just multiplying everything. 3 times 5 times 10 to the 8th times 10 to the 2nd."}, {"video_title": "Speed of light and distance from sun (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "This is equal to 3 times 5. I'm just recommuting and reassociating these numbers. So 3 times in this product, because we're just multiplying everything. 3 times 5 times 10 to the 8th times 10 to the 2nd. And then we're going to have meters per second. So we could write meters per second times seconds. And if you treated these like variables, this seconds would cancel out with that seconds right there."}, {"video_title": "Speed of light and distance from sun (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "3 times 5 times 10 to the 8th times 10 to the 2nd. And then we're going to have meters per second. So we could write meters per second times seconds. And if you treated these like variables, this seconds would cancel out with that seconds right there. And you'd just be left with the unit meters, which is good. Because we want a distance in meters, in just meters. So how does this simplify?"}, {"video_title": "Speed of light and distance from sun (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "And if you treated these like variables, this seconds would cancel out with that seconds right there. And you'd just be left with the unit meters, which is good. Because we want a distance in meters, in just meters. So how does this simplify? This gives us 3 times 5 is 15. 15 times 10 to the 8th times 10 squared. We have the same base."}, {"video_title": "Speed of light and distance from sun (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "So how does this simplify? This gives us 3 times 5 is 15. 15 times 10 to the 8th times 10 squared. We have the same base. We're taking the product, so we can add the exponents. So this is going to be 10 to the 8 plus 2 power, or 10 to the 10th power. Now you might be tempted to say that we're done, that we have this in scientific notation."}, {"video_title": "Speed of light and distance from sun (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "We have the same base. We're taking the product, so we can add the exponents. So this is going to be 10 to the 8 plus 2 power, or 10 to the 10th power. Now you might be tempted to say that we're done, that we have this in scientific notation. But remember, in scientific notation, this number here has to be greater than or equal to 1 and less than 10. This clearly is not less than 10. So how do we rewrite this?"}, {"video_title": "Speed of light and distance from sun (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "Now you might be tempted to say that we're done, that we have this in scientific notation. But remember, in scientific notation, this number here has to be greater than or equal to 1 and less than 10. This clearly is not less than 10. So how do we rewrite this? We can write 15 as 1.5. This clearly is greater than 1 and less than 10. And to get from 1.5 to 15, you have to multiply by 10."}, {"video_title": "Speed of light and distance from sun (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "So how do we rewrite this? We can write 15 as 1.5. This clearly is greater than 1 and less than 10. And to get from 1.5 to 15, you have to multiply by 10. One way to think about it is 15 is 15.0. And so you have a decimal here. If we're going to move the decimal 1 to the left to get us 1.5, we're essentially dividing by 5."}, {"video_title": "Speed of light and distance from sun (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "And to get from 1.5 to 15, you have to multiply by 10. One way to think about it is 15 is 15.0. And so you have a decimal here. If we're going to move the decimal 1 to the left to get us 1.5, we're essentially dividing by 5. If we're moving the decimal 1 to the left to make it 1.5, that's essentially dividing by 10. Moving the decimal to the left means you're dividing by 10. If we don't want to change the value of the number, we need to divide by 10 and then multiply by 10."}, {"video_title": "Speed of light and distance from sun (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "If we're going to move the decimal 1 to the left to get us 1.5, we're essentially dividing by 5. If we're moving the decimal 1 to the left to make it 1.5, that's essentially dividing by 10. Moving the decimal to the left means you're dividing by 10. If we don't want to change the value of the number, we need to divide by 10 and then multiply by 10. So this and that are the same number. Now 15 is 1.5 times 10. And then we have to multiply that times 10 to the 10th."}, {"video_title": "Speed of light and distance from sun (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "If we don't want to change the value of the number, we need to divide by 10 and then multiply by 10. So this and that are the same number. Now 15 is 1.5 times 10. And then we have to multiply that times 10 to the 10th. Not x to the 10th. Times 10 to the 10th power. This right over here."}, {"video_title": "Speed of light and distance from sun (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "And then we have to multiply that times 10 to the 10th. Not x to the 10th. Times 10 to the 10th power. This right over here. 10 is really just 10 to the first power. So we can just add the exponents. Same base, taking the product."}, {"video_title": "Speed of light and distance from sun (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "This right over here. 10 is really just 10 to the first power. So we can just add the exponents. Same base, taking the product. So this is equal to 1.5 times 10 to the 1 plus 10 power. Or 10 to the 11th power. And we are done."}, {"video_title": "Speed of light and distance from sun (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "Same base, taking the product. So this is equal to 1.5 times 10 to the 1 plus 10 power. Or 10 to the 11th power. And we are done. This is a huge, huge distance. Just so that you can, well, it's actually almost very hard to visualize. But anyway, hopefully you enjoyed that."}, {"video_title": "Factoring binomials common factor   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "So we have these two terms, and I want to figure out their greatest common monomial factor, and then I want to express this with that greatest common monomial factor factored out. So how can we tackle it? Well, one way to start is I can look at just the constant terms. I can look at, or not the constants, the coefficients, I should say. So I have the eight and the 12, and I can say, well, what is just the greatest common factor of eight and 12? The GCF of eight and 12. And there are a lot of common factors of eight and 12."}, {"video_title": "Factoring binomials common factor   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "I can look at, or not the constants, the coefficients, I should say. So I have the eight and the 12, and I can say, well, what is just the greatest common factor of eight and 12? The GCF of eight and 12. And there are a lot of common factors of eight and 12. They're both divisible by one, they're both divisible by two, they're both divisible by four, but the greatest of their common factors is going to be four. So that is equal to four. So let me just leave that there."}, {"video_title": "Factoring binomials common factor   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "And there are a lot of common factors of eight and 12. They're both divisible by one, they're both divisible by two, they're both divisible by four, but the greatest of their common factors is going to be four. So that is equal to four. So let me just leave that there. And then we can think about what is, well, let me actually write it right over here. I'll put a four here. And now we can move on to the powers of x."}, {"video_title": "Factoring binomials common factor   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "So let me just leave that there. And then we can think about what is, well, let me actually write it right over here. I'll put a four here. And now we can move on to the powers of x. We have an x squared and we have an x. And we can say, what is the largest power of x that is divisible into both x squared and x? Well, that's just going to be x. X squared is clearly divisible by x, and x is clearly divisible by x, but x isn't going to be, isn't going to have a larger power of x as a factor."}, {"video_title": "Factoring binomials common factor   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "And now we can move on to the powers of x. We have an x squared and we have an x. And we can say, what is the largest power of x that is divisible into both x squared and x? Well, that's just going to be x. X squared is clearly divisible by x, and x is clearly divisible by x, but x isn't going to be, isn't going to have a larger power of x as a factor. So this is the greatest, you could view this as the greatest common monomial factor of x squared and x. Now we do the same thing for the y's. So we have a y and a y squared."}, {"video_title": "Factoring binomials common factor   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "Well, that's just going to be x. X squared is clearly divisible by x, and x is clearly divisible by x, but x isn't going to be, isn't going to have a larger power of x as a factor. So this is the greatest, you could view this as the greatest common monomial factor of x squared and x. Now we do the same thing for the y's. So we have a y and a y squared. If we think in the same terms, the largest power of y that's divisible into both of these is going to be just y to the first power, or y. And so four xy is the greatest common monomial factor. And to see that, we can express each of these terms as a product of four xy and something else."}, {"video_title": "Factoring binomials common factor   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "So we have a y and a y squared. If we think in the same terms, the largest power of y that's divisible into both of these is going to be just y to the first power, or y. And so four xy is the greatest common monomial factor. And to see that, we can express each of these terms as a product of four xy and something else. So this first term right over here, so let me pick a color. So this term right over here, we could write as four xy, that one's actually, that color's hard to see, let me pick a darker color. We could write this right over here as four xy times what?"}, {"video_title": "Factoring binomials common factor   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "And to see that, we can express each of these terms as a product of four xy and something else. So this first term right over here, so let me pick a color. So this term right over here, we could write as four xy, that one's actually, that color's hard to see, let me pick a darker color. We could write this right over here as four xy times what? And I encourage you to pause the video and think about that. Let's see, four times what is equal, is going to get us to eight? Well, four times two is going to get us to eight."}, {"video_title": "Factoring binomials common factor   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "We could write this right over here as four xy times what? And I encourage you to pause the video and think about that. Let's see, four times what is equal, is going to get us to eight? Well, four times two is going to get us to eight. X times what is going to get us to x squared? Well, x times x is going to get us to x squared. And then y times what is going to get us to y?"}, {"video_title": "Factoring binomials common factor   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "Well, four times two is going to get us to eight. X times what is going to get us to x squared? Well, x times x is going to get us to x squared. And then y times what is going to get us to y? Well, it's just going to be y. So four xy times two x is actually going to give us this first term. So actually, let me just rewrite it a little bit differently."}, {"video_title": "Factoring binomials common factor   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "And then y times what is going to get us to y? Well, it's just going to be y. So four xy times two x is actually going to give us this first term. So actually, let me just rewrite it a little bit differently. So it's four xy times two x is this first term, and you can verify that. Four times two is going to be equal to eight. X times x is equal to x squared."}, {"video_title": "Factoring binomials common factor   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "So actually, let me just rewrite it a little bit differently. So it's four xy times two x is this first term, and you can verify that. Four times two is going to be equal to eight. X times x is equal to x squared. And then you just have the y. Now let's do the same thing with the second term. And I just want to do this to show you that this is their largest common monomial factor."}, {"video_title": "Factoring binomials common factor   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "X times x is equal to x squared. And then you just have the y. Now let's do the same thing with the second term. And I just want to do this to show you that this is their largest common monomial factor. So the second term, and I'll do this in a slightly different color, do it in blue. I want to write this as the product of four xy and another monomial. So four times what is 12?"}, {"video_title": "Factoring binomials common factor   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "And I just want to do this to show you that this is their largest common monomial factor. So the second term, and I'll do this in a slightly different color, do it in blue. I want to write this as the product of four xy and another monomial. So four times what is 12? Well, four times three is 12. X times what is x? Well, it's just going to be one, so we don't have to write up anything here."}, {"video_title": "Factoring binomials common factor   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "So four times what is 12? Well, four times three is 12. X times what is x? Well, it's just going to be one, so we don't have to write up anything here. And then y times what is y squared? It's going to be y times y is y squared. And you can verify."}, {"video_title": "Factoring binomials common factor   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "Well, it's just going to be one, so we don't have to write up anything here. And then y times what is y squared? It's going to be y times y is y squared. And you can verify. If you multiply these two, you're going to get 12xy squared. Four times three is 12. You get your x."}, {"video_title": "Factoring binomials common factor   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "And you can verify. If you multiply these two, you're going to get 12xy squared. Four times three is 12. You get your x. And then y times y is y squared. So so far, I've written this exact same expression, but I've taken each of those terms and I factored them into their greatest common monomial factor and then whatever is left over. And now I can factor the four xy out."}, {"video_title": "Factoring binomials common factor   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "You get your x. And then y times y is y squared. So so far, I've written this exact same expression, but I've taken each of those terms and I factored them into their greatest common monomial factor and then whatever is left over. And now I can factor the four xy out. I can actually factor it out. So this is going to be equal to, if I factor the four xys out, you could kind of say I undistribute the four xy. I factor it out."}, {"video_title": "Factoring binomials common factor   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "And now I can factor the four xy out. I can actually factor it out. So this is going to be equal to, if I factor the four xys out, you could kind of say I undistribute the four xy. I factor it out. This is going to be equal to four xy times 2x plus, when I factor four xy from here, I get the three y left over. So that's three y. And we're done."}, {"video_title": "Factoring binomials common factor   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "I factor it out. This is going to be equal to four xy times 2x plus, when I factor four xy from here, I get the three y left over. So that's three y. And we're done. And you can verify it. If you were to go the other way, if you were to distribute this four xy and multiply it times 2x, you'd get 8x squared y. And then when you distribute the four xy onto the three y, you get the 12xy squared."}, {"video_title": "Factoring binomials common factor   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "And we're done. And you can verify it. If you were to go the other way, if you were to distribute this four xy and multiply it times 2x, you'd get 8x squared y. And then when you distribute the four xy onto the three y, you get the 12xy squared. And so we're done. This right over here is our answer. The answer is going to be four xy, which is the greatest common monomial factor, times 2x plus three y."}, {"video_title": "Dividing by a multi-digit decimal   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "So every time you multiply something by 10, you're shifting the decimal over to the right once. So in this case, we want to shift it over to the right once and twice. So 0.25 times 10 twice is the same thing as 0.25 times 100, and we'll turn the 0.25 into 25. Now if you do that with the divisor, you also have to do that with the dividend, the number that you're dividing into. So we also have to multiply this by 10 twice. Or another way of doing it is shift the decimal over to the right twice. So we shift it over once, twice, it will sit right over here."}, {"video_title": "Dividing by a multi-digit decimal   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "Now if you do that with the divisor, you also have to do that with the dividend, the number that you're dividing into. So we also have to multiply this by 10 twice. Or another way of doing it is shift the decimal over to the right twice. So we shift it over once, twice, it will sit right over here. And to see why that makes sense, you just have to realize that this expression right here, this division problem, is the exact same thing as having 0.25 as having 1.03075 divided by 0.25. And so we're multiplying the 0.25 by 10 twice. We're essentially multiplying it by 100."}, {"video_title": "Dividing by a multi-digit decimal   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "So we shift it over once, twice, it will sit right over here. And to see why that makes sense, you just have to realize that this expression right here, this division problem, is the exact same thing as having 0.25 as having 1.03075 divided by 0.25. And so we're multiplying the 0.25 by 10 twice. We're essentially multiplying it by 100. Let me do that in a different color. We're multiplying it by 100 in the denominator. This is the divisor."}, {"video_title": "Dividing by a multi-digit decimal   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "We're essentially multiplying it by 100. Let me do that in a different color. We're multiplying it by 100 in the denominator. This is the divisor. We're multiplying it by 100. So we also have to do the same thing to the numerator if we don't want to change this expression. We don't want to change the number."}, {"video_title": "Dividing by a multi-digit decimal   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "This is the divisor. We're multiplying it by 100. So we also have to do the same thing to the numerator if we don't want to change this expression. We don't want to change the number. So we also have to multiply that by 100. And when you do that, this becomes 25, and this becomes 103.075. Now let me just rewrite this."}, {"video_title": "Dividing by a multi-digit decimal   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "We don't want to change the number. So we also have to multiply that by 100. And when you do that, this becomes 25, and this becomes 103.075. Now let me just rewrite this. Sometimes if you're doing this in a workbook or something, you don't have to rewrite it as long as you remember where the decimal is. But I'm going to rewrite it just so it's a little bit neater. So when we multiplied both the divisor and the dividend by 100, this problem becomes 25 divided into 103.075."}, {"video_title": "Dividing by a multi-digit decimal   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "Now let me just rewrite this. Sometimes if you're doing this in a workbook or something, you don't have to rewrite it as long as you remember where the decimal is. But I'm going to rewrite it just so it's a little bit neater. So when we multiplied both the divisor and the dividend by 100, this problem becomes 25 divided into 103.075. These are going to result in the exact same quotient. They're the exact same fraction if you want to view it that way. We just multiplied both the numerator and the denominator by 100 to shift the decimal over to the right twice."}, {"video_title": "Dividing by a multi-digit decimal   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "So when we multiplied both the divisor and the dividend by 100, this problem becomes 25 divided into 103.075. These are going to result in the exact same quotient. They're the exact same fraction if you want to view it that way. We just multiplied both the numerator and the denominator by 100 to shift the decimal over to the right twice. Now that we've done that, we're ready to divide. So the first thing, we have 25 here. And there's always a little bit of an art to dividing something by a multiple digit number."}, {"video_title": "Dividing by a multi-digit decimal   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "We just multiplied both the numerator and the denominator by 100 to shift the decimal over to the right twice. Now that we've done that, we're ready to divide. So the first thing, we have 25 here. And there's always a little bit of an art to dividing something by a multiple digit number. So we'll see how well we can do. So 25 does not go into 1. 25 does not go into 10."}, {"video_title": "Dividing by a multi-digit decimal   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "And there's always a little bit of an art to dividing something by a multiple digit number. So we'll see how well we can do. So 25 does not go into 1. 25 does not go into 10. 25 does go into 103. We know that 4 times 25 is 100. So 25 goes into 103 4 times."}, {"video_title": "Dividing by a multi-digit decimal   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "25 does not go into 10. 25 does go into 103. We know that 4 times 25 is 100. So 25 goes into 103 4 times. 4 times 5 is 20. 4 times 2 is 8, plus 2 is 100. We knew that."}, {"video_title": "Dividing by a multi-digit decimal   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "So 25 goes into 103 4 times. 4 times 5 is 20. 4 times 2 is 8, plus 2 is 100. We knew that. 4 times 4 quarters is $1. It's $0.10. And now we subtract."}, {"video_title": "Dividing by a multi-digit decimal   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "We knew that. 4 times 4 quarters is $1. It's $0.10. And now we subtract. 103 minus 100 is going to be 3. And now we can bring down this 0. So we bring down that 0 there."}, {"video_title": "Dividing by a multi-digit decimal   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "And now we subtract. 103 minus 100 is going to be 3. And now we can bring down this 0. So we bring down that 0 there. 25 goes into 30 one time. And if we want, we can immediately put this decimal here. We don't have to wait until the end of the problem."}, {"video_title": "Dividing by a multi-digit decimal   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "So we bring down that 0 there. 25 goes into 30 one time. And if we want, we can immediately put this decimal here. We don't have to wait until the end of the problem. This decimal sits right in that place. So we could always have that decimal sitting right there in our quotient or in our answer or our quotient. So anyway, we were at 25 goes into 30 one time."}, {"video_title": "Dividing by a multi-digit decimal   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "We don't have to wait until the end of the problem. This decimal sits right in that place. So we could always have that decimal sitting right there in our quotient or in our answer or our quotient. So anyway, we were at 25 goes into 30 one time. 1 times 25 is 25. And then we can subtract. 30 minus 25, well, that's just 5."}, {"video_title": "Dividing by a multi-digit decimal   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "So anyway, we were at 25 goes into 30 one time. 1 times 25 is 25. And then we can subtract. 30 minus 25, well, that's just 5. We could do all of this borrowing business or regrouping. This can become a 10. This becomes a 2."}, {"video_title": "Dividing by a multi-digit decimal   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "30 minus 25, well, that's just 5. We could do all of this borrowing business or regrouping. This can become a 10. This becomes a 2. 10 minus 5 is 5. 2 minus 2 is nothing. But anyway, 30 minus 25 is 5."}, {"video_title": "Dividing by a multi-digit decimal   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "This becomes a 2. 10 minus 5 is 5. 2 minus 2 is nothing. But anyway, 30 minus 25 is 5. Now we can bring down this 7. 25 goes into 57 two times. 25 times 2 is 50."}, {"video_title": "Dividing by a multi-digit decimal   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "But anyway, 30 minus 25 is 5. Now we can bring down this 7. 25 goes into 57 two times. 25 times 2 is 50. 25 goes into 57 two times. 2 times 25 is 50. And now we subtract again."}, {"video_title": "Dividing by a multi-digit decimal   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "25 times 2 is 50. 25 goes into 57 two times. 2 times 25 is 50. And now we subtract again. 57 minus 50 is 7. And now we're almost done. We can bring down this 5."}, {"video_title": "Dividing by a multi-digit decimal   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "And now we subtract again. 57 minus 50 is 7. And now we're almost done. We can bring down this 5. We bring down that 5 right over there. 25 goes into 75 three times. 3 times 25 is 75."}, {"video_title": "Dividing by a multi-digit decimal   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "We can bring down this 5. We bring down that 5 right over there. 25 goes into 75 three times. 3 times 25 is 75. We say 3 times 5 is 15. Regroup the 1. We can ignore that."}, {"video_title": "Dividing by a multi-digit decimal   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "3 times 25 is 75. We say 3 times 5 is 15. Regroup the 1. We can ignore that. That was from before. 3 times 2 is 6 plus 1 is 7. So you can see that."}, {"video_title": "Dividing by a multi-digit decimal   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "We can ignore that. That was from before. 3 times 2 is 6 plus 1 is 7. So you can see that. And then we subtract. And then we have no remainder. So 25 goes into 103.075 exactly 4.123 times, which makes sense because 25 goes into 100 about 4 times."}, {"video_title": "Dividing by a multi-digit decimal   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "So you can see that. And then we subtract. And then we have no remainder. So 25 goes into 103.075 exactly 4.123 times, which makes sense because 25 goes into 100 about 4 times. This is a little bit larger than 100, so it's going to be a little bit more than 4 times. And that's going to be the exact same answer as the number of times that 0.25 goes into 1.03075. This will also be 4.123."}, {"video_title": "Simplifying radicals examples.mp3", "Sentence": "Let's get some practice rewriting and simplifying radical expressions. So this first exercise, and these are all from Khan Academy, it says simplify the expression by removing all factors that are perfect squares from inside the radicals and combine the terms. If the expression cannot be simplified, enter it as given. All right, let's see what we can do here. So we have negative 40, the negative square root of 40, I should say. Let me write it a little bit bigger so you can see that. So the negative square root of 40 plus the square root of 90."}, {"video_title": "Simplifying radicals examples.mp3", "Sentence": "All right, let's see what we can do here. So we have negative 40, the negative square root of 40, I should say. Let me write it a little bit bigger so you can see that. So the negative square root of 40 plus the square root of 90. So let's see, what perfect squares are in 40? So what immediately jumps out at me is that it's divisible by four, and four is a perfect square. So this is the negative square root of four times 10 plus the square root of, well, what jumps out at me is that this is divisible by nine."}, {"video_title": "Simplifying radicals examples.mp3", "Sentence": "So the negative square root of 40 plus the square root of 90. So let's see, what perfect squares are in 40? So what immediately jumps out at me is that it's divisible by four, and four is a perfect square. So this is the negative square root of four times 10 plus the square root of, well, what jumps out at me is that this is divisible by nine. Nine is a perfect square, so nine times 10. And if we look at the 10s here, 10 does not have any perfect squares in it anymore. If you wanted to do a full factorization of 10, a full prime factorization, it would be two times five."}, {"video_title": "Simplifying radicals examples.mp3", "Sentence": "So this is the negative square root of four times 10 plus the square root of, well, what jumps out at me is that this is divisible by nine. Nine is a perfect square, so nine times 10. And if we look at the 10s here, 10 does not have any perfect squares in it anymore. If you wanted to do a full factorization of 10, a full prime factorization, it would be two times five. So there's no perfect squares in 10. And so we can work it out from here. This is the same thing as the negative of the square root of four times the square root of 10 plus the square root of nine times the square root of 10."}, {"video_title": "Simplifying radicals examples.mp3", "Sentence": "If you wanted to do a full factorization of 10, a full prime factorization, it would be two times five. So there's no perfect squares in 10. And so we can work it out from here. This is the same thing as the negative of the square root of four times the square root of 10 plus the square root of nine times the square root of 10. And when I say square root, I'm really saying principal root, the positive square root. So it's the negative of the positive square root of four. So that is, so let me do this in another color so it can be clear."}, {"video_title": "Simplifying radicals examples.mp3", "Sentence": "This is the same thing as the negative of the square root of four times the square root of 10 plus the square root of nine times the square root of 10. And when I say square root, I'm really saying principal root, the positive square root. So it's the negative of the positive square root of four. So that is, so let me do this in another color so it can be clear. So this right here is two. This right here is three. So it's going to be equal to negative two square roots of 10 plus three square roots of 10."}, {"video_title": "Simplifying radicals examples.mp3", "Sentence": "So that is, so let me do this in another color so it can be clear. So this right here is two. This right here is three. So it's going to be equal to negative two square roots of 10 plus three square roots of 10. So if I have negative two of something and I add three of that same something to it, that's going to be what? Well, that's going to be one square root of 10. Now, if this last step doesn't make full sense, actually, let me slow it down a little bit."}, {"video_title": "Simplifying radicals examples.mp3", "Sentence": "So it's going to be equal to negative two square roots of 10 plus three square roots of 10. So if I have negative two of something and I add three of that same something to it, that's going to be what? Well, that's going to be one square root of 10. Now, if this last step doesn't make full sense, actually, let me slow it down a little bit. I could rewrite it this way. I could write it as three square roots of 10 minus two square roots of 10. That might jump out at you a little bit clear."}, {"video_title": "Simplifying radicals examples.mp3", "Sentence": "Now, if this last step doesn't make full sense, actually, let me slow it down a little bit. I could rewrite it this way. I could write it as three square roots of 10 minus two square roots of 10. That might jump out at you a little bit clear. If I have three of something and I were to take away two of that something, in that case, the square roots of 10s, well, I'm gonna be left with just one of that something. I'm just gonna be left with one square root of 10, which we could just write as the square root of 10. Another way to think about it is we could factor out a square root of 10 here."}, {"video_title": "Simplifying radicals examples.mp3", "Sentence": "That might jump out at you a little bit clear. If I have three of something and I were to take away two of that something, in that case, the square roots of 10s, well, I'm gonna be left with just one of that something. I'm just gonna be left with one square root of 10, which we could just write as the square root of 10. Another way to think about it is we could factor out a square root of 10 here. So you undistribute it, do the distributive property in reverse. That would be the square root of 10 times three minus two, which is, of course, this is just one. So you're just left with the square root of 10."}, {"video_title": "Simplifying radicals examples.mp3", "Sentence": "Another way to think about it is we could factor out a square root of 10 here. So you undistribute it, do the distributive property in reverse. That would be the square root of 10 times three minus two, which is, of course, this is just one. So you're just left with the square root of 10. So all of this simplifies to square root of 10. Let's do a few more of these. So this says, simplify the expression by removing all factors that are perfect squares from inside the radicals and combining the terms."}, {"video_title": "Simplifying radicals examples.mp3", "Sentence": "So you're just left with the square root of 10. So all of this simplifies to square root of 10. Let's do a few more of these. So this says, simplify the expression by removing all factors that are perfect squares from inside the radicals and combining the terms. So essentially, the same idea. All right, let's see what we can do. So this is interesting."}, {"video_title": "Simplifying radicals examples.mp3", "Sentence": "So this says, simplify the expression by removing all factors that are perfect squares from inside the radicals and combining the terms. So essentially, the same idea. All right, let's see what we can do. So this is interesting. We have a square root of 1 1\u20442. So can I, well, actually, what could be interesting is if I have a square root of something times the square root of something else, so the square root of 180 times the square root of 1 1\u20442. This is the same thing as the square root of 180 times 1 1\u20442."}, {"video_title": "Simplifying radicals examples.mp3", "Sentence": "So this is interesting. We have a square root of 1 1\u20442. So can I, well, actually, what could be interesting is if I have a square root of something times the square root of something else, so the square root of 180 times the square root of 1 1\u20442. This is the same thing as the square root of 180 times 1 1\u20442. And this just comes straight out of our exponent properties. It might look a little bit more familiar if I wrote it as 180 to the 1 1\u20442 power times 1 1\u20442 to the 1 1\u20442 power is going to be equal to 180 times 1 1\u20442 to the 1 1\u20442 power. Taking the square root, the principal root, is the same thing as raising something to the 1 1\u20442 power."}, {"video_title": "Simplifying radicals examples.mp3", "Sentence": "This is the same thing as the square root of 180 times 1 1\u20442. And this just comes straight out of our exponent properties. It might look a little bit more familiar if I wrote it as 180 to the 1 1\u20442 power times 1 1\u20442 to the 1 1\u20442 power is going to be equal to 180 times 1 1\u20442 to the 1 1\u20442 power. Taking the square root, the principal root, is the same thing as raising something to the 1 1\u20442 power. And so this is the square root of 80 times 1 1\u20442, which is going to be the square root of 90, which is equal to the square root of 9 times 10. And we just simplified square root of 90 in the last problem. That's equal to the square root of 9 times the square root or the principal root of 10, which is equal to 3 times the square root of 10."}, {"video_title": "Simplifying radicals examples.mp3", "Sentence": "Taking the square root, the principal root, is the same thing as raising something to the 1 1\u20442 power. And so this is the square root of 80 times 1 1\u20442, which is going to be the square root of 90, which is equal to the square root of 9 times 10. And we just simplified square root of 90 in the last problem. That's equal to the square root of 9 times the square root or the principal root of 10, which is equal to 3 times the square root of 10. 3 times the square root of 10. All right, let's keep going. So I have one more of these examples."}, {"video_title": "Simplifying radicals examples.mp3", "Sentence": "That's equal to the square root of 9 times the square root or the principal root of 10, which is equal to 3 times the square root of 10. 3 times the square root of 10. All right, let's keep going. So I have one more of these examples. And like always, pause the video and see if you can work through these on your own before I work it out with you. Simplify the expression by removing all factors that are perfect squares. Okay, these are the same directions that we've seen the last few times."}, {"video_title": "Simplifying radicals examples.mp3", "Sentence": "So I have one more of these examples. And like always, pause the video and see if you can work through these on your own before I work it out with you. Simplify the expression by removing all factors that are perfect squares. Okay, these are the same directions that we've seen the last few times. And so let's see. If I wanted to simplify this, this is equal to the square root of, well, 64 times 2 is 128, and 64 is a perfect square. So I'm gonna write it as 64 times 2 over, 27 is 9 times 3."}, {"video_title": "Simplifying radicals examples.mp3", "Sentence": "Okay, these are the same directions that we've seen the last few times. And so let's see. If I wanted to simplify this, this is equal to the square root of, well, 64 times 2 is 128, and 64 is a perfect square. So I'm gonna write it as 64 times 2 over, 27 is 9 times 3. 9 is a perfect square. So this is going to be the same thing. And there's a couple of ways that we could think about it."}, {"video_title": "Simplifying radicals examples.mp3", "Sentence": "So I'm gonna write it as 64 times 2 over, 27 is 9 times 3. 9 is a perfect square. So this is going to be the same thing. And there's a couple of ways that we could think about it. We could say this is the same thing as the square root of 64 times 2 over the square root of 9 times 3, which is the same thing as the square root of 64 times the square root of 2 over the square root of 9 times the square root of 3, which is equal to, this is 8, this is 3. So it would be 8 times the square root of 2 over 3 times the square root of 3. That's one way to say it."}, {"video_title": "Simplifying radicals examples.mp3", "Sentence": "And there's a couple of ways that we could think about it. We could say this is the same thing as the square root of 64 times 2 over the square root of 9 times 3, which is the same thing as the square root of 64 times the square root of 2 over the square root of 9 times the square root of 3, which is equal to, this is 8, this is 3. So it would be 8 times the square root of 2 over 3 times the square root of 3. That's one way to say it. Or we could even view the square root of 2 over the square root of 3 as the square root of 2 thirds. So we could say this is 8 over 3 times the square root of 2 thirds. So these are all possible ways of trying to tackle this."}, {"video_title": "Simplifying radicals examples.mp3", "Sentence": "That's one way to say it. Or we could even view the square root of 2 over the square root of 3 as the square root of 2 thirds. So we could say this is 8 over 3 times the square root of 2 thirds. So these are all possible ways of trying to tackle this. So we could just write it, let's see, have we removed all factors that are perfect squares? Yes, from inside the radicals. And we've combined terms."}, {"video_title": "Simplifying radicals examples.mp3", "Sentence": "So these are all possible ways of trying to tackle this. So we could just write it, let's see, have we removed all factors that are perfect squares? Yes, from inside the radicals. And we've combined terms. We weren't doing any adding or subtracting here. So it's really just removing the perfect squares from inside the radicals. And I think we've done that."}, {"video_title": "Simplifying radicals examples.mp3", "Sentence": "And we've combined terms. We weren't doing any adding or subtracting here. So it's really just removing the perfect squares from inside the radicals. And I think we've done that. So we could say this is going to be 8 thirds times the square root of 2 thirds. And there's other ways that you could express this that would be equivalent. But hopefully this makes some sense."}, {"video_title": "Proof Sum of measures of angles in a triangle are 180   Geometry   Khan Academy.mp3", "Sentence": "And what I want to prove is that the sum of the measures of the interior angles of a triangle, that x plus y plus z, is equal to 180 degrees. And the way that I'm going to do it is using our knowledge of parallel lines or transversals of parallel lines and corresponding angles. And to do that, I'm going to extend each of these sides of the triangle, which right now are line segments, but extend them into lines. So this side down here, if I keep going on and on forever in the same directions, then now all of a sudden I have an orange line. And what I want to do is construct another line that is parallel to the orange line that goes through this vertex of the triangle right over here. And I can always do that. I can just start from this point and go in the same direction as this line, and I will never intersect."}, {"video_title": "Proof Sum of measures of angles in a triangle are 180   Geometry   Khan Academy.mp3", "Sentence": "So this side down here, if I keep going on and on forever in the same directions, then now all of a sudden I have an orange line. And what I want to do is construct another line that is parallel to the orange line that goes through this vertex of the triangle right over here. And I can always do that. I can just start from this point and go in the same direction as this line, and I will never intersect. I'm not getting any closer or further away from that line. So I'm never going to intersect that line. So these two lines right over here are parallel."}, {"video_title": "Proof Sum of measures of angles in a triangle are 180   Geometry   Khan Academy.mp3", "Sentence": "I can just start from this point and go in the same direction as this line, and I will never intersect. I'm not getting any closer or further away from that line. So I'm never going to intersect that line. So these two lines right over here are parallel. This is parallel to that. Now I'm going to go to the other two sides of my original triangle and extend them into lines. So I'm going to extend this one into a line."}, {"video_title": "Proof Sum of measures of angles in a triangle are 180   Geometry   Khan Academy.mp3", "Sentence": "So these two lines right over here are parallel. This is parallel to that. Now I'm going to go to the other two sides of my original triangle and extend them into lines. So I'm going to extend this one into a line. So I'll do that as neatly as I can. So I'm going to extend that into a line. And you see that this is clearly a transversal of these two parallel lines."}, {"video_title": "Proof Sum of measures of angles in a triangle are 180   Geometry   Khan Academy.mp3", "Sentence": "So I'm going to extend this one into a line. So I'll do that as neatly as I can. So I'm going to extend that into a line. And you see that this is clearly a transversal of these two parallel lines. Now, if we have a transversal here of two parallel lines, then we must have some corresponding angles. And we see that this angle is formed when the transversal intersects the bottom orange line. Well, what's the corresponding angle when the transversal intersects this top blue line?"}, {"video_title": "Proof Sum of measures of angles in a triangle are 180   Geometry   Khan Academy.mp3", "Sentence": "And you see that this is clearly a transversal of these two parallel lines. Now, if we have a transversal here of two parallel lines, then we must have some corresponding angles. And we see that this angle is formed when the transversal intersects the bottom orange line. Well, what's the corresponding angle when the transversal intersects this top blue line? What's the angle on the top right of the intersection? Angle on the top right of the intersection must also be x. The other thing that pops out at you is there's another vertical angle with x, another angle that must be equivalent."}, {"video_title": "Proof Sum of measures of angles in a triangle are 180   Geometry   Khan Academy.mp3", "Sentence": "Well, what's the corresponding angle when the transversal intersects this top blue line? What's the angle on the top right of the intersection? Angle on the top right of the intersection must also be x. The other thing that pops out at you is there's another vertical angle with x, another angle that must be equivalent. On the opposite side of this intersection, you have this angle right over here. These two angles are vertical. So if this has measure x, then this one must have measure x as well."}, {"video_title": "Proof Sum of measures of angles in a triangle are 180   Geometry   Khan Academy.mp3", "Sentence": "The other thing that pops out at you is there's another vertical angle with x, another angle that must be equivalent. On the opposite side of this intersection, you have this angle right over here. These two angles are vertical. So if this has measure x, then this one must have measure x as well. Let's do the same thing with the last side of the triangle that we have not extended into a line yet. So let's do that. So if we take this one, so we just keep going."}, {"video_title": "Proof Sum of measures of angles in a triangle are 180   Geometry   Khan Academy.mp3", "Sentence": "So if this has measure x, then this one must have measure x as well. Let's do the same thing with the last side of the triangle that we have not extended into a line yet. So let's do that. So if we take this one, so we just keep going. So it becomes a line. So now it becomes a transversal of the two parallel lines, just like the magenta line did. And we say, hey, look, this angle y right over here, this angle is formed from the intersection of the transversal and the bottom parallel line."}, {"video_title": "Proof Sum of measures of angles in a triangle are 180   Geometry   Khan Academy.mp3", "Sentence": "So if we take this one, so we just keep going. So it becomes a line. So now it becomes a transversal of the two parallel lines, just like the magenta line did. And we say, hey, look, this angle y right over here, this angle is formed from the intersection of the transversal and the bottom parallel line. What angle does it correspond to up here? Well, this is kind of on the left side of the intersection. It corresponds to this angle right over here, where the green line, the green transversal intersects the blue parallel line."}, {"video_title": "Proof Sum of measures of angles in a triangle are 180   Geometry   Khan Academy.mp3", "Sentence": "And we say, hey, look, this angle y right over here, this angle is formed from the intersection of the transversal and the bottom parallel line. What angle does it correspond to up here? Well, this is kind of on the left side of the intersection. It corresponds to this angle right over here, where the green line, the green transversal intersects the blue parallel line. Well, what angle is vertical to it? Well, this angle. So this is going to have measure y as well."}, {"video_title": "Proof Sum of measures of angles in a triangle are 180   Geometry   Khan Academy.mp3", "Sentence": "It corresponds to this angle right over here, where the green line, the green transversal intersects the blue parallel line. Well, what angle is vertical to it? Well, this angle. So this is going to have measure y as well. So now we're really at the home stretch of our proof, because we will see that the measure, we have this angle and this angle. This has measure angle x. This has measure z."}, {"video_title": "Proof Sum of measures of angles in a triangle are 180   Geometry   Khan Academy.mp3", "Sentence": "So this is going to have measure y as well. So now we're really at the home stretch of our proof, because we will see that the measure, we have this angle and this angle. This has measure angle x. This has measure z. They're both adjacent angles. If we take the two outer rays that form the angle, and we think about this angle right over here, what's this measure of this wide angle right over there? Well, it's going to be x plus z."}, {"video_title": "Proof Sum of measures of angles in a triangle are 180   Geometry   Khan Academy.mp3", "Sentence": "This has measure z. They're both adjacent angles. If we take the two outer rays that form the angle, and we think about this angle right over here, what's this measure of this wide angle right over there? Well, it's going to be x plus z. And that angle is supplementary to this angle right over here that has measure y. So the measure of this wide angle, which is x plus z, plus the measure of this magenta angle, which is y, must be equal to 180 degrees, because these two angles are supplementary. So the measure of the wide angle, x plus z plus the measure of the magenta angle, which is supplementary to the wide angle, it must be equal to 180 degrees, because they are supplementary."}, {"video_title": "Proof Sum of measures of angles in a triangle are 180   Geometry   Khan Academy.mp3", "Sentence": "Well, it's going to be x plus z. And that angle is supplementary to this angle right over here that has measure y. So the measure of this wide angle, which is x plus z, plus the measure of this magenta angle, which is y, must be equal to 180 degrees, because these two angles are supplementary. So the measure of the wide angle, x plus z plus the measure of the magenta angle, which is supplementary to the wide angle, it must be equal to 180 degrees, because they are supplementary. Well, we could just reorder this if we want to put it in alphabetical order, but we've just completed our proof. The measure of the interior angles of the triangle, x plus z plus y. We could write this as x plus y plus z, if the lack of alphabetical order is making you uncomfortable."}, {"video_title": "How to solve equations of the form x a = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Solve for x and check your solution. We have x divided by 3 is equal to 14. So to solve for x, to figure out what the variable x must be equal to, we really just have to isolate it on the left-hand side of this equation. It's already sitting there. We have x divided by 3 is equal to 14. We could also write this as 1 3rd x is equal to 14. Obviously x times 1 3rd is going to be x over 3."}, {"video_title": "How to solve equations of the form x a = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "It's already sitting there. We have x divided by 3 is equal to 14. We could also write this as 1 3rd x is equal to 14. Obviously x times 1 3rd is going to be x over 3. These are equivalent. So how can we just end up with an x on the left-hand side of either of these equations? These are really the same thing."}, {"video_title": "How to solve equations of the form x a = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Obviously x times 1 3rd is going to be x over 3. These are equivalent. So how can we just end up with an x on the left-hand side of either of these equations? These are really the same thing. Or another way, how can we just have a 1 in front of the x, a 1x, which is really just saying x over here? Well, I'm dividing it by 3 right now. So if I were to multiply both sides of this equation by 3, that would isolate the x."}, {"video_title": "How to solve equations of the form x a = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "These are really the same thing. Or another way, how can we just have a 1 in front of the x, a 1x, which is really just saying x over here? Well, I'm dividing it by 3 right now. So if I were to multiply both sides of this equation by 3, that would isolate the x. And the reason that would work is if I multiply this by 3 over here, I'm multiplying by 3 and dividing by 3. That's equivalent to multiplying or dividing by 1. These guys cancel out."}, {"video_title": "How to solve equations of the form x a = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So if I were to multiply both sides of this equation by 3, that would isolate the x. And the reason that would work is if I multiply this by 3 over here, I'm multiplying by 3 and dividing by 3. That's equivalent to multiplying or dividing by 1. These guys cancel out. But remember, if you do it to the left-hand side, you also have to do it to the right-hand side. And actually, I'll do both of these equations at the same time because they're really the exact same equation. So what are we going to get over here on the left-hand side?"}, {"video_title": "How to solve equations of the form x a = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "These guys cancel out. But remember, if you do it to the left-hand side, you also have to do it to the right-hand side. And actually, I'll do both of these equations at the same time because they're really the exact same equation. So what are we going to get over here on the left-hand side? 3 times anything divided by 3 is going to be that anything. We're just going to have an x left over on the left-hand side. And on the right-hand side, what's 14 times 3?"}, {"video_title": "How to solve equations of the form x a = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So what are we going to get over here on the left-hand side? 3 times anything divided by 3 is going to be that anything. We're just going to have an x left over on the left-hand side. And on the right-hand side, what's 14 times 3? 3 times 10 is 30, 3 times 4 is 12, so it's going to be 42. So we get x is equal to 42. And the same thing would happen here."}, {"video_title": "How to solve equations of the form x a = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "And on the right-hand side, what's 14 times 3? 3 times 10 is 30, 3 times 4 is 12, so it's going to be 42. So we get x is equal to 42. And the same thing would happen here. 3 times 1 third is just 1. So you get 1x is equal to 14 times 3, which is 42. Now let's just check our answer."}, {"video_title": "How to solve equations of the form x a = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "And the same thing would happen here. 3 times 1 third is just 1. So you get 1x is equal to 14 times 3, which is 42. Now let's just check our answer. Let's substitute 42 into our original equation. So we have 42 in place for x over 3 is equal to 14. So what's 42 divided by 3?"}, {"video_title": "How to solve equations of the form x a = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Now let's just check our answer. Let's substitute 42 into our original equation. So we have 42 in place for x over 3 is equal to 14. So what's 42 divided by 3? And we could do a little bit of, I guess we'd call it medium-long division. It's not really long division. 3 into 4, 3 goes into 4 one time."}, {"video_title": "How to solve equations of the form x a = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So what's 42 divided by 3? And we could do a little bit of, I guess we'd call it medium-long division. It's not really long division. 3 into 4, 3 goes into 4 one time. 1 times 3 is 3. You subtract. 4 minus 3 is 1."}, {"video_title": "How to solve equations of the form x a = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "3 into 4, 3 goes into 4 one time. 1 times 3 is 3. You subtract. 4 minus 3 is 1. Bring down the 2. 3 goes into 12 four times. So 3 goes into 42 14 times."}, {"video_title": "How to solve equations of the form x a = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "4 minus 3 is 1. Bring down the 2. 3 goes into 12 four times. So 3 goes into 42 14 times. So this right over here simplifies to 14. And it all checks out. So we're done."}, {"video_title": "Adding numbers with different signs (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "So let's just do 15, 15, plus, plus negative, let me do that in a different color, plus negative 46, I'm doing that orange, plus negative 46. Let me draw a number line here, just so we can properly visualize what is going on. So that's my number line. That's my number line. We're starting at 15, so let me put, let's draw a zero over here. And so we are starting at 15, 15 could be right over here, so this is 15. Maybe just draw, let me draw a big fat arrow to signify this is 15."}, {"video_title": "Adding numbers with different signs (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "That's my number line. We're starting at 15, so let me put, let's draw a zero over here. And so we are starting at 15, 15 could be right over here, so this is 15. Maybe just draw, let me draw a big fat arrow to signify this is 15. 15 has an absolute value of 15, so the length of this arrow would be 15. Now, we're adding negative 46 to that 15. That is the same thing."}, {"video_title": "Adding numbers with different signs (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "Maybe just draw, let me draw a big fat arrow to signify this is 15. 15 has an absolute value of 15, so the length of this arrow would be 15. Now, we're adding negative 46 to that 15. That is the same thing. This is equivalent to 15 minus 46. Minus 46, which means we are going to move 46 spots to the left of 15. Negative sign, or minus, means we're moving to the left on the number line."}, {"video_title": "Adding numbers with different signs (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "That is the same thing. This is equivalent to 15 minus 46. Minus 46, which means we are going to move 46 spots to the left of 15. Negative sign, or minus, means we're moving to the left on the number line. So we're going to move 46 to the left, so we're starting at 15, and we are going to move 46, we are going to move 46 to the left. So the length of this arrow right here, the length of this arrow is going to be 46. And we're moving to the left."}, {"video_title": "Adding numbers with different signs (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "Negative sign, or minus, means we're moving to the left on the number line. So we're going to move 46 to the left, so we're starting at 15, and we are going to move 46, we are going to move 46 to the left. So the length of this arrow right here, the length of this arrow is going to be 46. And we're moving to the left. This is the negative 46 that we're adding to the 15. So we're going to end up, we're going to end up at some point over here. That point is clearly zero, because we were 15 to the right."}, {"video_title": "Adding numbers with different signs (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "And we're moving to the left. This is the negative 46 that we're adding to the 15. So we're going to end up, we're going to end up at some point over here. That point is clearly zero, because we were 15 to the right. Now we're going to move 46 to the left, so we're definitely going to be to the left of zero, so it's definitely going to be a negative number. And we can even think about, we can even think about the absolute value of that negative number. We can just visualize it."}, {"video_title": "Adding numbers with different signs (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "That point is clearly zero, because we were 15 to the right. Now we're going to move 46 to the left, so we're definitely going to be to the left of zero, so it's definitely going to be a negative number. And we can even think about, we can even think about the absolute value of that negative number. We can just visualize it. We moved this yellow arrow as a length of 15, this orange arrow has a length of 46, the blue arrow that I'm about to draw, which is the sum of these two, is going to be, is going to be right, is going to have this length right over here. And just visually, how could we figure out the length of this blue part, if we know the length of this orange part and we know the length of this yellow part? What's this going to be the difference?"}, {"video_title": "Adding numbers with different signs (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "We can just visualize it. We moved this yellow arrow as a length of 15, this orange arrow has a length of 46, the blue arrow that I'm about to draw, which is the sum of these two, is going to be, is going to be right, is going to have this length right over here. And just visually, how could we figure out the length of this blue part, if we know the length of this orange part and we know the length of this yellow part? What's this going to be the difference? It's just going to be the difference of these two. So the absolute value of the sum is going to be the difference between this length, 46 and 15. So this is 46, let me just figure that out, 46 minus 15, 6 minus 5 is 1, 4 minus 1 is 3."}, {"video_title": "Adding numbers with different signs (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "What's this going to be the difference? It's just going to be the difference of these two. So the absolute value of the sum is going to be the difference between this length, 46 and 15. So this is 46, let me just figure that out, 46 minus 15, 6 minus 5 is 1, 4 minus 1 is 3. So the length of this is going to be 31, and it's going to be 31 to the left of 0. So this is going to be negative 31 right over here. So we know that this first part over here is negative 31, and then to that we are going to add 29."}, {"video_title": "Adding numbers with different signs (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "So this is 46, let me just figure that out, 46 minus 15, 6 minus 5 is 1, 4 minus 1 is 3. So the length of this is going to be 31, and it's going to be 31 to the left of 0. So this is going to be negative 31 right over here. So we know that this first part over here is negative 31, and then to that we are going to add 29. So we're going to add 29, let me do that in another color. So what does that mean? That means that we're going to start at negative 31, and we're going to move 29 to the right, we're adding 29."}, {"video_title": "Adding numbers with different signs (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "So we know that this first part over here is negative 31, and then to that we are going to add 29. So we're going to add 29, let me do that in another color. So what does that mean? That means that we're going to start at negative 31, and we're going to move 29 to the right, we're adding 29. So we're going to move 29 to the right, so maybe that gets us right about there. Try and draw an arrow of length 29. I can draw a cleaner looking arrow than that."}, {"video_title": "Adding numbers with different signs (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "That means that we're going to start at negative 31, and we're going to move 29 to the right, we're adding 29. So we're going to move 29 to the right, so maybe that gets us right about there. Try and draw an arrow of length 29. I can draw a cleaner looking arrow than that. I'll do it right over here actually. So then we're going to move 29 over to the right. That's the 29 part."}, {"video_title": "Adding numbers with different signs (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "I can draw a cleaner looking arrow than that. I'll do it right over here actually. So then we're going to move 29 over to the right. That's the 29 part. So now this is a positive 29. So how do we figure out what this is? So this is going to land us, so this number, this is 29 right here that we're adding, this is going to land us right over here on the number line."}, {"video_title": "Adding numbers with different signs (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "That's the 29 part. So now this is a positive 29. So how do we figure out what this is? So this is going to land us, so this number, this is 29 right here that we're adding, this is going to land us right over here on the number line. So how do we figure out what number that is? Well, once again we can just visualize it, and eventually you won't have to draw number lines and stuff, but I think it'll be useful here. The number that we're going to do, we're starting at negative 31, we're adding 29 to it, so it's going to make it less negative, but we're still adding less than 31, so we're not going to get all the way back to 0, but we're still going to have a negative number."}, {"video_title": "Adding numbers with different signs (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "So this is going to land us, so this number, this is 29 right here that we're adding, this is going to land us right over here on the number line. So how do we figure out what number that is? Well, once again we can just visualize it, and eventually you won't have to draw number lines and stuff, but I think it'll be useful here. The number that we're going to do, we're starting at negative 31, we're adding 29 to it, so it's going to make it less negative, but we're still adding less than 31, so we're not going to get all the way back to 0, but we're still going to have a negative number. So we're still going to have a negative number, but how can we figure out the absolute value of that negative number, its distance? Well, once again, that little white part right there, that white part plus this 29 is going to equal 31 if you just think of absolute value, if we don't think about the signs, if we just think about the length. Or another way to think about it, 31 minus 29 will give us the length of that white part, and of course it's going to be negative because the negative number here is larger than the positive number, and we're adding the two."}, {"video_title": "Adding numbers with different signs (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "The number that we're going to do, we're starting at negative 31, we're adding 29 to it, so it's going to make it less negative, but we're still adding less than 31, so we're not going to get all the way back to 0, but we're still going to have a negative number. So we're still going to have a negative number, but how can we figure out the absolute value of that negative number, its distance? Well, once again, that little white part right there, that white part plus this 29 is going to equal 31 if you just think of absolute value, if we don't think about the signs, if we just think about the length. Or another way to think about it, 31 minus 29 will give us the length of that white part, and of course it's going to be negative because the negative number here is larger than the positive number, and we're adding the two. So if we do 31 minus 29, you could borrow in all of that, but that's clearly just going to be equal to 2. You could say that's 11, this is a 2, you subtract, this is equal to 2, but since it's negative 31 plus 29, it's going to be a negative 2. We're still going to stay negative."}, {"video_title": "Adding numbers with different signs (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "Or another way to think about it, 31 minus 29 will give us the length of that white part, and of course it's going to be negative because the negative number here is larger than the positive number, and we're adding the two. So if we do 31 minus 29, you could borrow in all of that, but that's clearly just going to be equal to 2. You could say that's 11, this is a 2, you subtract, this is equal to 2, but since it's negative 31 plus 29, it's going to be a negative 2. We're still going to stay negative. We haven't moved far to the right enough to pass 0, so this right here is going to be negative 2. Or another way to think about it, the length of this white bar, the absolute value is going to be 2. 2 plus 29 is 31, but we're operating to the left of 0, so it's negative 31."}, {"video_title": "Adding numbers with different signs (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "We're still going to stay negative. We haven't moved far to the right enough to pass 0, so this right here is going to be negative 2. Or another way to think about it, the length of this white bar, the absolute value is going to be 2. 2 plus 29 is 31, but we're operating to the left of 0, so it's negative 31. This is negative 2. Anyway, hopefully you found that useful. Well, anyway, let me make it clear."}, {"video_title": "Finding the inequality representing the graph example   Algebra I   Khan Academy.mp3", "Sentence": "So if you give us any x, and let me label the axes here, so this is the x-axis, this is the y-axis. So this is saying, you give me an x. So let's say we take x is equal to 1 right there. 3 times 1 plus 5, so 3 times x plus 5. So 3 times 1 is 3, plus 5 is 8. So 1, 2, 3, 4, 5, 6, 7, 8. This is saying that y will be less than 8. y will be less than 3 times 1 plus 5."}, {"video_title": "Finding the inequality representing the graph example   Algebra I   Khan Academy.mp3", "Sentence": "3 times 1 plus 5, so 3 times x plus 5. So 3 times 1 is 3, plus 5 is 8. So 1, 2, 3, 4, 5, 6, 7, 8. This is saying that y will be less than 8. y will be less than 3 times 1 plus 5. So the y values that satisfy this constraint for that x are going to be all of these values down here. Let me do it in a lighter color. It'll be all of these values."}, {"video_title": "Finding the inequality representing the graph example   Algebra I   Khan Academy.mp3", "Sentence": "This is saying that y will be less than 8. y will be less than 3 times 1 plus 5. So the y values that satisfy this constraint for that x are going to be all of these values down here. Let me do it in a lighter color. It'll be all of these values. For x is equal to 1, it'll be all the values down here. And it would not include y is equal to 8. y has to be less than 8. Now, if we kept doing that, we would essentially just graph the line of y is equal to 3x plus 5, but we wouldn't include it."}, {"video_title": "Finding the inequality representing the graph example   Algebra I   Khan Academy.mp3", "Sentence": "It'll be all of these values. For x is equal to 1, it'll be all the values down here. And it would not include y is equal to 8. y has to be less than 8. Now, if we kept doing that, we would essentially just graph the line of y is equal to 3x plus 5, but we wouldn't include it. We would just include everything below it, just like we did right here. So we know how to graph just y is equal to 3x plus 5. Let me write it over here."}, {"video_title": "Finding the inequality representing the graph example   Algebra I   Khan Academy.mp3", "Sentence": "Now, if we kept doing that, we would essentially just graph the line of y is equal to 3x plus 5, but we wouldn't include it. We would just include everything below it, just like we did right here. So we know how to graph just y is equal to 3x plus 5. Let me write it over here. So if I were to write y is equal to 3x plus 5, we'd say, OK, 3 is the slope. Slope is equal to 3. And then 5 is the y-intercept."}, {"video_title": "Finding the inequality representing the graph example   Algebra I   Khan Academy.mp3", "Sentence": "Let me write it over here. So if I were to write y is equal to 3x plus 5, we'd say, OK, 3 is the slope. Slope is equal to 3. And then 5 is the y-intercept. Now, I could just graph the line, but because that won't be included in the y's that satisfy this constraint, I'm going to graph it as a dotted line. So we'll start with the y-intercept of 5. So 1, 2, 3, 4, 5."}, {"video_title": "Finding the inequality representing the graph example   Algebra I   Khan Academy.mp3", "Sentence": "And then 5 is the y-intercept. Now, I could just graph the line, but because that won't be included in the y's that satisfy this constraint, I'm going to graph it as a dotted line. So we'll start with the y-intercept of 5. So 1, 2, 3, 4, 5. That's the y-intercept. And the slope is 3. So if you go over the 1, you go up 3."}, {"video_title": "Finding the inequality representing the graph example   Algebra I   Khan Academy.mp3", "Sentence": "So 1, 2, 3, 4, 5. That's the y-intercept. And the slope is 3. So if you go over the 1, you go up 3. Let me do that in that darker purple color. So it will look like this. It will look like that."}, {"video_title": "Finding the inequality representing the graph example   Algebra I   Khan Academy.mp3", "Sentence": "So if you go over the 1, you go up 3. Let me do that in that darker purple color. So it will look like this. It will look like that. And you see that point would be on it. That point would be on it. If you go back, you're going to go down by 3."}, {"video_title": "Finding the inequality representing the graph example   Algebra I   Khan Academy.mp3", "Sentence": "It will look like that. And you see that point would be on it. That point would be on it. If you go back, you're going to go down by 3. So that point will be on it, that point and that point. And then I'll just connect the dots with a dotted line. So it will look."}, {"video_title": "Finding the inequality representing the graph example   Algebra I   Khan Academy.mp3", "Sentence": "If you go back, you're going to go down by 3. So that point will be on it, that point and that point. And then I'll just connect the dots with a dotted line. So it will look. So that dotted line is the graph of y is equal to 3x plus 5. But we're not going to include it. So that's why I made it a dotted line."}, {"video_title": "Finding the inequality representing the graph example   Algebra I   Khan Academy.mp3", "Sentence": "So it will look. So that dotted line is the graph of y is equal to 3x plus 5. But we're not going to include it. So that's why I made it a dotted line. So we're going to include the y's that are less than that. So for any x, so you pick an x. Let's say x is equal to negative 1."}, {"video_title": "Finding the inequality representing the graph example   Algebra I   Khan Academy.mp3", "Sentence": "So that's why I made it a dotted line. So we're going to include the y's that are less than that. So for any x, so you pick an x. Let's say x is equal to negative 1. If you evaluate 3x plus 5 for that x, you'd get here. But we only care about the y's that are strictly less than that. So you don't include the line."}, {"video_title": "Finding the inequality representing the graph example   Algebra I   Khan Academy.mp3", "Sentence": "Let's say x is equal to negative 1. If you evaluate 3x plus 5 for that x, you'd get here. But we only care about the y's that are strictly less than that. So you don't include the line. It's everything below it. So for any x you pick, it's going to be below that line. You take the x, go up to that line, and everything below it."}, {"video_title": "Finding the inequality representing the graph example   Algebra I   Khan Academy.mp3", "Sentence": "So you don't include the line. It's everything below it. So for any x you pick, it's going to be below that line. You take the x, go up to that line, and everything below it. So for all of the x's, it's going to be this entire area. It's going to be this entire area. It's going to be this entire area that's under the line."}, {"video_title": "Red blood cells in human body (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "A human body has five liters of blood, 40% of which is red blood cells. Each red blood cell has a volume of approximately 90 times 10 to the negative 15 liters. How many red blood cells are there in a human body? Write your answer in scientific notation and round to two decimal places. So they tell us the total volume of blood in the human body. We have five liters. So we have five liters."}, {"video_title": "Red blood cells in human body (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "Write your answer in scientific notation and round to two decimal places. So they tell us the total volume of blood in the human body. We have five liters. So we have five liters. And then they tell us that 40% of that is red blood cells. So if we take the five liters and we multiply by 40%, this expression right here gives us the total volume of the red blood cells, 40% of our total volume of blood. Now, if this is a total volume of red blood cells and we divide by the volume of each red blood cell, then we're going to get the number of red blood cells."}, {"video_title": "Red blood cells in human body (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "So we have five liters. And then they tell us that 40% of that is red blood cells. So if we take the five liters and we multiply by 40%, this expression right here gives us the total volume of the red blood cells, 40% of our total volume of blood. Now, if this is a total volume of red blood cells and we divide by the volume of each red blood cell, then we're going to get the number of red blood cells. So let's do that. Let's divide by the volume of each red blood cell. So the volume of each red blood cell is 90 times 10 to the negative 15 liters."}, {"video_title": "Red blood cells in human body (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "Now, if this is a total volume of red blood cells and we divide by the volume of each red blood cell, then we're going to get the number of red blood cells. So let's do that. Let's divide by the volume of each red blood cell. So the volume of each red blood cell is 90 times 10 to the negative 15 liters. So let's see if we can simplify it. So one thing that we can feel good about is that the units actually do cancel out. We have liters in the numerator, liters in the denominator."}, {"video_title": "Red blood cells in human body (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "So the volume of each red blood cell is 90 times 10 to the negative 15 liters. So let's see if we can simplify it. So one thing that we can feel good about is that the units actually do cancel out. We have liters in the numerator, liters in the denominator. So we're going to get just a pure number, which is what we want. We just want how many red blood cells there are actually in the body. So let's just focus on the numbers here."}, {"video_title": "Red blood cells in human body (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "We have liters in the numerator, liters in the denominator. So we're going to get just a pure number, which is what we want. We just want how many red blood cells there are actually in the body. So let's just focus on the numbers here. So 5 times 40%. Well, 40% is the same thing as 0.4. So let me write that down."}, {"video_title": "Red blood cells in human body (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "So let's just focus on the numbers here. So 5 times 40%. Well, 40% is the same thing as 0.4. So let me write that down. This is the same thing as 0.4. 5 times 0.4 is 2. So our numerator simplifies to 2."}, {"video_title": "Red blood cells in human body (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "So let me write that down. This is the same thing as 0.4. 5 times 0.4 is 2. So our numerator simplifies to 2. And in the denominator, we have 90 times 10 to the negative 15, which isn't quite in scientific notation. Or actually, it definitely is not in scientific notation. It looks like it."}, {"video_title": "Red blood cells in human body (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "So our numerator simplifies to 2. And in the denominator, we have 90 times 10 to the negative 15, which isn't quite in scientific notation. Or actually, it definitely is not in scientific notation. It looks like it. But remember, in order to be in scientific notation, this number has to be greater than or equal to 1 and less than 10. It's clearly not less than 10. But we can convert this to scientific notation very easily."}, {"video_title": "Red blood cells in human body (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "It looks like it. But remember, in order to be in scientific notation, this number has to be greater than or equal to 1 and less than 10. It's clearly not less than 10. But we can convert this to scientific notation very easily. 90 is the same thing as 9 times 10. Or you could even say 9 times 10 to the first. And then you multiply that times 10 to the negative 15."}, {"video_title": "Red blood cells in human body (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "But we can convert this to scientific notation very easily. 90 is the same thing as 9 times 10. Or you could even say 9 times 10 to the first. And then you multiply that times 10 to the negative 15. And then this simplifies to 9 times 10 to the, well, let's add these two exponents, 10 to the negative 14. And now we can actually divide. And let's simplify this division a little bit."}, {"video_title": "Red blood cells in human body (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "And then you multiply that times 10 to the negative 15. And then this simplifies to 9 times 10 to the, well, let's add these two exponents, 10 to the negative 14. And now we can actually divide. And let's simplify this division a little bit. This is going to be the same thing as 2 over 9 times 1 over 10 to the negative 14. Well, what's 1 over 10 to the negative 14? Well, that's just 10 to the 14."}, {"video_title": "Red blood cells in human body (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "And let's simplify this division a little bit. This is going to be the same thing as 2 over 9 times 1 over 10 to the negative 14. Well, what's 1 over 10 to the negative 14? Well, that's just 10 to the 14. So this right over here is just the same thing as 10 to the 14. Now, you might say, OK, we just have to figure out what 2 ninths is. And we're done."}, {"video_title": "Red blood cells in human body (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "Well, that's just 10 to the 14. So this right over here is just the same thing as 10 to the 14. Now, you might say, OK, we just have to figure out what 2 ninths is. And we're done. We've written this in scientific notation. But you might already realize, look, 2 ninths is not greater than or equal to 1. How can we make this greater than or equal to 1?"}, {"video_title": "Red blood cells in human body (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "And we're done. We've written this in scientific notation. But you might already realize, look, 2 ninths is not greater than or equal to 1. How can we make this greater than or equal to 1? Well, we could multiply it by 10. If we multiply this by 10, then we've got to divide this by 10 to not change the value of this expression. But let's do that."}, {"video_title": "Red blood cells in human body (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "How can we make this greater than or equal to 1? Well, we could multiply it by 10. If we multiply this by 10, then we've got to divide this by 10 to not change the value of this expression. But let's do that. So this is equal to, if we multiply, so I'm going to multiply this by 10. And I'm going to divide this by 10. So I haven't changed."}, {"video_title": "Red blood cells in human body (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "But let's do that. So this is equal to, if we multiply, so I'm going to multiply this by 10. And I'm going to divide this by 10. So I haven't changed. I've multiplied and divided by 10. So this is equal to 20 over 9 times 10 to the 14th divided by 10 is 10 to the 13th power. So what's 20 over 9?"}, {"video_title": "Red blood cells in human body (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "So I haven't changed. I've multiplied and divided by 10. So this is equal to 20 over 9 times 10 to the 14th divided by 10 is 10 to the 13th power. So what's 20 over 9? This is going to give us a number that is greater than or equal to 1 and less than 10. So let's figure it out. And they wanted us to round, I think they said, round our answer to two decimal places."}, {"video_title": "Red blood cells in human body (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "So what's 20 over 9? This is going to give us a number that is greater than or equal to 1 and less than 10. So let's figure it out. And they wanted us to round, I think they said, round our answer to two decimal places. So let's do that. So 20 divided by 9. 9 doesn't go into 2."}, {"video_title": "Red blood cells in human body (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "And they wanted us to round, I think they said, round our answer to two decimal places. So let's do that. So 20 divided by 9. 9 doesn't go into 2. It does go into 20. 2 times 9 is 18. Subtract."}, {"video_title": "Red blood cells in human body (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "9 doesn't go into 2. It does go into 20. 2 times 9 is 18. Subtract. Get a remainder of 2. I think you see where this show is going to go. Get a remainder of 2."}, {"video_title": "Red blood cells in human body (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "Subtract. Get a remainder of 2. I think you see where this show is going to go. Get a remainder of 2. 9 goes into 20 2 times. 2 times 9 is 18. We're just going to keep getting 2's."}, {"video_title": "Red blood cells in human body (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "Get a remainder of 2. 9 goes into 20 2 times. 2 times 9 is 18. We're just going to keep getting 2's. So we get another 2. Bring down a 0. 9 goes into 20 2 times."}, {"video_title": "Red blood cells in human body (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "We're just going to keep getting 2's. So we get another 2. Bring down a 0. 9 goes into 20 2 times. So this thing right over here is really 2.2 repeating. But they said round to two decimal places. So this is going to be equal to 2.22 times 10 to the 13th power."}, {"video_title": "Compound inequalities 2   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So we have to find all the y's that meet both of these constraints. So let's just solve for y in each of the constraints. And just remember that this and is here. So we have 3y plus 7 is less than 2y. So let's isolate the y's on the left-hand side. So let's get rid of this 2y on the right-hand side. And we can do that by subtracting 2y from both sides."}, {"video_title": "Compound inequalities 2   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So we have 3y plus 7 is less than 2y. So let's isolate the y's on the left-hand side. So let's get rid of this 2y on the right-hand side. And we can do that by subtracting 2y from both sides. So we're going to subtract 2y from both sides. The left-hand side, we have 3y minus 2y, which is just y, plus 7 is less than 2y minus 2y. And there's nothing else there."}, {"video_title": "Compound inequalities 2   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "And we can do that by subtracting 2y from both sides. So we're going to subtract 2y from both sides. The left-hand side, we have 3y minus 2y, which is just y, plus 7 is less than 2y minus 2y. And there's nothing else there. That's just going to be 0. And then we can get rid of this 7 here by subtracting 7 from both sides. So let's subtract 7 from both sides."}, {"video_title": "Compound inequalities 2   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "And there's nothing else there. That's just going to be 0. And then we can get rid of this 7 here by subtracting 7 from both sides. So let's subtract 7 from both sides. Left-hand side, y plus 7 minus 7, those cancel out. We just have a y, is less than 0 minus 7, which is negative 7. So that's one of the constraints."}, {"video_title": "Compound inequalities 2   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So let's subtract 7 from both sides. Left-hand side, y plus 7 minus 7, those cancel out. We just have a y, is less than 0 minus 7, which is negative 7. So that's one of the constraints. That's this constraint right over here. Now let's work on this constraint. We have 4y plus 8 is greater than negative 48."}, {"video_title": "Compound inequalities 2   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So that's one of the constraints. That's this constraint right over here. Now let's work on this constraint. We have 4y plus 8 is greater than negative 48. So let's get rid of the 8 from the left-hand side. So we can subtract 8 from both sides. The left-hand side, we're just left with a 4y, because these guys cancel out."}, {"video_title": "Compound inequalities 2   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "We have 4y plus 8 is greater than negative 48. So let's get rid of the 8 from the left-hand side. So we can subtract 8 from both sides. The left-hand side, we're just left with a 4y, because these guys cancel out. 4y is greater than negative 48 minus 8. So we're going to go another 8 negative. So 48 plus 8 would be 56."}, {"video_title": "Compound inequalities 2   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "The left-hand side, we're just left with a 4y, because these guys cancel out. 4y is greater than negative 48 minus 8. So we're going to go another 8 negative. So 48 plus 8 would be 56. So this is going to be negative 56. And now to isolate the y, we can divide both sides by positive 4. And we don't have to swap the inequality, since we're dividing by a positive number."}, {"video_title": "Compound inequalities 2   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So 48 plus 8 would be 56. So this is going to be negative 56. And now to isolate the y, we can divide both sides by positive 4. And we don't have to swap the inequality, since we're dividing by a positive number. So let's divide both sides by 4 over here. So we get y is greater than, what is 56 over 4? Negative 56 over 4."}, {"video_title": "Compound inequalities 2   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "And we don't have to swap the inequality, since we're dividing by a positive number. So let's divide both sides by 4 over here. So we get y is greater than, what is 56 over 4? Negative 56 over 4. Let's see, 40 is 10 times 4. And then we have another 16 to worry about. So it's 14 times 4."}, {"video_title": "Compound inequalities 2   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Negative 56 over 4. Let's see, 40 is 10 times 4. And then we have another 16 to worry about. So it's 14 times 4. So y is greater than negative 14. Is that right? 4 times 10 is 40."}, {"video_title": "Compound inequalities 2   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So it's 14 times 4. So y is greater than negative 14. Is that right? 4 times 10 is 40. 4 times 4 is 16. Yep, 56. So y is greater than negative 14."}, {"video_title": "Compound inequalities 2   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "4 times 10 is 40. 4 times 4 is 16. Yep, 56. So y is greater than negative 14. And let's remember, we have this and here. And y is less than negative 7. So we have to meet both of these constraints over here."}, {"video_title": "Compound inequalities 2   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So y is greater than negative 14. And let's remember, we have this and here. And y is less than negative 7. So we have to meet both of these constraints over here. So let's draw them on the number line. So I have my number line over here. And let's say negative 14 is over here."}, {"video_title": "Compound inequalities 2   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So we have to meet both of these constraints over here. So let's draw them on the number line. So I have my number line over here. And let's say negative 14 is over here. So you have negative 13, 12, 11, 10, 9, 8, 7. That's negative 7. And then negative 6, 5, 4, 3, 2, 1."}, {"video_title": "Compound inequalities 2   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "And let's say negative 14 is over here. So you have negative 13, 12, 11, 10, 9, 8, 7. That's negative 7. And then negative 6, 5, 4, 3, 2, 1. This would be 0. Then you could keep going up more positive. And so we're looking for all of the y's that are less than negative 7."}, {"video_title": "Compound inequalities 2   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "And then negative 6, 5, 4, 3, 2, 1. This would be 0. Then you could keep going up more positive. And so we're looking for all of the y's that are less than negative 7. So let's look at this. Less than negative 7. So not including negative 7."}, {"video_title": "Compound inequalities 2   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "And so we're looking for all of the y's that are less than negative 7. So let's look at this. Less than negative 7. So not including negative 7. So we'll do an open circle around negative 7. And less than negative 7. And if that was the only constraint, we would keep going to the left."}, {"video_title": "Compound inequalities 2   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So not including negative 7. So we'll do an open circle around negative 7. And less than negative 7. And if that was the only constraint, we would keep going to the left. But we have this other constraint. And y has to be greater than negative 14. And y is greater than negative 14."}, {"video_title": "Compound inequalities 2   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "And if that was the only constraint, we would keep going to the left. But we have this other constraint. And y has to be greater than negative 14. And y is greater than negative 14. So you make a circle around negative 14 and everything that's greater than that. And if you didn't have this other constraint, you would keep going. But the y's that satisfy both of them are all of the y's in between."}, {"video_title": "Compound inequalities 2   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "And y is greater than negative 14. So you make a circle around negative 14 and everything that's greater than that. And if you didn't have this other constraint, you would keep going. But the y's that satisfy both of them are all of the y's in between. These are the y's that are both less than negative 7 and greater than negative 14. We can verify that things here work. So let's try some values out."}, {"video_title": "Compound inequalities 2   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "But the y's that satisfy both of them are all of the y's in between. These are the y's that are both less than negative 7 and greater than negative 14. We can verify that things here work. So let's try some values out. So a value that would work, well, let me just do negative 10 is right here. 8, 9. This is negative 10."}, {"video_title": "Compound inequalities 2   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So let's try some values out. So a value that would work, well, let me just do negative 10 is right here. 8, 9. This is negative 10. That should work. So let's try it out. So we'd have 3 times negative 10 plus 7 should be less than 2 times negative 10."}, {"video_title": "Compound inequalities 2   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "This is negative 10. That should work. So let's try it out. So we'd have 3 times negative 10 plus 7 should be less than 2 times negative 10. So this is negative 30 plus 7 is negative 23, which is indeed less than negative 20. So that works. And negative 10 has to work for this one as well."}, {"video_title": "Compound inequalities 2   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So we'd have 3 times negative 10 plus 7 should be less than 2 times negative 10. So this is negative 30 plus 7 is negative 23, which is indeed less than negative 20. So that works. And negative 10 has to work for this one as well. So you have 4 times negative 10, which is negative 40, plus 8. Negative 40 plus 8 should be greater than negative 48. Well, negative 40 plus 8 is negative 32."}, {"video_title": "Compound inequalities 2   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "And negative 10 has to work for this one as well. So you have 4 times negative 10, which is negative 40, plus 8. Negative 40 plus 8 should be greater than negative 48. Well, negative 40 plus 8 is negative 32. We're going 8 in the positive direction. So we're getting less negative. And negative 32 is greater than negative 48."}, {"video_title": "Compound inequalities 2   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Well, negative 40 plus 8 is negative 32. We're going 8 in the positive direction. So we're getting less negative. And negative 32 is greater than negative 48. It's less negative. So this works. So negative 10 works."}, {"video_title": "Compound inequalities 2   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "And negative 32 is greater than negative 48. It's less negative. So this works. So negative 10 works. Now let's just verify things that shouldn't work. So 0 should not work. It's not in the solution set."}, {"video_title": "Compound inequalities 2   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So negative 10 works. Now let's just verify things that shouldn't work. So 0 should not work. It's not in the solution set. So let's try it out. We fit 3 times 0 plus 7. That would be 7."}, {"video_title": "Compound inequalities 2   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "It's not in the solution set. So let's try it out. We fit 3 times 0 plus 7. That would be 7. And 7 is not less than 0. So it would violate this condition right over here if you put a 0 over here. If you put a negative 15 over here, it should violate this condition right over here because it wasn't in this guy's solution set."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "What is the equation of the line? So let's just try to visualize this. So that is my x-axis. And you don't have to draw it to do this problem, but it always helps to visualize. That is my y-axis. And the first point is (-1, 6). So (-1, 1, 2, 3, 4, 5, 6)."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "And you don't have to draw it to do this problem, but it always helps to visualize. That is my y-axis. And the first point is (-1, 6). So (-1, 1, 2, 3, 4, 5, 6). So it's this point right over there. It's (-1, 6). And the other point is (-5, 4)."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "So (-1, 1, 2, 3, 4, 5, 6). So it's this point right over there. It's (-1, 6). And the other point is (-5, 4). So 1, 2, 3, 4, 5. And then we go down 4. So 1, 2, 3, 4."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "And the other point is (-5, 4). So 1, 2, 3, 4, 5. And then we go down 4. So 1, 2, 3, 4. So it's right over there. And so the line that connects them will look something like this. The line will draw a rough approximation."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "So 1, 2, 3, 4. So it's right over there. And so the line that connects them will look something like this. The line will draw a rough approximation. I could draw a straighter line than that. I'll draw a little dotted line maybe. Easier to do dotted lines."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "The line will draw a rough approximation. I could draw a straighter line than that. I'll draw a little dotted line maybe. Easier to do dotted lines. So the line will look something like that. So let's find its equation. So a good place to start is we could find its slope."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "Easier to do dotted lines. So the line will look something like that. So let's find its equation. So a good place to start is we could find its slope. Remember, we can find equation y is equal to mx plus b. This is a slope-intercept form where m is a slope and b is a y-intercept. We can first try to solve for m. We could find the slope of this line."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "So a good place to start is we could find its slope. Remember, we can find equation y is equal to mx plus b. This is a slope-intercept form where m is a slope and b is a y-intercept. We can first try to solve for m. We could find the slope of this line. So m, or the slope, is the change in y over the change in x. Or we could view it as the y values of our end point minus the y value of our starting point over the x values of our end point minus the x values of our starting point. Let me make that clear."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "We can first try to solve for m. We could find the slope of this line. So m, or the slope, is the change in y over the change in x. Or we could view it as the y values of our end point minus the y value of our starting point over the x values of our end point minus the x values of our starting point. Let me make that clear. So this is equal to change in y over change in x, which is the same thing as rise over run, which is the same thing as the y value of your ending point minus the y value of your starting point. This is the same exact thing as change in y. And that over the x value of your ending point minus the x value of your starting point."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "Let me make that clear. So this is equal to change in y over change in x, which is the same thing as rise over run, which is the same thing as the y value of your ending point minus the y value of your starting point. This is the same exact thing as change in y. And that over the x value of your ending point minus the x value of your starting point. This is the exact same thing as change in x. And you just have to pick one of these as a starting point and one as the ending point. So let's just make this over here our starting point and make that our ending point."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "And that over the x value of your ending point minus the x value of your starting point. This is the exact same thing as change in x. And you just have to pick one of these as a starting point and one as the ending point. So let's just make this over here our starting point and make that our ending point. So what is our change in y? So our change in y, we started at y is equal to 6. We started at y is equal to 6."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "So let's just make this over here our starting point and make that our ending point. So what is our change in y? So our change in y, we started at y is equal to 6. We started at y is equal to 6. And we go down all the way to y is equal to negative 4. So this is right here. That is our change in y."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "We started at y is equal to 6. And we go down all the way to y is equal to negative 4. So this is right here. That is our change in y. You could look at the graph. You say, well, if I start at 6 and I go to negative 4, I went down 10. Or if you just want to use this formula here, it'll give you the same thing."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "That is our change in y. You could look at the graph. You say, well, if I start at 6 and I go to negative 4, I went down 10. Or if you just want to use this formula here, it'll give you the same thing. We finished at negative 4. We finished at negative 4. And from that, we want to subtract 6."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "Or if you just want to use this formula here, it'll give you the same thing. We finished at negative 4. We finished at negative 4. And from that, we want to subtract 6. This right here is y2. This is our ending y. And this is our beginning y."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "And from that, we want to subtract 6. This right here is y2. This is our ending y. And this is our beginning y. This is y1. So y2, negative 4, minus y1. Negative 6."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "And this is our beginning y. This is y1. So y2, negative 4, minus y1. Negative 6. So negative 4 minus 6. That is equal to negative 10. And all that's doing is telling us the change in y."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "Negative 6. So negative 4 minus 6. That is equal to negative 10. And all that's doing is telling us the change in y. To go from this point to that point, we had to go down. Our rise was negative. We had to go down 10."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "And all that's doing is telling us the change in y. To go from this point to that point, we had to go down. Our rise was negative. We had to go down 10. That's where the negative 10 comes from. Now we just have to find our change in x. So we can look at this graph over here."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "We had to go down 10. That's where the negative 10 comes from. Now we just have to find our change in x. So we can look at this graph over here. We started at x is equal to negative 1. And we go all the way to x is equal to 5. So we start at x is equal to negative 1."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "So we can look at this graph over here. We started at x is equal to negative 1. And we go all the way to x is equal to 5. So we start at x is equal to negative 1. And we go all the way to x is equal to 5. So it takes us 1 to get to 0, and then 5 more. So our change in x is 6."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "So we start at x is equal to negative 1. And we go all the way to x is equal to 5. So it takes us 1 to get to 0, and then 5 more. So our change in x is 6. You can look at it visually there. Or you could use this formula. Same exact idea."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "So our change in x is 6. You can look at it visually there. Or you could use this formula. Same exact idea. Our ending x value is 5. And our starting x value is negative 1. 5 minus negative 1."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "Same exact idea. Our ending x value is 5. And our starting x value is negative 1. 5 minus negative 1. 5 minus negative 1 is the same thing as 5 plus 1. So it is 6. So our slope here is negative 10 over 6, which is the exact same thing as negative 5 thirds."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "5 minus negative 1. 5 minus negative 1 is the same thing as 5 plus 1. So it is 6. So our slope here is negative 10 over 6, which is the exact same thing as negative 5 thirds. As negative 5 over 3. Divide the numerator and denominator by 2. So we now know our equation will be y is equal to negative 5 thirds, that's our slope, x plus b."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "So our slope here is negative 10 over 6, which is the exact same thing as negative 5 thirds. As negative 5 over 3. Divide the numerator and denominator by 2. So we now know our equation will be y is equal to negative 5 thirds, that's our slope, x plus b. So we still need to solve for our y-intercept to get our equation. And to do that, we can use the information that we know. Or we have several points of information."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "So we now know our equation will be y is equal to negative 5 thirds, that's our slope, x plus b. So we still need to solve for our y-intercept to get our equation. And to do that, we can use the information that we know. Or we have several points of information. But we can use the fact that the line goes through the point negative 1, 6. We could use the other point as well. But we know that when x is equal to negative 1, so y is equal to 6."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "Or we have several points of information. But we can use the fact that the line goes through the point negative 1, 6. We could use the other point as well. But we know that when x is equal to negative 1, so y is equal to 6. So y is equal to 6 when x is equal to negative 1. So negative 5 thirds times x. When x is equal to negative 1, y is equal to 6."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "But we know that when x is equal to negative 1, so y is equal to 6. So y is equal to 6 when x is equal to negative 1. So negative 5 thirds times x. When x is equal to negative 1, y is equal to 6. So we literally just substitute this x and y value back into this. And now we can solve for b. So let's see."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "When x is equal to negative 1, y is equal to 6. So we literally just substitute this x and y value back into this. And now we can solve for b. So let's see. This negative 1 times negative 5 thirds. So we get 6 is equal to positive 5 thirds plus b. And now we can subtract 5 thirds from both sides of this equation."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "So let's see. This negative 1 times negative 5 thirds. So we get 6 is equal to positive 5 thirds plus b. And now we can subtract 5 thirds from both sides of this equation. So we have subtract the left-hand side from the left-hand side. And subtract from the right-hand side. And then we get, what's 6 minus 5 thirds?"}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "And now we can subtract 5 thirds from both sides of this equation. So we have subtract the left-hand side from the left-hand side. And subtract from the right-hand side. And then we get, what's 6 minus 5 thirds? So that's going to be, let me do it over here. We could take a common denominator. So 6 is the same thing as, let me just do it over here."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "And then we get, what's 6 minus 5 thirds? So that's going to be, let me do it over here. We could take a common denominator. So 6 is the same thing as, let me just do it over here. So 6 minus 5 over 3 is the same thing as 6 is 18 over 3 minus 5 over 3. That's 6 is 18 over 3. And this is just 13 over 3."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "So 6 is the same thing as, let me just do it over here. So 6 minus 5 over 3 is the same thing as 6 is 18 over 3 minus 5 over 3. That's 6 is 18 over 3. And this is just 13 over 3. So this is 13 over 3. And then, of course, these cancel out. So we get b is equal to 13 thirds."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "And this is just 13 over 3. So this is 13 over 3. And then, of course, these cancel out. So we get b is equal to 13 thirds. So we're done. We know the slope and we know the y-intercept. The equation of our line is y is equal to negative 5 thirds x plus our y-intercept, which is 13 over 3."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "So we get b is equal to 13 thirds. So we're done. We know the slope and we know the y-intercept. The equation of our line is y is equal to negative 5 thirds x plus our y-intercept, which is 13 over 3. And we could write these as mixed numbers if it's easier to visualize. 13 over 3 is 4 and 1 third. So this y-intercept right over here, that's 0 comma 13 over 3 or 0 comma 4 and 1 third."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "The equation of our line is y is equal to negative 5 thirds x plus our y-intercept, which is 13 over 3. And we could write these as mixed numbers if it's easier to visualize. 13 over 3 is 4 and 1 third. So this y-intercept right over here, that's 0 comma 13 over 3 or 0 comma 4 and 1 third. And even with my very roughly drawn diagram, it does look like this. And this slope, negative 5 thirds, that's the same thing as negative 1 and 2 thirds. And you can see here, the slope is downward sloping."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "So this y-intercept right over here, that's 0 comma 13 over 3 or 0 comma 4 and 1 third. And even with my very roughly drawn diagram, it does look like this. And this slope, negative 5 thirds, that's the same thing as negative 1 and 2 thirds. And you can see here, the slope is downward sloping. It's negative. And it's a little bit steeper than a slope of 1. It's not quite a negative 2."}, {"video_title": "Introduction to vertex form of a quadratic.mp3", "Sentence": "It might not be obvious when you look at these three equations, but they're the exact same equation. They've just been algebraically manipulated. They are in different forms. This is the equation in sometimes called standard form for a quadratic. This is the quadratic in factored form. Notice this has been factored right over here. And this last form is what we're going to focus on in this video."}, {"video_title": "Introduction to vertex form of a quadratic.mp3", "Sentence": "This is the equation in sometimes called standard form for a quadratic. This is the quadratic in factored form. Notice this has been factored right over here. And this last form is what we're going to focus on in this video. This is sometimes known as vertex form. And we're not gonna focus on how do you get from one of these other forms to vertex form in this video. We'll do that in future videos."}, {"video_title": "Introduction to vertex form of a quadratic.mp3", "Sentence": "And this last form is what we're going to focus on in this video. This is sometimes known as vertex form. And we're not gonna focus on how do you get from one of these other forms to vertex form in this video. We'll do that in future videos. But what we're going to do is appreciate why this is called vertex, vertex form. Now to start, let's just remind ourselves what a vertex is. So as you might remember from other videos, if we have a quadratic, if we're graphing y is equal to some quadratic expression in terms of x, the graph of that will be a parabola."}, {"video_title": "Introduction to vertex form of a quadratic.mp3", "Sentence": "We'll do that in future videos. But what we're going to do is appreciate why this is called vertex, vertex form. Now to start, let's just remind ourselves what a vertex is. So as you might remember from other videos, if we have a quadratic, if we're graphing y is equal to some quadratic expression in terms of x, the graph of that will be a parabola. And it might be an upward opening parabola or a downward opening parabola. So this one in particular is going to be an upward opening parabola. And so it might look something like this."}, {"video_title": "Introduction to vertex form of a quadratic.mp3", "Sentence": "So as you might remember from other videos, if we have a quadratic, if we're graphing y is equal to some quadratic expression in terms of x, the graph of that will be a parabola. And it might be an upward opening parabola or a downward opening parabola. So this one in particular is going to be an upward opening parabola. And so it might look something like this. So it might look something, something like this right over here. And for an upward opening parabola like this, the vertex is this point right over here. You could view it as this minimum point."}, {"video_title": "Introduction to vertex form of a quadratic.mp3", "Sentence": "And so it might look something like this. So it might look something, something like this right over here. And for an upward opening parabola like this, the vertex is this point right over here. You could view it as this minimum point. You have your x coordinate of the vertex right over there, and you have your y coordinate of the vertex right over here. Now the reason why this is called vertex form is it's fairly straightforward to pick out the coordinates of this vertex from this form. How do we do that?"}, {"video_title": "Introduction to vertex form of a quadratic.mp3", "Sentence": "You could view it as this minimum point. You have your x coordinate of the vertex right over there, and you have your y coordinate of the vertex right over here. Now the reason why this is called vertex form is it's fairly straightforward to pick out the coordinates of this vertex from this form. How do we do that? Well to do that, we just have to appreciate the structure that's in this expression. Let me just rewrite it again. We have y is equal to three times x plus two squared minus 27."}, {"video_title": "Introduction to vertex form of a quadratic.mp3", "Sentence": "How do we do that? Well to do that, we just have to appreciate the structure that's in this expression. Let me just rewrite it again. We have y is equal to three times x plus two squared minus 27. The important thing to realize is that this part of the expression is never going to be negative. No matter what you have here, if you square it, you're never going to get a negative value. And so if this is never going to be negative, and we're multiplying it by a positive right over here, this whole thing right over here is going to be greater than or equal to zero."}, {"video_title": "Introduction to vertex form of a quadratic.mp3", "Sentence": "We have y is equal to three times x plus two squared minus 27. The important thing to realize is that this part of the expression is never going to be negative. No matter what you have here, if you square it, you're never going to get a negative value. And so if this is never going to be negative, and we're multiplying it by a positive right over here, this whole thing right over here is going to be greater than or equal to zero. So another way to think about it, it's only going to be additive to negative 27. So your minimum point for this curve right over here, for your parabola, is going to happen when this expression is equal to zero, when you're not adding anything to negative 27. And so when will this equal zero?"}, {"video_title": "Introduction to vertex form of a quadratic.mp3", "Sentence": "And so if this is never going to be negative, and we're multiplying it by a positive right over here, this whole thing right over here is going to be greater than or equal to zero. So another way to think about it, it's only going to be additive to negative 27. So your minimum point for this curve right over here, for your parabola, is going to happen when this expression is equal to zero, when you're not adding anything to negative 27. And so when will this equal zero? Well it's going to be equal to zero when x plus two is going to be equal to zero. So you could just say, if you want to find the x-coordinate of the vertex, well for what x value does x plus two equal zero? And of course we can subtract two from both sides and you get x is equal to negative two."}, {"video_title": "Introduction to vertex form of a quadratic.mp3", "Sentence": "And so when will this equal zero? Well it's going to be equal to zero when x plus two is going to be equal to zero. So you could just say, if you want to find the x-coordinate of the vertex, well for what x value does x plus two equal zero? And of course we can subtract two from both sides and you get x is equal to negative two. So we know that this x-coordinate right over here is negative two. And then what's the y-coordinate of the vertex? You could say, hey, what is the minimum y that this curve takes on?"}, {"video_title": "Introduction to vertex form of a quadratic.mp3", "Sentence": "And of course we can subtract two from both sides and you get x is equal to negative two. So we know that this x-coordinate right over here is negative two. And then what's the y-coordinate of the vertex? You could say, hey, what is the minimum y that this curve takes on? Well, when x is equal to negative two, this whole thing is zero and y is equal to negative 27. Y is equal to negative 27. So this right over here is negative 27."}, {"video_title": "Introduction to vertex form of a quadratic.mp3", "Sentence": "You could say, hey, what is the minimum y that this curve takes on? Well, when x is equal to negative two, this whole thing is zero and y is equal to negative 27. Y is equal to negative 27. So this right over here is negative 27. And so the coordinates of the vertex here are negative two comma negative 27. And you were able to pick that out just by looking at the quadratic in vertex form. Now let's get a few more examples under our belt so that we can really get good at picking out the vertex when a quadratic is written in vertex form."}, {"video_title": "Introduction to vertex form of a quadratic.mp3", "Sentence": "So this right over here is negative 27. And so the coordinates of the vertex here are negative two comma negative 27. And you were able to pick that out just by looking at the quadratic in vertex form. Now let's get a few more examples under our belt so that we can really get good at picking out the vertex when a quadratic is written in vertex form. So let's say, let's pick a scenario where we have a downward opening parabola where y is equal to, let's just say, negative two times x plus five. Actually, let me make it x minus five. X minus five squared, and then let's say plus 10."}, {"video_title": "Introduction to vertex form of a quadratic.mp3", "Sentence": "Now let's get a few more examples under our belt so that we can really get good at picking out the vertex when a quadratic is written in vertex form. So let's say, let's pick a scenario where we have a downward opening parabola where y is equal to, let's just say, negative two times x plus five. Actually, let me make it x minus five. X minus five squared, and then let's say plus 10. Well, here, this is going to be downward opening, and let's appreciate why that is. So here, this part is still always going to be non-negative, but it's being multiplied by a negative two. So it's actually always going to be non-positive."}, {"video_title": "Introduction to vertex form of a quadratic.mp3", "Sentence": "X minus five squared, and then let's say plus 10. Well, here, this is going to be downward opening, and let's appreciate why that is. So here, this part is still always going to be non-negative, but it's being multiplied by a negative two. So it's actually always going to be non-positive. So this whole thing right over here is going to be less than or equal to zero for all x's. So it can only take away from the 10. So where do we hit a maximum point?"}, {"video_title": "Introduction to vertex form of a quadratic.mp3", "Sentence": "So it's actually always going to be non-positive. So this whole thing right over here is going to be less than or equal to zero for all x's. So it can only take away from the 10. So where do we hit a maximum point? Well, we hit a maximum point when x minus five is equal to zero, when we're not taking anything away from the 10. And so x minus five is equal to zero. Well, that, of course, is going to happen when x is equal to five."}, {"video_title": "Introduction to vertex form of a quadratic.mp3", "Sentence": "So where do we hit a maximum point? Well, we hit a maximum point when x minus five is equal to zero, when we're not taking anything away from the 10. And so x minus five is equal to zero. Well, that, of course, is going to happen when x is equal to five. And that, indeed, is the x-coordinate for the vertex. And what's the y-coordinate for the vertex? Well, if x is equal to five, this thing is zero."}, {"video_title": "Introduction to vertex form of a quadratic.mp3", "Sentence": "Well, that, of course, is going to happen when x is equal to five. And that, indeed, is the x-coordinate for the vertex. And what's the y-coordinate for the vertex? Well, if x is equal to five, this thing is zero. You're not going to be taking anything away from the 10, and so y is going to be equal to 10. And so the vertex here is x equals five, which, and I'm just gonna eyeball it, maybe it's right over here, x equals five, and y is equal to 10. If this is negative 27, this would be positive 27, 10 would be something like this, not using the same scales for the x and y-axis, but there you have it."}, {"video_title": "Introduction to vertex form of a quadratic.mp3", "Sentence": "Well, if x is equal to five, this thing is zero. You're not going to be taking anything away from the 10, and so y is going to be equal to 10. And so the vertex here is x equals five, which, and I'm just gonna eyeball it, maybe it's right over here, x equals five, and y is equal to 10. If this is negative 27, this would be positive 27, 10 would be something like this, not using the same scales for the x and y-axis, but there you have it. So it's five comma 10, and our curve is gonna look something like this. I don't know exactly where it intersects the x-axis, but it's going to be a downward-opening parabola. Let's do one more example, just so that we get really fluent at identifying the vertex from vertex form."}, {"video_title": "Introduction to vertex form of a quadratic.mp3", "Sentence": "If this is negative 27, this would be positive 27, 10 would be something like this, not using the same scales for the x and y-axis, but there you have it. So it's five comma 10, and our curve is gonna look something like this. I don't know exactly where it intersects the x-axis, but it's going to be a downward-opening parabola. Let's do one more example, just so that we get really fluent at identifying the vertex from vertex form. So let's say, I'm just gonna make this up, we have y is equal to negative pi times x minus 2.8 squared plus, I don't know, plus 7.1. What is the vertex of the parabola here? Well, the x-coordinate is going to be the x-value that makes this equal to zero, which is 2.8, and then if this is equal to zero, then this whole thing is going to be equal to zero, and y is going to be 7.1."}, {"video_title": "Clarifying standard form rules.mp3", "Sentence": "We've talked about the idea of standard form of a linear equation in other videos, and the point of this video is to clarify something and resolve some differences that you might see in different classes in terms of what standard form is. So everyone agrees that standard form is generally a linear equation where you have some number times x plus some number times y is equal to some number. So things that are in standard form would include things like three x plus four y is equal to 10, or two x plus five y is equal to negative 10. Everyone would agree that these are standard form, and everyone would agree that the following are not standard form. So if I were to write three x is equal to negative four y plus 10, even though these are equivalent equations, this is just not in standard form. Similarly, if I wrote that y is equal to three times x plus seven, this is also not in standard form. Now, the place where some people might disagree is if you were to see something like six x plus eight y is equal to 20."}, {"video_title": "Clarifying standard form rules.mp3", "Sentence": "Everyone would agree that these are standard form, and everyone would agree that the following are not standard form. So if I were to write three x is equal to negative four y plus 10, even though these are equivalent equations, this is just not in standard form. Similarly, if I wrote that y is equal to three times x plus seven, this is also not in standard form. Now, the place where some people might disagree is if you were to see something like six x plus eight y is equal to 20. Now, why would some folks argue that this is not standard form? Well, for some folks, they would say standard form, the coefficients on x and y and our constant terms, so our a, b, and c, can't share any common factors. And here, six, eight, and 20, they're all divisible by two."}, {"video_title": "Clarifying standard form rules.mp3", "Sentence": "Now, the place where some people might disagree is if you were to see something like six x plus eight y is equal to 20. Now, why would some folks argue that this is not standard form? Well, for some folks, they would say standard form, the coefficients on x and y and our constant terms, so our a, b, and c, can't share any common factors. And here, six, eight, and 20, they're all divisible by two. And so some folks would argue that this is not standard form and to get it into standard form, you would divide all of these by two. And if you did, you would get this equation here. Now, that's useful because then you only have one unique equation, but on Khan Academy, we do not restrict in that way."}, {"video_title": "Clarifying standard form rules.mp3", "Sentence": "And here, six, eight, and 20, they're all divisible by two. And so some folks would argue that this is not standard form and to get it into standard form, you would divide all of these by two. And if you did, you would get this equation here. Now, that's useful because then you only have one unique equation, but on Khan Academy, we do not restrict in that way. And that is also a very popular way of thinking about it. We just want you to think about it in this form, ax plus by is equal to c. When you do the exercises on Khan Academy, it's not going to be checking whether these coefficients a, b, and c are divisible, have a common factor. So for Khan Academy purposes, this is considered standard form, although don't be surprised if you encounter some folks who say, no, we would rather you remove any common factors."}, {"video_title": "Clarifying standard form rules.mp3", "Sentence": "Now, that's useful because then you only have one unique equation, but on Khan Academy, we do not restrict in that way. And that is also a very popular way of thinking about it. We just want you to think about it in this form, ax plus by is equal to c. When you do the exercises on Khan Academy, it's not going to be checking whether these coefficients a, b, and c are divisible, have a common factor. So for Khan Academy purposes, this is considered standard form, although don't be surprised if you encounter some folks who say, no, we would rather you remove any common factors. Now, another example would be something like negative three x minus four y is equal to negative 10. So some folks would argue that this is not standard form because they wanna see this first coefficient right over here, the a being greater than zero, while here it is less than zero. For our purposes on Khan Academy, we do consider this standard form."}, {"video_title": "Clarifying standard form rules.mp3", "Sentence": "So for Khan Academy purposes, this is considered standard form, although don't be surprised if you encounter some folks who say, no, we would rather you remove any common factors. Now, another example would be something like negative three x minus four y is equal to negative 10. So some folks would argue that this is not standard form because they wanna see this first coefficient right over here, the a being greater than zero, while here it is less than zero. For our purposes on Khan Academy, we do consider this standard form. But I'm just letting you know because some folks might not because this leading coefficient is not greater than zero. Now, another example that some people might be on the edge with would be something like 1.25x plus 5.50y is equal to 10.5. And the reason why some people might not consider this standard form is that a, b, and c are not integers."}, {"video_title": "Clarifying standard form rules.mp3", "Sentence": "For our purposes on Khan Academy, we do consider this standard form. But I'm just letting you know because some folks might not because this leading coefficient is not greater than zero. Now, another example that some people might be on the edge with would be something like 1.25x plus 5.50y is equal to 10.5. And the reason why some people might not consider this standard form is that a, b, and c are not integers. Some folks would say to be in standard form, a, b, and c need to be integers. And you could multiply both sides of the equation by some value that will give you integers for a, b, and c. But for Khan Academy purposes, we do consider this to be in standard form. And we think this is important, actually not just being able to have non-integers as a, b, or c, but also being able to have a negative a right over there, this negative three is our a, and also having coefficients, having our a's, b's, and c's, having shared factors, we think all of that's important because sometimes the equation itself has meaning when you write it that way."}, {"video_title": "Clarifying standard form rules.mp3", "Sentence": "And the reason why some people might not consider this standard form is that a, b, and c are not integers. Some folks would say to be in standard form, a, b, and c need to be integers. And you could multiply both sides of the equation by some value that will give you integers for a, b, and c. But for Khan Academy purposes, we do consider this to be in standard form. And we think this is important, actually not just being able to have non-integers as a, b, or c, but also being able to have a negative a right over there, this negative three is our a, and also having coefficients, having our a's, b's, and c's, having shared factors, we think all of that's important because sometimes the equation itself has meaning when you write it that way. And we'll see that when we do some word problems, when we actually go into real life and we try to construct equations. And based on the information in the equation, it's easier to understand if you keep it in this form. So for Khan Academy purposes, this is all standard form, but it's good to be aware in your mathematical lives that some folks might wanna see the restriction of no common factors between a, b, and c, that a is greater than zero, and that a, b, and c need to all be integers, but Khan Academy does not hold you to that."}, {"video_title": "How to determine the domain of a modeling function   Functions   Algebra I   Khan Academy.mp3", "Sentence": "Alright, I input B, the number of candy bars I want to buy, and P of B will tell me, well, what's the purchase price? It's really just gonna take the number of candy bars and multiply it by 50 cents, but we don't have to worry about that just yet. Which number type is more appropriate for the domain of the function? So let's just remind ourselves, what is a domain of a function? A domain is a set of all inputs over which the function is defined. So it's a set of all Bs, it's a set of all inputs over which P of B will produce a defined response. So let's think about it."}, {"video_title": "How to determine the domain of a modeling function   Functions   Algebra I   Khan Academy.mp3", "Sentence": "So let's just remind ourselves, what is a domain of a function? A domain is a set of all inputs over which the function is defined. So it's a set of all Bs, it's a set of all inputs over which P of B will produce a defined response. So let's think about it. Is it integers or real numbers? So I could buy, B could be zero candy bars, one candy bars, two candy bars, all the way up to 400 candy bars. Could I have a fractional, could B be.372 of a candy bar?"}, {"video_title": "How to determine the domain of a modeling function   Functions   Algebra I   Khan Academy.mp3", "Sentence": "So let's think about it. Is it integers or real numbers? So I could buy, B could be zero candy bars, one candy bars, two candy bars, all the way up to 400 candy bars. Could I have a fractional, could B be.372 of a candy bar? Well, if this is a normal candy shop, each candy bar is gonna be in its own packet. It's going to be in a discrete chunk. You're not going to be able to buy.372 of a candy bar."}, {"video_title": "How to determine the domain of a modeling function   Functions   Algebra I   Khan Academy.mp3", "Sentence": "Could I have a fractional, could B be.372 of a candy bar? Well, if this is a normal candy shop, each candy bar is gonna be in its own packet. It's going to be in a discrete chunk. You're not going to be able to buy.372 of a candy bar. You're either gonna buy one more or none more. So you're gonna buy zero, one, two, three, all the way up to 400. So I would say integers, that the domain of this function is going to be a subset of integers."}, {"video_title": "How to determine the domain of a modeling function   Functions   Algebra I   Khan Academy.mp3", "Sentence": "You're not going to be able to buy.372 of a candy bar. You're either gonna buy one more or none more. So you're gonna buy zero, one, two, three, all the way up to 400. So I would say integers, that the domain of this function is going to be a subset of integers. You can't have a real, all real number. Inputs integers are obviously a subset of real numbers, but you can't say, hey, I'm gonna buy pi candy bars, or I'm gonna buy the square root of two candy bars. You're gonna buy integer number of candy bars."}, {"video_title": "How to determine the domain of a modeling function   Functions   Algebra I   Khan Academy.mp3", "Sentence": "So I would say integers, that the domain of this function is going to be a subset of integers. You can't have a real, all real number. Inputs integers are obviously a subset of real numbers, but you can't say, hey, I'm gonna buy pi candy bars, or I'm gonna buy the square root of two candy bars. You're gonna buy integer number of candy bars. Now they say define the interval of the domain. So the fewest candy bars I could buy are zero candy bars, and I have to decide, do I put a bracket or do I put a parentheses? So I could actually buy zero candy bars, so I'm gonna put a bracket."}, {"video_title": "How to determine the domain of a modeling function   Functions   Algebra I   Khan Academy.mp3", "Sentence": "You're gonna buy integer number of candy bars. Now they say define the interval of the domain. So the fewest candy bars I could buy are zero candy bars, and I have to decide, do I put a bracket or do I put a parentheses? So I could actually buy zero candy bars, so I'm gonna put a bracket. If I put a parentheses, that means I could have values above zero but not including zero. But I want to include zero, so I'm gonna put the bracket there. So the least I could buy is zero, and then the most I could buy, the store has 400 candy bars, so that's the most I can buy."}, {"video_title": "How to determine the domain of a modeling function   Functions   Algebra I   Khan Academy.mp3", "Sentence": "So I could actually buy zero candy bars, so I'm gonna put a bracket. If I put a parentheses, that means I could have values above zero but not including zero. But I want to include zero, so I'm gonna put the bracket there. So the least I could buy is zero, and then the most I could buy, the store has 400 candy bars, so that's the most I can buy. The most I could buy are 400 candy bars, and I can buy 400. So I would put brackets there as well. So the interval of the domain, I would want to select integers, so b is a member of integers such that b is also a member of this interval."}, {"video_title": "Find measure of supplementary angles   Angles and intersecting lines   Geometry   Khan Academy.mp3", "Sentence": "We're told that the measure of angle QPR, so QPR, so that's this angle right over here, is 2x plus 122. And I'll assume that these are in degrees. So it's 2x plus 122 degrees. And the measure of angle RPS, so that's this angle right over here, is 2x plus 22 degrees. And they ask us to find the measure of angle RPS. So we need to figure out this right over here. So we would be able to figure that out if we just knew what x is."}, {"video_title": "Find measure of supplementary angles   Angles and intersecting lines   Geometry   Khan Academy.mp3", "Sentence": "And the measure of angle RPS, so that's this angle right over here, is 2x plus 22 degrees. And they ask us to find the measure of angle RPS. So we need to figure out this right over here. So we would be able to figure that out if we just knew what x is. And lucky for us, we can use the information given to solve for x and then figure out what 2 times x plus 22 is. And the big idea here, the thing that pops out here, is that the outside rays for both of these angles form a line. These two angles form a line."}, {"video_title": "Find measure of supplementary angles   Angles and intersecting lines   Geometry   Khan Academy.mp3", "Sentence": "So we would be able to figure that out if we just knew what x is. And lucky for us, we can use the information given to solve for x and then figure out what 2 times x plus 22 is. And the big idea here, the thing that pops out here, is that the outside rays for both of these angles form a line. These two angles form a line. You could say that they are supplementary. Those of these angles are supplementary. 2x plus 22 plus another 2x plus 122 is going to add up to 180."}, {"video_title": "Find measure of supplementary angles   Angles and intersecting lines   Geometry   Khan Academy.mp3", "Sentence": "These two angles form a line. You could say that they are supplementary. Those of these angles are supplementary. 2x plus 22 plus another 2x plus 122 is going to add up to 180. We know that this entire angle right over here is 180 degrees. So we can say that the measure of angle QPR, this angle right over here, 2x plus 122, plus the green angle, plus angle RPS, so plus 2x plus 22, is going to be equal to 180 degrees. Is going to be equal to 180."}, {"video_title": "Find measure of supplementary angles   Angles and intersecting lines   Geometry   Khan Academy.mp3", "Sentence": "2x plus 22 plus another 2x plus 122 is going to add up to 180. We know that this entire angle right over here is 180 degrees. So we can say that the measure of angle QPR, this angle right over here, 2x plus 122, plus the green angle, plus angle RPS, so plus 2x plus 22, is going to be equal to 180 degrees. Is going to be equal to 180. And now we can start simplifying this. We have 2x's. We have another 2x's."}, {"video_title": "Find measure of supplementary angles   Angles and intersecting lines   Geometry   Khan Academy.mp3", "Sentence": "Is going to be equal to 180. And now we can start simplifying this. We have 2x's. We have another 2x's. So those are going to add up to be 4x. And then we have 122 plus 22. So that's going to be 144."}, {"video_title": "Find measure of supplementary angles   Angles and intersecting lines   Geometry   Khan Academy.mp3", "Sentence": "We have another 2x's. So those are going to add up to be 4x. And then we have 122 plus 22. So that's going to be 144. And the sum of those two are going to be equal to 180 degrees. We can subtract 144 from both sides. On the left-hand side, we're just going to be left with a 4x, this 4x right here."}, {"video_title": "Find measure of supplementary angles   Angles and intersecting lines   Geometry   Khan Academy.mp3", "Sentence": "So that's going to be 144. And the sum of those two are going to be equal to 180 degrees. We can subtract 144 from both sides. On the left-hand side, we're just going to be left with a 4x, this 4x right here. And on the right-hand side, we're going to have, let's see, if we were subtracting 140, we would have 40 left. And then we have to subtract another 4. So it's going to be 36."}, {"video_title": "Find measure of supplementary angles   Angles and intersecting lines   Geometry   Khan Academy.mp3", "Sentence": "On the left-hand side, we're just going to be left with a 4x, this 4x right here. And on the right-hand side, we're going to have, let's see, if we were subtracting 140, we would have 40 left. And then we have to subtract another 4. So it's going to be 36. Divide both sides by 4. And we get x is equal to 9. Now remember, we're not done yet."}, {"video_title": "Find measure of supplementary angles   Angles and intersecting lines   Geometry   Khan Academy.mp3", "Sentence": "So it's going to be 36. Divide both sides by 4. And we get x is equal to 9. Now remember, we're not done yet. They didn't say solve for x. They said find the measure of angle RPS, which is 2 times x plus 22, or 2 times 9 plus 22, which is 18 plus 22, which is equal to 40. So the measure of angle RPS is 40 degrees."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "But all it means is a second degree polynomial. So something that's going to have a variable raised to the second power. In this case, and all of the examples will do, it'll be x. So let's say I have the quadratic expression, x squared plus 10x plus 9. And I want to factor it into the product of two binomials. How do we do that? Well, let's just think about what happens if we were to take x plus a and multiply that by x plus b."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "So let's say I have the quadratic expression, x squared plus 10x plus 9. And I want to factor it into the product of two binomials. How do we do that? Well, let's just think about what happens if we were to take x plus a and multiply that by x plus b. If we were to multiply these two things, what happens? Well, we have a little bit of experience doing this. This will be x times x, which is x squared, plus x times b, which is bx, plus a times x, plus a times b, plus ab."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "Well, let's just think about what happens if we were to take x plus a and multiply that by x plus b. If we were to multiply these two things, what happens? Well, we have a little bit of experience doing this. This will be x times x, which is x squared, plus x times b, which is bx, plus a times x, plus a times b, plus ab. Or if we want to add these two in the middle right here, because they have the same, they're both coefficients of x, we could write this as x squared plus, I could write it as b plus a, or a plus bx, plus ab. So in general, if we assume that this is the product of two binomials, we see that this middle coefficient on the x term, or you could say the first degree coefficient there, that's going to be the sum of our a and b. This is going to be the sum of our a and b."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "This will be x times x, which is x squared, plus x times b, which is bx, plus a times x, plus a times b, plus ab. Or if we want to add these two in the middle right here, because they have the same, they're both coefficients of x, we could write this as x squared plus, I could write it as b plus a, or a plus bx, plus ab. So in general, if we assume that this is the product of two binomials, we see that this middle coefficient on the x term, or you could say the first degree coefficient there, that's going to be the sum of our a and b. This is going to be the sum of our a and b. And then the constant term is going to be the product of our a and b. Notice, this would map to this, and this would map to this. And of course, this is the same thing as this."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "This is going to be the sum of our a and b. And then the constant term is going to be the product of our a and b. Notice, this would map to this, and this would map to this. And of course, this is the same thing as this. So can we somehow pattern match this to that? Is there some a and b where a plus b is equal to 10, and a times b is equal to 9? Well, let's just think about it a little bit."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "And of course, this is the same thing as this. So can we somehow pattern match this to that? Is there some a and b where a plus b is equal to 10, and a times b is equal to 9? Well, let's just think about it a little bit. What are the factors of 9? What are the things that a and b could be equal to? And we're assuming that everything is an integer, and normally when we're factoring, especially when we're beginning to factor, we're dealing with integer numbers."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "Well, let's just think about it a little bit. What are the factors of 9? What are the things that a and b could be equal to? And we're assuming that everything is an integer, and normally when we're factoring, especially when we're beginning to factor, we're dealing with integer numbers. So what are the factors of 9? They're 1, 3, and 9. So this could be a 3 and a 3, or it could be a 1 and a 9."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "And we're assuming that everything is an integer, and normally when we're factoring, especially when we're beginning to factor, we're dealing with integer numbers. So what are the factors of 9? They're 1, 3, and 9. So this could be a 3 and a 3, or it could be a 1 and a 9. Now, if it's a 3 and a 3, then you'll have 3 plus 3. That doesn't equal 10. But if it's a 1 and a 9, 1 times 9 is 9."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "So this could be a 3 and a 3, or it could be a 1 and a 9. Now, if it's a 3 and a 3, then you'll have 3 plus 3. That doesn't equal 10. But if it's a 1 and a 9, 1 times 9 is 9. 1 plus 9 is 10. So it does work. So a could be equal to 1, and b could be equal to 9."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "But if it's a 1 and a 9, 1 times 9 is 9. 1 plus 9 is 10. So it does work. So a could be equal to 1, and b could be equal to 9. So we could factor this as being x plus 1 times x plus 9. And if you multiply these two out using the skills we developed in the last few videos, you'll see that it is indeed x squared plus 10x plus 9. So when you see something like this, when the coefficient on the x squared term, or the leading coefficient on this quadratic, is a 1, you can just say, all right, what two numbers add up to this coefficient right here?"}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "So a could be equal to 1, and b could be equal to 9. So we could factor this as being x plus 1 times x plus 9. And if you multiply these two out using the skills we developed in the last few videos, you'll see that it is indeed x squared plus 10x plus 9. So when you see something like this, when the coefficient on the x squared term, or the leading coefficient on this quadratic, is a 1, you can just say, all right, what two numbers add up to this coefficient right here? And what two numbers add up? And those same two numbers, when you take their product, have to be equal to 9. And of course, this has to be in standard form."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "So when you see something like this, when the coefficient on the x squared term, or the leading coefficient on this quadratic, is a 1, you can just say, all right, what two numbers add up to this coefficient right here? And what two numbers add up? And those same two numbers, when you take their product, have to be equal to 9. And of course, this has to be in standard form. Or if it's not in standard form, you should put it in that form so that you can always say, OK, whatever's on the first degree coefficient, my a and b have to add to that. Whatever's my constant term, my a times b, the product has to be that. Let's do several more examples."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "And of course, this has to be in standard form. Or if it's not in standard form, you should put it in that form so that you can always say, OK, whatever's on the first degree coefficient, my a and b have to add to that. Whatever's my constant term, my a times b, the product has to be that. Let's do several more examples. And I think the more examples we do, the more sense this will make. Let's say we had x squared plus 10x plus, well, I already did 10x. Let's do a different number."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "Let's do several more examples. And I think the more examples we do, the more sense this will make. Let's say we had x squared plus 10x plus, well, I already did 10x. Let's do a different number. x squared plus 15x plus 50. We want to factor this. Well, same drill."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "Let's do a different number. x squared plus 15x plus 50. We want to factor this. Well, same drill. We have an x squared term, x squared term. We have a first degree term. This should be, this right here, should be the sum of two numbers, and then this term, the constant term right here, should be the product of two numbers."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "Well, same drill. We have an x squared term, x squared term. We have a first degree term. This should be, this right here, should be the sum of two numbers, and then this term, the constant term right here, should be the product of two numbers. So we need to think of two numbers that when I multiply them, I get 50, and when I add them, I get 15. And this is going to be a bit of an art that you're going to develop, but the more practice you do, you're going to see that it'll start to come naturally. So what could a and b be?"}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "This should be, this right here, should be the sum of two numbers, and then this term, the constant term right here, should be the product of two numbers. So we need to think of two numbers that when I multiply them, I get 50, and when I add them, I get 15. And this is going to be a bit of an art that you're going to develop, but the more practice you do, you're going to see that it'll start to come naturally. So what could a and b be? Let's think about the factors of 50. It could be 1 times 50, 2 times 25. See, 4 doesn't go into 50."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "So what could a and b be? Let's think about the factors of 50. It could be 1 times 50, 2 times 25. See, 4 doesn't go into 50. It could be 5 times 10. I think that's all of them. Let's try out these numbers and see if any of these add up to 15."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "See, 4 doesn't go into 50. It could be 5 times 10. I think that's all of them. Let's try out these numbers and see if any of these add up to 15. So 1 plus 50 does not add up to 15. 2 plus 25 does not add up to 15. But 5 plus 10 does add up to 15."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "Let's try out these numbers and see if any of these add up to 15. So 1 plus 50 does not add up to 15. 2 plus 25 does not add up to 15. But 5 plus 10 does add up to 15. So this could be 5 plus 10, and this could be 5 times 10. So if we were to factor this, this would be equal to x plus 5 times x plus 10. And multiply it out."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "But 5 plus 10 does add up to 15. So this could be 5 plus 10, and this could be 5 times 10. So if we were to factor this, this would be equal to x plus 5 times x plus 10. And multiply it out. I encourage you to multiply this out and see that this is indeed x squared plus 15x plus 10. In fact, let's do it. x times x, x squared."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "And multiply it out. I encourage you to multiply this out and see that this is indeed x squared plus 15x plus 10. In fact, let's do it. x times x, x squared. x times 10 plus 10x. 5 times x plus 5x. 5 times 10 plus 50."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "x times x, x squared. x times 10 plus 10x. 5 times x plus 5x. 5 times 10 plus 50. Notice the 5 times 10 gave us the 50. The 5x plus the 10x is giving us the 15x in between. So it's x squared plus 15x plus 50. x squared plus 15x plus 50."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "5 times 10 plus 50. Notice the 5 times 10 gave us the 50. The 5x plus the 10x is giving us the 15x in between. So it's x squared plus 15x plus 50. x squared plus 15x plus 50. Let's up the stakes a little bit, introduce some negative signs in here. Let's say I had x squared minus 11x plus 24. Now, it's the exact same principle."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "So it's x squared plus 15x plus 50. x squared plus 15x plus 50. Let's up the stakes a little bit, introduce some negative signs in here. Let's say I had x squared minus 11x plus 24. Now, it's the exact same principle. I need to think of two numbers that when I add them need to be equal to negative 11. a plus b need to be equal to negative 11. And a times b need to be equal to 24. Now, there's something for you to think about."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "Now, it's the exact same principle. I need to think of two numbers that when I add them need to be equal to negative 11. a plus b need to be equal to negative 11. And a times b need to be equal to 24. Now, there's something for you to think about. If when I multiply both of these numbers, I'm getting a positive number. I'm getting a 24. That means that both of these need to be positive or both of these need to be negative."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "Now, there's something for you to think about. If when I multiply both of these numbers, I'm getting a positive number. I'm getting a 24. That means that both of these need to be positive or both of these need to be negative. That's the only way I'm going to get a positive number here. Now, if when I add them I get a negative number, if these were positive, there's no way I can add two positive numbers and get a negative number. So the fact that their sum is negative and the fact that their product is positive tells me that both a and b are negative."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "That means that both of these need to be positive or both of these need to be negative. That's the only way I'm going to get a positive number here. Now, if when I add them I get a negative number, if these were positive, there's no way I can add two positive numbers and get a negative number. So the fact that their sum is negative and the fact that their product is positive tells me that both a and b are negative. a and b have to be negative. Remember, one can't be negative and the other one can't be positive because then the product would be negative. And they both can't be positive because then this, when you add them, it would get you a positive number."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "So the fact that their sum is negative and the fact that their product is positive tells me that both a and b are negative. a and b have to be negative. Remember, one can't be negative and the other one can't be positive because then the product would be negative. And they both can't be positive because then this, when you add them, it would get you a positive number. So let's just think about what a and b can be. So two negative numbers. So let's think about the factors of 24."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "And they both can't be positive because then this, when you add them, it would get you a positive number. So let's just think about what a and b can be. So two negative numbers. So let's think about the factors of 24. And we'll kind of have to think of the negative factors. But let me see, it could be 1 times 24, 2 times 11, 3 times 8, or 4 times 6. Now, which of these, when I multiply these, well, obviously when I multiply 1 times 24 I get 24."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "So let's think about the factors of 24. And we'll kind of have to think of the negative factors. But let me see, it could be 1 times 24, 2 times 11, 3 times 8, or 4 times 6. Now, which of these, when I multiply these, well, obviously when I multiply 1 times 24 I get 24. When I get 2 times 12 I get 24. So we know that all of these, the products are 24. But which two of these, which two factors, when I add them, should I get 11?"}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "Now, which of these, when I multiply these, well, obviously when I multiply 1 times 24 I get 24. When I get 2 times 12 I get 24. So we know that all of these, the products are 24. But which two of these, which two factors, when I add them, should I get 11? And then we could say, let's take the negative of both of those. So when you look at these, 3 and 8 jump out. 3 times 8 is equal to 24."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "But which two of these, which two factors, when I add them, should I get 11? And then we could say, let's take the negative of both of those. So when you look at these, 3 and 8 jump out. 3 times 8 is equal to 24. 3 plus 8 is equal to 11. But that doesn't quite work out, right? Because we have a negative 11 here."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "3 times 8 is equal to 24. 3 plus 8 is equal to 11. But that doesn't quite work out, right? Because we have a negative 11 here. But what if we did negative 3 and negative 8? Negative 3 times negative 8 is equal to positive 24. Negative 3 minus 11, or sorry, negative 3 plus negative 8 is equal to negative 11."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "Because we have a negative 11 here. But what if we did negative 3 and negative 8? Negative 3 times negative 8 is equal to positive 24. Negative 3 minus 11, or sorry, negative 3 plus negative 8 is equal to negative 11. So negative 3 and negative 8 work. So if we factor this, this is going to x squared minus 11x plus 24 is going to be equal to x minus 3 times x minus 8. Let's do another one like that."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "Negative 3 minus 11, or sorry, negative 3 plus negative 8 is equal to negative 11. So negative 3 and negative 8 work. So if we factor this, this is going to x squared minus 11x plus 24 is going to be equal to x minus 3 times x minus 8. Let's do another one like that. Actually, let's mix it up a little bit. Let's say I had x squared plus 5x minus 14. So here we have a different situation."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "Let's do another one like that. Actually, let's mix it up a little bit. Let's say I had x squared plus 5x minus 14. So here we have a different situation. The product of my two numbers is negative. a times b is equal to negative 14. My product is negative."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "So here we have a different situation. The product of my two numbers is negative. a times b is equal to negative 14. My product is negative. That tells me that one of them is positive and one of them is negative. And when I add the two, a plus b, I get it being equal to 5. So let's think about the factors of 14 and what combinations of them, when I add them, if one is positive and one is negative, or I'm really kind of taking the difference of the two, do I get 5?"}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "My product is negative. That tells me that one of them is positive and one of them is negative. And when I add the two, a plus b, I get it being equal to 5. So let's think about the factors of 14 and what combinations of them, when I add them, if one is positive and one is negative, or I'm really kind of taking the difference of the two, do I get 5? So if I take 1 and 14, I'm just going to try out things. Negative 1 plus 14 is negative 13. So let me write all of the combinations that I could do."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "So let's think about the factors of 14 and what combinations of them, when I add them, if one is positive and one is negative, or I'm really kind of taking the difference of the two, do I get 5? So if I take 1 and 14, I'm just going to try out things. Negative 1 plus 14 is negative 13. So let me write all of the combinations that I could do. And eventually your brain will just zone in on it. So you could have negative 1 plus 14 is equal to 13. And 1 plus negative 14 is equal to negative 13."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "So let me write all of the combinations that I could do. And eventually your brain will just zone in on it. So you could have negative 1 plus 14 is equal to 13. And 1 plus negative 14 is equal to negative 13. So those don't work. That doesn't equal 5. Now what about 2 and 7?"}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "And 1 plus negative 14 is equal to negative 13. So those don't work. That doesn't equal 5. Now what about 2 and 7? If I do negative 2 plus 7, that is equal to 5. We're done. That worked."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "Now what about 2 and 7? If I do negative 2 plus 7, that is equal to 5. We're done. That worked. I mean, we could have tried 2 plus negative 7, but that would have equaled negative 5. So that wouldn't have worked. But negative 2 plus 7 works."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "That worked. I mean, we could have tried 2 plus negative 7, but that would have equaled negative 5. So that wouldn't have worked. But negative 2 plus 7 works. And negative 2 times 7 is negative 14. So there we have it. We know it's x minus 2 times x plus 7."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "But negative 2 plus 7 works. And negative 2 times 7 is negative 14. So there we have it. We know it's x minus 2 times x plus 7. That's pretty neat. Negative 2 times 7 is negative 14. Negative 2 plus 7 is positive 5."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "We know it's x minus 2 times x plus 7. That's pretty neat. Negative 2 times 7 is negative 14. Negative 2 plus 7 is positive 5. Let's do several more of these, just to really get well honed the skill. So let's say we have x squared minus x minus 56. So the product of the two numbers have to be minus 56, have to be negative 56."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "Negative 2 plus 7 is positive 5. Let's do several more of these, just to really get well honed the skill. So let's say we have x squared minus x minus 56. So the product of the two numbers have to be minus 56, have to be negative 56. And their difference, because 1 is going to be positive and 1 is going to be negative, their difference has to be negative 1. And the numbers that immediately jump out in my brain, and I don't know if they jump out in your brain, we just learned this in the times tables, 56 is 8 times 7. I mean, there's other numbers."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "So the product of the two numbers have to be minus 56, have to be negative 56. And their difference, because 1 is going to be positive and 1 is going to be negative, their difference has to be negative 1. And the numbers that immediately jump out in my brain, and I don't know if they jump out in your brain, we just learned this in the times tables, 56 is 8 times 7. I mean, there's other numbers. It's also 28 times 2. It's all sorts of things. But 8 times 7 really jumped out into my brain because they're very close to each other."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "I mean, there's other numbers. It's also 28 times 2. It's all sorts of things. But 8 times 7 really jumped out into my brain because they're very close to each other. We need numbers that are very close to each other. And one of these has to be positive and one of these has to be negative. Now, the fact that when their sum is negative tells me that the larger of these two should probably be negative."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "But 8 times 7 really jumped out into my brain because they're very close to each other. We need numbers that are very close to each other. And one of these has to be positive and one of these has to be negative. Now, the fact that when their sum is negative tells me that the larger of these two should probably be negative. So if we take negative 8 times 7, that's equal to negative 56, and then if we take negative 8 plus 7, that is equal to negative 1, which is exactly the coefficient right there. So when I factor this, this is going to be x minus 8 times x plus 7. This is often one of the hardest concepts people learn in algebra because it is a bit of an art."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "Now, the fact that when their sum is negative tells me that the larger of these two should probably be negative. So if we take negative 8 times 7, that's equal to negative 56, and then if we take negative 8 plus 7, that is equal to negative 1, which is exactly the coefficient right there. So when I factor this, this is going to be x minus 8 times x plus 7. This is often one of the hardest concepts people learn in algebra because it is a bit of an art. You have to look at all of the factors here, play with the positive and negative signs, see which of those factors, when 1 is positive and 1 is negative, add up to the coefficient on the x term. But as you do more and more practice, you'll see that it'll become a bit of second nature. Now let's step up the stakes a little bit more."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "This is often one of the hardest concepts people learn in algebra because it is a bit of an art. You have to look at all of the factors here, play with the positive and negative signs, see which of those factors, when 1 is positive and 1 is negative, add up to the coefficient on the x term. But as you do more and more practice, you'll see that it'll become a bit of second nature. Now let's step up the stakes a little bit more. Let's say we had negative x squared. Everything we've done so far had a positive coefficient, a positive 1 coefficient on the x squared term. But let's say we had a negative x squared minus 5x plus 24."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "Now let's step up the stakes a little bit more. Let's say we had negative x squared. Everything we've done so far had a positive coefficient, a positive 1 coefficient on the x squared term. But let's say we had a negative x squared minus 5x plus 24. How do we do this? Well, the easiest way I can think of doing it is factor out a negative 1, and then it becomes just like the problems we've been doing before. So this is the same thing as negative 1 times positive x squared plus 5x minus 24."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "But let's say we had a negative x squared minus 5x plus 24. How do we do this? Well, the easiest way I can think of doing it is factor out a negative 1, and then it becomes just like the problems we've been doing before. So this is the same thing as negative 1 times positive x squared plus 5x minus 24. I just factored a negative 1 out. You can multiply negative 1 times all of these, and you'll see it becomes this. Or you could factor the negative 1 out and divide all of these by negative 1, and you get that right there."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "So this is the same thing as negative 1 times positive x squared plus 5x minus 24. I just factored a negative 1 out. You can multiply negative 1 times all of these, and you'll see it becomes this. Or you could factor the negative 1 out and divide all of these by negative 1, and you get that right there. Now, same game as before. I need two numbers that when I take their product, I get negative 24. So one will be positive, one will be negative."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "Or you could factor the negative 1 out and divide all of these by negative 1, and you get that right there. Now, same game as before. I need two numbers that when I take their product, I get negative 24. So one will be positive, one will be negative. And when I take their sum, it's going to be 5. So let's think about 24. Is 1 and 24."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "So one will be positive, one will be negative. And when I take their sum, it's going to be 5. So let's think about 24. Is 1 and 24. Let's see, if this is negative 1 and 24, it's negative 23. Otherwise, if it's negative 1 and 24, it would be positive 23. If it was the other way around, it would be negative 23."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "Is 1 and 24. Let's see, if this is negative 1 and 24, it's negative 23. Otherwise, if it's negative 1 and 24, it would be positive 23. If it was the other way around, it would be negative 23. It doesn't work. What about 2 and 12? Well, if this is negative, if the 2 is negative, remember, one of these have to be negative."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "If it was the other way around, it would be negative 23. It doesn't work. What about 2 and 12? Well, if this is negative, if the 2 is negative, remember, one of these have to be negative. If the 2 is negative, their sum would be 10. If the 12 is negative, their sum would be negative 10. Still doesn't work."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "Well, if this is negative, if the 2 is negative, remember, one of these have to be negative. If the 2 is negative, their sum would be 10. If the 12 is negative, their sum would be negative 10. Still doesn't work. 3 and 8. If the 3 is negative, their sum will be 5. So it works."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "Still doesn't work. 3 and 8. If the 3 is negative, their sum will be 5. So it works. So if we pick negative 3 and 8, negative 3 and 8 work. So if we use it because negative 3 plus 8 is 5, negative 3 times 8 is negative 24. So this is going to be equal to, can't forget that negative 1 out front."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "So it works. So if we pick negative 3 and 8, negative 3 and 8 work. So if we use it because negative 3 plus 8 is 5, negative 3 times 8 is negative 24. So this is going to be equal to, can't forget that negative 1 out front. And then we factor the inside. Negative 1 times x minus 3 times x plus 8. And if you really wanted to, you could multiply the negative 1 times this."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "So this is going to be equal to, can't forget that negative 1 out front. And then we factor the inside. Negative 1 times x minus 3 times x plus 8. And if you really wanted to, you could multiply the negative 1 times this. You would get 3 minus x if you did, or you don't have to. Let's do one more of these. The more practice, the better, I think."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "And if you really wanted to, you could multiply the negative 1 times this. You would get 3 minus x if you did, or you don't have to. Let's do one more of these. The more practice, the better, I think. All right, let's say I had negative x squared plus 18x minus 72. So once again, I like to factor out the negative 1. So this is equal to negative 1 times x squared minus 18x plus 72."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "The more practice, the better, I think. All right, let's say I had negative x squared plus 18x minus 72. So once again, I like to factor out the negative 1. So this is equal to negative 1 times x squared minus 18x plus 72. Now we just have to think of two numbers that when I multiply them, I get positive 72. So they have to be the same sign. And that makes it easier in our head, at least in my head."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "So this is equal to negative 1 times x squared minus 18x plus 72. Now we just have to think of two numbers that when I multiply them, I get positive 72. So they have to be the same sign. And that makes it easier in our head, at least in my head. When I multiply them, I get positive 72. When I add them, I get negative 18. So if they're the same sign and their sum is a negative number, they both must be negative."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "And that makes it easier in our head, at least in my head. When I multiply them, I get positive 72. When I add them, I get negative 18. So if they're the same sign and their sum is a negative number, they both must be negative. So they're both negative. And we could go through all of the factors of 72, but the one that springs up, maybe you think of 8 times 9, but 8 times 9, or negative 8 minus 9, or negative 8 plus negative 9 doesn't work. That turns into 17."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "So if they're the same sign and their sum is a negative number, they both must be negative. So they're both negative. And we could go through all of the factors of 72, but the one that springs up, maybe you think of 8 times 9, but 8 times 9, or negative 8 minus 9, or negative 8 plus negative 9 doesn't work. That turns into 17. That was close. Let me show you that. Negative 9 plus negative 8, that is equal to negative 17."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "That turns into 17. That was close. Let me show you that. Negative 9 plus negative 8, that is equal to negative 17. Close, but no cigar. So what other ones are? We have 6 and 12."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "Negative 9 plus negative 8, that is equal to negative 17. Close, but no cigar. So what other ones are? We have 6 and 12. That actually seems pretty good. If we have negative 6 plus negative 12, that is equal to negative 18. Notice, it's a bit of an art."}, {"video_title": "Negative exponent intuition   Pre-Algebra   Khan Academy.mp3", "Sentence": "I have been asked for some intuition as to why, let's say, a to the minus b is equal to 1 over a to the b. And before I give you the intuition, I want you to just realize that this really is a definition. I don't know, the inventor of mathematics wasn't one person. It was a convention that arose. But they defined this, and they defined this for the reasons that I'm going to show you. Well, what I'm going to show you is one of the reasons. And we'll see that this is a good definition, because once you learned exponent rules, all of the other exponent rules stay consistent for negative exponents and when you raise something to the 0th power."}, {"video_title": "Negative exponent intuition   Pre-Algebra   Khan Academy.mp3", "Sentence": "It was a convention that arose. But they defined this, and they defined this for the reasons that I'm going to show you. Well, what I'm going to show you is one of the reasons. And we'll see that this is a good definition, because once you learned exponent rules, all of the other exponent rules stay consistent for negative exponents and when you raise something to the 0th power. So let's take the positive exponents. Those are pretty intuitive, I think. So the positive exponents, so you have a to the 1, a squared, a cubed, a to the 4th."}, {"video_title": "Negative exponent intuition   Pre-Algebra   Khan Academy.mp3", "Sentence": "And we'll see that this is a good definition, because once you learned exponent rules, all of the other exponent rules stay consistent for negative exponents and when you raise something to the 0th power. So let's take the positive exponents. Those are pretty intuitive, I think. So the positive exponents, so you have a to the 1, a squared, a cubed, a to the 4th. What's a to the 1? a to the 1, we said, was a. And then to get to a squared, what did we do?"}, {"video_title": "Negative exponent intuition   Pre-Algebra   Khan Academy.mp3", "Sentence": "So the positive exponents, so you have a to the 1, a squared, a cubed, a to the 4th. What's a to the 1? a to the 1, we said, was a. And then to get to a squared, what did we do? We multiplied by a, right? a squared is just a times a. And then to get to a cubed, what did we do?"}, {"video_title": "Negative exponent intuition   Pre-Algebra   Khan Academy.mp3", "Sentence": "And then to get to a squared, what did we do? We multiplied by a, right? a squared is just a times a. And then to get to a cubed, what did we do? We multiplied by a again. And then to get to a to the 4th, what did we do? We multiplied by a again."}, {"video_title": "Negative exponent intuition   Pre-Algebra   Khan Academy.mp3", "Sentence": "And then to get to a cubed, what did we do? We multiplied by a again. And then to get to a to the 4th, what did we do? We multiplied by a again. Or the other way you could imagine is when you decrease the exponent, what are we doing? We are multiplying by 1 over a, or dividing by a. Similarly, decrease again, you're dividing by a."}, {"video_title": "Negative exponent intuition   Pre-Algebra   Khan Academy.mp3", "Sentence": "We multiplied by a again. Or the other way you could imagine is when you decrease the exponent, what are we doing? We are multiplying by 1 over a, or dividing by a. Similarly, decrease again, you're dividing by a. And go from a squared to a to the 1st, you're dividing by a. So let's use this progression to figure out what a to the 0 is. So this is the first hard one."}, {"video_title": "Negative exponent intuition   Pre-Algebra   Khan Academy.mp3", "Sentence": "Similarly, decrease again, you're dividing by a. And go from a squared to a to the 1st, you're dividing by a. So let's use this progression to figure out what a to the 0 is. So this is the first hard one. So a to the 0. So you're the inventor, the founding mother of mathematics, and you need to define what a to the 0 is. And maybe it's 17, maybe it's pi, I don't know."}, {"video_title": "Negative exponent intuition   Pre-Algebra   Khan Academy.mp3", "Sentence": "So this is the first hard one. So a to the 0. So you're the inventor, the founding mother of mathematics, and you need to define what a to the 0 is. And maybe it's 17, maybe it's pi, I don't know. It's up to you to decide what a to the 0 is. But wouldn't it be nice if a to the 0 retained this pattern? That every time you decrease the exponent, you're dividing by a, right?"}, {"video_title": "Negative exponent intuition   Pre-Algebra   Khan Academy.mp3", "Sentence": "And maybe it's 17, maybe it's pi, I don't know. It's up to you to decide what a to the 0 is. But wouldn't it be nice if a to the 0 retained this pattern? That every time you decrease the exponent, you're dividing by a, right? So if you're going from a to the 1st to a to the 0, wouldn't it be nice if we just divided by a? So let's do that. So if we go from a to the 1st, which is just a, and divide by a, so we're just going to divide it by a."}, {"video_title": "Negative exponent intuition   Pre-Algebra   Khan Academy.mp3", "Sentence": "That every time you decrease the exponent, you're dividing by a, right? So if you're going from a to the 1st to a to the 0, wouldn't it be nice if we just divided by a? So let's do that. So if we go from a to the 1st, which is just a, and divide by a, so we're just going to divide it by a. What is a divided by a? Well, it's just 1. So that's where the definition, or that's one of the intuitions behind why something to the 0th power is equal to 1."}, {"video_title": "Negative exponent intuition   Pre-Algebra   Khan Academy.mp3", "Sentence": "So if we go from a to the 1st, which is just a, and divide by a, so we're just going to divide it by a. What is a divided by a? Well, it's just 1. So that's where the definition, or that's one of the intuitions behind why something to the 0th power is equal to 1. Because when you take that number and you divide it by itself one more time, you just get 1. So that's pretty reasonable. But now let's go into the negative domain."}, {"video_title": "Negative exponent intuition   Pre-Algebra   Khan Academy.mp3", "Sentence": "So that's where the definition, or that's one of the intuitions behind why something to the 0th power is equal to 1. Because when you take that number and you divide it by itself one more time, you just get 1. So that's pretty reasonable. But now let's go into the negative domain. So what should a to the negative 1 equal? a to the negative 1. Well, once again, it's nice if we can retain this pattern, where every time we decrease the exponent, we're dividing by a."}, {"video_title": "Negative exponent intuition   Pre-Algebra   Khan Academy.mp3", "Sentence": "But now let's go into the negative domain. So what should a to the negative 1 equal? a to the negative 1. Well, once again, it's nice if we can retain this pattern, where every time we decrease the exponent, we're dividing by a. So let's divide by a again. So 1 over a. So we're going to take a to the 0 and divide it by a. a to the 0 is 1."}, {"video_title": "Negative exponent intuition   Pre-Algebra   Khan Academy.mp3", "Sentence": "Well, once again, it's nice if we can retain this pattern, where every time we decrease the exponent, we're dividing by a. So let's divide by a again. So 1 over a. So we're going to take a to the 0 and divide it by a. a to the 0 is 1. So what's 1 divided by a? It's 1 over a. And let's do it one more time, and then I think you're going to get the pattern."}, {"video_title": "Negative exponent intuition   Pre-Algebra   Khan Academy.mp3", "Sentence": "So we're going to take a to the 0 and divide it by a. a to the 0 is 1. So what's 1 divided by a? It's 1 over a. And let's do it one more time, and then I think you're going to get the pattern. Well, I think you probably already got the pattern. What's a to the minus 2? Well, it'd be silly now to change this pattern."}, {"video_title": "Negative exponent intuition   Pre-Algebra   Khan Academy.mp3", "Sentence": "And let's do it one more time, and then I think you're going to get the pattern. Well, I think you probably already got the pattern. What's a to the minus 2? Well, it'd be silly now to change this pattern. Every time we decrease the exponent, we're dividing by a. So to go from a to the minus 1 to a to the minus 2, let's just divide by a again. And what do we get?"}, {"video_title": "Negative exponent intuition   Pre-Algebra   Khan Academy.mp3", "Sentence": "Well, it'd be silly now to change this pattern. Every time we decrease the exponent, we're dividing by a. So to go from a to the minus 1 to a to the minus 2, let's just divide by a again. And what do we get? We get, if you take 1 over a and divide by a, you get 1 over a squared. And you could just keep doing this pattern all the way to the left, and you would get a to the minus b is equal to 1 over a to the b. Hopefully that gave you a little intuition as to why, well, first of all, the big mystery is, well, something to a 0th power, why does that equal 1? First, keep in mind that that's just a definition."}, {"video_title": "Negative exponent intuition   Pre-Algebra   Khan Academy.mp3", "Sentence": "And what do we get? We get, if you take 1 over a and divide by a, you get 1 over a squared. And you could just keep doing this pattern all the way to the left, and you would get a to the minus b is equal to 1 over a to the b. Hopefully that gave you a little intuition as to why, well, first of all, the big mystery is, well, something to a 0th power, why does that equal 1? First, keep in mind that that's just a definition. Someone decided that it should be equal to 1, but they had a good reason. And their good reason was they wanted to keep this pattern going. And that's the same reason why they defined negative exponents in this way."}, {"video_title": "Negative exponent intuition   Pre-Algebra   Khan Academy.mp3", "Sentence": "First, keep in mind that that's just a definition. Someone decided that it should be equal to 1, but they had a good reason. And their good reason was they wanted to keep this pattern going. And that's the same reason why they defined negative exponents in this way. And what's extra cool about it is not only does it retain this pattern of when you decrease exponents, you're dividing by a, or when you're increasing exponents, you're multiplying by a. But as you'll see in the exponent rules videos, all of the exponent rules hold. All of the exponent rules are consistent with this definition of something to the 0th power and this definition of something to the negative power."}, {"video_title": "Examples of simplifying expressions involving rational numbers  7th grade   Khan Academy.mp3", "Sentence": "And like always, try to pause this video and see if you can simplify this expression before I take a stab at it. Alright, I'm assuming you have attempted it. Now let's look at it. We have negative 5.55 minus 8.55c plus 4.35c. So the first thing I want to do is, can I combine these c terms? And I definitely can. This is if you, we can add negative 8.55c to 4.35c first, and then that would be, let's see, that would be negative 8.55 plus 4.35, I'm just adding the coefficients, times c, and of course we still have that negative 5.55 out front."}, {"video_title": "Examples of simplifying expressions involving rational numbers  7th grade   Khan Academy.mp3", "Sentence": "We have negative 5.55 minus 8.55c plus 4.35c. So the first thing I want to do is, can I combine these c terms? And I definitely can. This is if you, we can add negative 8.55c to 4.35c first, and then that would be, let's see, that would be negative 8.55 plus 4.35, I'm just adding the coefficients, times c, and of course we still have that negative 5.55 out front. Negative 5.55, and I'll just put a plus there. Now how do we calculate negative 8.55 plus 4.35? Well there's a couple of ways to think about it or visualize it."}, {"video_title": "Examples of simplifying expressions involving rational numbers  7th grade   Khan Academy.mp3", "Sentence": "This is if you, we can add negative 8.55c to 4.35c first, and then that would be, let's see, that would be negative 8.55 plus 4.35, I'm just adding the coefficients, times c, and of course we still have that negative 5.55 out front. Negative 5.55, and I'll just put a plus there. Now how do we calculate negative 8.55 plus 4.35? Well there's a couple of ways to think about it or visualize it. One way is to say, well this is the same thing as the negative of 8.55 minus 4.35. And 8.55 minus 4.35, let's see, 8 minus 4 is gonna be the negative, 8 minus 4 is 4, 55 hundredths minus 35 hundredths is 20 hundredths, so I can write 4.20, which is really just the same thing as 4.2. So all of this, all of this can be replaced with a negative 4.2."}, {"video_title": "Examples of simplifying expressions involving rational numbers  7th grade   Khan Academy.mp3", "Sentence": "Well there's a couple of ways to think about it or visualize it. One way is to say, well this is the same thing as the negative of 8.55 minus 4.35. And 8.55 minus 4.35, let's see, 8 minus 4 is gonna be the negative, 8 minus 4 is 4, 55 hundredths minus 35 hundredths is 20 hundredths, so I can write 4.20, which is really just the same thing as 4.2. So all of this, all of this can be replaced with a negative 4.2. So my entire expression has simplified to negative 5.55, and instead of saying plus negative 4.2c, I can just write it as minus 4.2, 4.2c, and we're done. We can't simplify this anymore. We can't add this term that doesn't involve the variable to this term that does involve the variable."}, {"video_title": "Examples of simplifying expressions involving rational numbers  7th grade   Khan Academy.mp3", "Sentence": "So all of this, all of this can be replaced with a negative 4.2. So my entire expression has simplified to negative 5.55, and instead of saying plus negative 4.2c, I can just write it as minus 4.2, 4.2c, and we're done. We can't simplify this anymore. We can't add this term that doesn't involve the variable to this term that does involve the variable. So this is about as simple as we're gonna get. So let's do another example. So here I have these, I have some more hairy numbers involved, these are all expressed as fractions."}, {"video_title": "Examples of simplifying expressions involving rational numbers  7th grade   Khan Academy.mp3", "Sentence": "We can't add this term that doesn't involve the variable to this term that does involve the variable. So this is about as simple as we're gonna get. So let's do another example. So here I have these, I have some more hairy numbers involved, these are all expressed as fractions. And so let's see, I have 2 fifths m, minus 4 fifths, minus 3 fifths m. So how can I simplify? Well I could add all the m terms together. So let me just change the order, I could rewrite this as 2 fifths m, minus 3 fifths m, minus 4 fifths."}, {"video_title": "Examples of simplifying expressions involving rational numbers  7th grade   Khan Academy.mp3", "Sentence": "So here I have these, I have some more hairy numbers involved, these are all expressed as fractions. And so let's see, I have 2 fifths m, minus 4 fifths, minus 3 fifths m. So how can I simplify? Well I could add all the m terms together. So let me just change the order, I could rewrite this as 2 fifths m, minus 3 fifths m, minus 4 fifths. All I did is I changed the order, and we can see that I have these 2 m terms, I can add those two together. So this is going to be 2 fifths, minus 3 fifths, times m, and then I have the minus 4 fifths still on the right hand side. Now what's 2 fifths minus 3 fifths?"}, {"video_title": "Examples of simplifying expressions involving rational numbers  7th grade   Khan Academy.mp3", "Sentence": "So let me just change the order, I could rewrite this as 2 fifths m, minus 3 fifths m, minus 4 fifths. All I did is I changed the order, and we can see that I have these 2 m terms, I can add those two together. So this is going to be 2 fifths, minus 3 fifths, times m, and then I have the minus 4 fifths still on the right hand side. Now what's 2 fifths minus 3 fifths? Well that's gonna be negative 1 fifth. It's gonna be negative 1 fifth. So I have negative 1 fifth m, minus 4 fifths."}, {"video_title": "Examples of simplifying expressions involving rational numbers  7th grade   Khan Academy.mp3", "Sentence": "Now what's 2 fifths minus 3 fifths? Well that's gonna be negative 1 fifth. It's gonna be negative 1 fifth. So I have negative 1 fifth m, minus 4 fifths. Minus 4 fifths. And once again I'm done, I can't simplify it anymore, I can't add this term that involves m somehow to this negative 4 fifths. So we are done here."}, {"video_title": "Examples of simplifying expressions involving rational numbers  7th grade   Khan Academy.mp3", "Sentence": "So I have negative 1 fifth m, minus 4 fifths. Minus 4 fifths. And once again I'm done, I can't simplify it anymore, I can't add this term that involves m somehow to this negative 4 fifths. So we are done here. Let's do one more example. So here, this is interesting, I have a parentheses and all the rest. And like always, pause the video, see if you can simplify this."}, {"video_title": "Examples of simplifying expressions involving rational numbers  7th grade   Khan Academy.mp3", "Sentence": "So we are done here. Let's do one more example. So here, this is interesting, I have a parentheses and all the rest. And like always, pause the video, see if you can simplify this. Alright, let's work through it together. Now the first thing that I want to do is, let's distribute this 2 so that we just have three terms that are just being added and subtracted. So if we distribute this 2, we're gonna get 2 times 1 fifth m is 2 fifths m. Let me make sure you see that m, m is right here."}, {"video_title": "Examples of simplifying expressions involving rational numbers  7th grade   Khan Academy.mp3", "Sentence": "And like always, pause the video, see if you can simplify this. Alright, let's work through it together. Now the first thing that I want to do is, let's distribute this 2 so that we just have three terms that are just being added and subtracted. So if we distribute this 2, we're gonna get 2 times 1 fifth m is 2 fifths m. Let me make sure you see that m, m is right here. 2 times negative 2 fifths is negative 4 fifths, and then I have plus 3 fifths. Now how can we simplify this more? Well I have these two terms here that don't involve the variable, those are just numbers, I can add them to each other."}, {"video_title": "Examples of simplifying expressions involving rational numbers  7th grade   Khan Academy.mp3", "Sentence": "So if we distribute this 2, we're gonna get 2 times 1 fifth m is 2 fifths m. Let me make sure you see that m, m is right here. 2 times negative 2 fifths is negative 4 fifths, and then I have plus 3 fifths. Now how can we simplify this more? Well I have these two terms here that don't involve the variable, those are just numbers, I can add them to each other. So I have negative 4 fifths plus 3 fifths. So what's negative 4 plus 3? That's gonna be negative 1."}, {"video_title": "Examples of simplifying expressions involving rational numbers  7th grade   Khan Academy.mp3", "Sentence": "Well I have these two terms here that don't involve the variable, those are just numbers, I can add them to each other. So I have negative 4 fifths plus 3 fifths. So what's negative 4 plus 3? That's gonna be negative 1. So this is gonna be negative 1 fifth. What we have in yellow here, and I still have the 2 over 5 m. 2 fifths m minus 1 fifth. And we're done, we've simplified that as much as we can."}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "One are these x masses, and we know that they have the same identical mass, and we'll call that identical. Each of them have a mass of x. But then we have this other, this blue thing, and that has a mass of y, which isn't necessarily going to be the same as a mass of x. But when I have two of these x's and a y, it seems like their total mass, y is the case, that their total mass balances out these 8 kilograms right over here. Each of these are 1-kilogram blocks. It balances them out. So the first question I'm going to ask you is, can you express this mathematically?"}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "But when I have two of these x's and a y, it seems like their total mass, y is the case, that their total mass balances out these 8 kilograms right over here. Each of these are 1-kilogram blocks. It balances them out. So the first question I'm going to ask you is, can you express this mathematically? Can you express what we're seeing here, the fact that this total mass balances out with this total mass? Can you express that mathematically? Well, let's just think about our total mass on this side."}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So the first question I'm going to ask you is, can you express this mathematically? Can you express what we're seeing here, the fact that this total mass balances out with this total mass? Can you express that mathematically? Well, let's just think about our total mass on this side. We have two masses of mass x, so those two are going to total at 2x. And then you have a mass of y, so then you're going to have another y. So the total mass on the left-hand side, and let me write it a little bit closer to the center, just so it doesn't get too spread out."}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Well, let's just think about our total mass on this side. We have two masses of mass x, so those two are going to total at 2x. And then you have a mass of y, so then you're going to have another y. So the total mass on the left-hand side, and let me write it a little bit closer to the center, just so it doesn't get too spread out. On the left-hand side, I have 2x plus a mass of y. That's the total mass. The total mass on the left-hand side is 2x plus y."}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So the total mass on the left-hand side, and let me write it a little bit closer to the center, just so it doesn't get too spread out. On the left-hand side, I have 2x plus a mass of y. That's the total mass. The total mass on the left-hand side is 2x plus y. The total mass on the right-hand side is just 8. 1, 2, 3, 4, 5, 6, 7, 8. It is equal to 8."}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "The total mass on the left-hand side is 2x plus y. The total mass on the right-hand side is just 8. 1, 2, 3, 4, 5, 6, 7, 8. It is equal to 8. And since we see that the scale is balanced, this total mass must be equal to this total mass, so we can write an equal sign there. Now my question to you. Is there anything we can do, just based on the information that we have here, to solve for either the mass x or for the mass y?"}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "It is equal to 8. And since we see that the scale is balanced, this total mass must be equal to this total mass, so we can write an equal sign there. Now my question to you. Is there anything we can do, just based on the information that we have here, to solve for either the mass x or for the mass y? Is there anything that we can do? Well, the simple answer is, just with this information here, there is actually very little. You might say, well, let me take a y from both sides."}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Is there anything we can do, just based on the information that we have here, to solve for either the mass x or for the mass y? Is there anything that we can do? Well, the simple answer is, just with this information here, there is actually very little. You might say, well, let me take a y from both sides. You might take this y block up. But if you take this y block up, you have to take away y from this side, and you don't know what y is, and if you think about it algebraically, you might get rid of the y here, subtracting y, but then you're going to have to subtract y from this side too. So you're not going to get rid of the y."}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "You might say, well, let me take a y from both sides. You might take this y block up. But if you take this y block up, you have to take away y from this side, and you don't know what y is, and if you think about it algebraically, you might get rid of the y here, subtracting y, but then you're going to have to subtract y from this side too. So you're not going to get rid of the y. Same thing with the x's. You actually don't have enough information. y depends on what x is, and x depends on what y is."}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So you're not going to get rid of the y. Same thing with the x's. You actually don't have enough information. y depends on what x is, and x depends on what y is. Lucky for us, however, we do have some more of these blocks laying around. And what we do is, we take one of these x blocks, and I stick it over here, and I also take one of the y blocks, and I stick them over here, and I stick it right over there. And then I keep adding these ones until I balance this thing out."}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "y depends on what x is, and x depends on what y is. Lucky for us, however, we do have some more of these blocks laying around. And what we do is, we take one of these x blocks, and I stick it over here, and I also take one of the y blocks, and I stick them over here, and I stick it right over there. And then I keep adding these ones until I balance this thing out. So I keep adding these ones. So obviously, if I just place this, this will go down, because there's nothing on that side. But I keep adding these blocks until it all balances out."}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "And then I keep adding these ones until I balance this thing out. So I keep adding these ones. So obviously, if I just place this, this will go down, because there's nothing on that side. But I keep adding these blocks until it all balances out. And I find that my scale balances once I have 5 kilograms on the right-hand side. So once again, let me ask you this information, that now having an x and a y on the left-hand side, and a 5 kilograms on the right-hand side, and the fact that they're balanced, how can we represent that mathematically? Well, our total mass on the left-hand side is x plus y, and our total mass on the right-hand side is x plus y, and on the right-hand side I have 5 kilograms."}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "But I keep adding these blocks until it all balances out. And I find that my scale balances once I have 5 kilograms on the right-hand side. So once again, let me ask you this information, that now having an x and a y on the left-hand side, and a 5 kilograms on the right-hand side, and the fact that they're balanced, how can we represent that mathematically? Well, our total mass on the left-hand side is x plus y, and our total mass on the right-hand side is x plus y, and on the right-hand side I have 5 kilograms. And we know this is what actually balanced the scale. So these total masses must be equal to each other. And this information by itself, once again, there's nothing I can do with it."}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Well, our total mass on the left-hand side is x plus y, and our total mass on the right-hand side is x plus y, and on the right-hand side I have 5 kilograms. And we know this is what actually balanced the scale. So these total masses must be equal to each other. And this information by itself, once again, there's nothing I can do with it. I don't know what x and y are. If y is 4, maybe x is 1. Maybe x is 4, y is 1. Who knows what these are?"}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "And this information by itself, once again, there's nothing I can do with it. I don't know what x and y are. If y is 4, maybe x is 1. Maybe x is 4, y is 1. Who knows what these are? The interesting thing is we can actually use both of this information to figure out what x and y actually is. And I'll give you a few seconds to think about how we can approach this situation. Well, think about it this way."}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Maybe x is 4, y is 1. Who knows what these are? The interesting thing is we can actually use both of this information to figure out what x and y actually is. And I'll give you a few seconds to think about how we can approach this situation. Well, think about it this way. We know that x plus y is equal to 5. So if we were to get rid of an x and a y on this side, on the left-hand side of the equation, well, what would we have to get rid of on the right-hand side of the, if we got rid of an x and a y on the left-hand side of the scale, what would we have to get rid of on the right-hand side of the scale to take away the same mass? Well, if we take away an x and a y on the left-hand side, we know that an x plus a y is 5 kilograms."}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Well, think about it this way. We know that x plus y is equal to 5. So if we were to get rid of an x and a y on this side, on the left-hand side of the equation, well, what would we have to get rid of on the right-hand side of the, if we got rid of an x and a y on the left-hand side of the scale, what would we have to get rid of on the right-hand side of the scale to take away the same mass? Well, if we take away an x and a y on the left-hand side, we know that an x plus a y is 5 kilograms. So we would just have to take 5 kilograms from the right-hand side. So let's think about what that would do. Well, then I would just have an x left over here, and I would just have some of these masses left over there, and I wouldn't know what x is."}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Well, if we take away an x and a y on the left-hand side, we know that an x plus a y is 5 kilograms. So we would just have to take 5 kilograms from the right-hand side. So let's think about what that would do. Well, then I would just have an x left over here, and I would just have some of these masses left over there, and I wouldn't know what x is. Let's think about how we can represent that algebraically. Essentially, if we're taking an x and a y from the left-hand side, if I'm taking an x and a y from the left-hand side, I'm subtracting an x, I'm subtracting an x, and I am subtracting, actually, let me think of it this way. I am subtracting, I am subtracting an x, I am subtracting an x plus y. I'm subtracting an x and a y from the left-hand side, but then what am I gonna have to do on the right-hand side?"}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Well, then I would just have an x left over here, and I would just have some of these masses left over there, and I wouldn't know what x is. Let's think about how we can represent that algebraically. Essentially, if we're taking an x and a y from the left-hand side, if I'm taking an x and a y from the left-hand side, I'm subtracting an x, I'm subtracting an x, and I am subtracting, actually, let me think of it this way. I am subtracting, I am subtracting an x, I am subtracting an x plus y. I'm subtracting an x and a y from the left-hand side, but then what am I gonna have to do on the right-hand side? Well, an x and a y we know has a mass of five, so we can subtract five from the right-hand side. And the only way I'm able to do this is because of the information that we got from the second scale. And so I can take away five, so this is going to be equal to taking away five."}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "I am subtracting, I am subtracting an x, I am subtracting an x plus y. I'm subtracting an x and a y from the left-hand side, but then what am I gonna have to do on the right-hand side? Well, an x and a y we know has a mass of five, so we can subtract five from the right-hand side. And the only way I'm able to do this is because of the information that we got from the second scale. And so I can take away five, so this is going to be equal to taking away five. Taking away an x and a y is the same thing as taking away five, and we know that because an x and a y is equal to five kilograms. And so if we take away an x and a y on the left-hand side, what are we left with? Well, this is going to be the same thing."}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "And so I can take away five, so this is going to be equal to taking away five. Taking away an x and a y is the same thing as taking away five, and we know that because an x and a y is equal to five kilograms. And so if we take away an x and a y on the left-hand side, what are we left with? Well, this is going to be the same thing. Let me rewrite, let me rewrite this part. This taking away an x and a y is the same thing, is the same thing if you distribute the negative sign as taking away an x, taking away an x and taking away a y, and taking away a y. And so on the left-hand side, we are left with, on the left-hand side, we are left with just two x, and we take away one of the x's."}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Well, this is going to be the same thing. Let me rewrite, let me rewrite this part. This taking away an x and a y is the same thing, is the same thing if you distribute the negative sign as taking away an x, taking away an x and taking away a y, and taking away a y. And so on the left-hand side, we are left with, on the left-hand side, we are left with just two x, and we take away one of the x's. We're left with just an x. And we had a y and we took away one of the y's, so we're left with no y's left. And we see that here visually."}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "And so on the left-hand side, we are left with, on the left-hand side, we are left with just two x, and we take away one of the x's. We're left with just an x. And we had a y and we took away one of the y's, so we're left with no y's left. And we see that here visually. We have just an x here. And what do we have on the right-hand side? We had eight, and we knew that an x and a y is equal to five so we took away five, to keep the scale balanced."}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "And we see that here visually. We have just an x here. And what do we have on the right-hand side? We had eight, and we knew that an x and a y is equal to five so we took away five, to keep the scale balanced. And so eight minus five is going to be three. Eight minus five is equal to three. And just like that, using this extra information, we were able to figure out that the mass of x is equal to, the mass of x is equal to three."}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "We had eight, and we knew that an x and a y is equal to five so we took away five, to keep the scale balanced. And so eight minus five is going to be three. Eight minus five is equal to three. And just like that, using this extra information, we were able to figure out that the mass of x is equal to, the mass of x is equal to three. Now, one final question. We were able to figure out what the mass of x is, but can you now figure out what the mass of y is? Well, we can go back to either one of these scales, but it'll probably be simpler to go back to this one."}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "And just like that, using this extra information, we were able to figure out that the mass of x is equal to, the mass of x is equal to three. Now, one final question. We were able to figure out what the mass of x is, but can you now figure out what the mass of y is? Well, we can go back to either one of these scales, but it'll probably be simpler to go back to this one. We know that the mass of x plus the mass of y is equal to five. So you could say, so one thing we know, that x is now equal to three. We know that this is now a three kilogram mass."}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Well, we can go back to either one of these scales, but it'll probably be simpler to go back to this one. We know that the mass of x plus the mass of y is equal to five. So you could say, so one thing we know, that x is now equal to three. We know that this is now a three kilogram mass. We can rewrite this as three, three plus y, three plus y is equal to, is equal to five. Well now we say, well, we could take three away from both sides. If I take three away from the left-hand side, I just have to take three away from the right-hand side to keep my scale balanced, and I'll be left with that the mass of y is balanced with a mass of two, or y is equal to two."}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "We know that this is now a three kilogram mass. We can rewrite this as three, three plus y, three plus y is equal to, is equal to five. Well now we say, well, we could take three away from both sides. If I take three away from the left-hand side, I just have to take three away from the right-hand side to keep my scale balanced, and I'll be left with that the mass of y is balanced with a mass of two, or y is equal to two. That's analogous to taking three from both sides of this equation. And on the left-hand side, I'm just left with a y. And on the right-hand side, I am just left with a two."}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "If I take three away from the left-hand side, I just have to take three away from the right-hand side to keep my scale balanced, and I'll be left with that the mass of y is balanced with a mass of two, or y is equal to two. That's analogous to taking three from both sides of this equation. And on the left-hand side, I'm just left with a y. And on the right-hand side, I am just left with a two. So x is equal to three kilograms, and y is equal to two kilograms. And what I encourage you to do is verify that it made sense right up here. Figure out what the total mass on the left-hand and the right-hand, or verify what the total mass right over here really was eight to begin with."}, {"video_title": "Number of solutions to linear equations ex 2   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So no solution linear, or equation with no solutions is going to be one where I don't care how you manipulate it, the thing on the left can never be equal to the thing on the right. And so let's see what options they give us. So one, they want us to, we can pick the coefficient on the x term, and then we can pick the constant. So if we made this negative 11x, so now we have a negative 11x on both sides. Here on the left hand side we have negative 11x plus 4. If we do something other than 4 here, so if we did like say negative 11x minus 11, then here we're not going to have any solutions. And you're saying, hey Sal, how did you come up with that?"}, {"video_title": "Number of solutions to linear equations ex 2   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So if we made this negative 11x, so now we have a negative 11x on both sides. Here on the left hand side we have negative 11x plus 4. If we do something other than 4 here, so if we did like say negative 11x minus 11, then here we're not going to have any solutions. And you're saying, hey Sal, how did you come up with that? Well, think about it right over here. We have a negative 11x here, we have a negative 11x there. If you wanted to solve it algebraically, you could add 11x to both sides, and both of these terms would cancel out with each other."}, {"video_title": "Number of solutions to linear equations ex 2   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "And you're saying, hey Sal, how did you come up with that? Well, think about it right over here. We have a negative 11x here, we have a negative 11x there. If you wanted to solve it algebraically, you could add 11x to both sides, and both of these terms would cancel out with each other. And all you would be left with is a 4 is equal to a negative 11, which is not possible for any x that you pick. Another way that you think about it is, here we have negative 11 times some number, and we're adding 4 to it. And here we're taking negative 11 times that same number, and we're subtracting 11 from it."}, {"video_title": "The beauty of algebra   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Before we get into the meat of algebra, I wanted to give you a quote from one of the greatest minds in human history, Galileo Galilei, because I think this quote encapsulates the true point of algebra and really mathematics in general. He said, philosophy is written in that great book which ever lies before our eyes. I mean the universe. But we cannot understand it if we do not first learn the language and grasp the symbols in which it is written. This book is written in the mathematical language without which one wanders in vain through a dark labyrinth. So very dramatic, but very deep. And this really is the point of mathematics."}, {"video_title": "The beauty of algebra   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "But we cannot understand it if we do not first learn the language and grasp the symbols in which it is written. This book is written in the mathematical language without which one wanders in vain through a dark labyrinth. So very dramatic, but very deep. And this really is the point of mathematics. And what we'll see as we start getting deeper and deeper into algebra is that we're going to start abstracting things. And we're going to start getting to core ideas that start explaining really how the universe is structured. Sure, these ideas can be applied to things like economics and finance and physics and chemistry."}, {"video_title": "The beauty of algebra   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And this really is the point of mathematics. And what we'll see as we start getting deeper and deeper into algebra is that we're going to start abstracting things. And we're going to start getting to core ideas that start explaining really how the universe is structured. Sure, these ideas can be applied to things like economics and finance and physics and chemistry. But at their core, they're the same idea. And so they're even more fundamental, more pure than any one of those applications. And to see what I mean by getting down to the root idea, let's go with a, I guess we started with the very grand, the philosophy of the universe is written in mathematics."}, {"video_title": "The beauty of algebra   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Sure, these ideas can be applied to things like economics and finance and physics and chemistry. But at their core, they're the same idea. And so they're even more fundamental, more pure than any one of those applications. And to see what I mean by getting down to the root idea, let's go with a, I guess we started with the very grand, the philosophy of the universe is written in mathematics. But let's start with a very concrete, simple idea. But we'll keep abstracting. And we'll see how the same idea connects across many domains in our universe."}, {"video_title": "The beauty of algebra   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And to see what I mean by getting down to the root idea, let's go with a, I guess we started with the very grand, the philosophy of the universe is written in mathematics. But let's start with a very concrete, simple idea. But we'll keep abstracting. And we'll see how the same idea connects across many domains in our universe. So let's just say we're at a store. And we're going to buy something. And there's a sale."}, {"video_title": "The beauty of algebra   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And we'll see how the same idea connects across many domains in our universe. So let's just say we're at a store. And we're going to buy something. And there's a sale. The sale says that it is 30% off. And I'm interested. I don't shop at too fancy stores."}, {"video_title": "The beauty of algebra   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And there's a sale. The sale says that it is 30% off. And I'm interested. I don't shop at too fancy stores. So let's say I'm interested in a pair of pants. And the pair of pants before the sale even is about $20. And that is about how much I spend on my pants."}, {"video_title": "The beauty of algebra   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "I don't shop at too fancy stores. So let's say I'm interested in a pair of pants. And the pair of pants before the sale even is about $20. And that is about how much I spend on my pants. So I'm interested in a $20 pair of pants. But it's even better. There's a 30% off sale on these pants."}, {"video_title": "The beauty of algebra   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And that is about how much I spend on my pants. So I'm interested in a $20 pair of pants. But it's even better. There's a 30% off sale on these pants. Well, how would I think about how much I'm going to get off of that $20? And this isn't algebra yet. This is something that you've probably had exposure to."}, {"video_title": "The beauty of algebra   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "There's a 30% off sale on these pants. Well, how would I think about how much I'm going to get off of that $20? And this isn't algebra yet. This is something that you've probably had exposure to. You would multiply the 30% times the $20. So you would say your discount is equal to, you could write it as 30% times $20. I'll do the $20 in purple."}, {"video_title": "The beauty of algebra   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "This is something that you've probably had exposure to. You would multiply the 30% times the $20. So you would say your discount is equal to, you could write it as 30% times $20. I'll do the $20 in purple. Or you could write it, if you wanted to write this as a decimal, you could write this as 0.30 times $20. And if you were to do the math, you would get $6. So nothing new over there."}, {"video_title": "The beauty of algebra   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "I'll do the $20 in purple. Or you could write it, if you wanted to write this as a decimal, you could write this as 0.30 times $20. And if you were to do the math, you would get $6. So nothing new over there. But what if I want to generalize a little bit? That's the discount on this particular pair of pants. But what if I wanted to know the discount on anything in the store?"}, {"video_title": "The beauty of algebra   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "So nothing new over there. But what if I want to generalize a little bit? That's the discount on this particular pair of pants. But what if I wanted to know the discount on anything in the store? Well, then I could say, well, let x be the price. Let me do this in a different color. So I'm just going to make a symbol."}, {"video_title": "The beauty of algebra   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "But what if I wanted to know the discount on anything in the store? Well, then I could say, well, let x be the price. Let me do this in a different color. So I'm just going to make a symbol. Let x be the price of the product I want to buy. Price, the non-discount price of the product in the store. So now all of a sudden, we can say that our discount is equal to 30% times x."}, {"video_title": "The beauty of algebra   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "So I'm just going to make a symbol. Let x be the price of the product I want to buy. Price, the non-discount price of the product in the store. So now all of a sudden, we can say that our discount is equal to 30% times x. Or if we wanted to write it as a decimal, we could write, if we wanted to write 30% as a decimal, we could write 0.30 times x. Now this is interesting. Now you give me the price of any product in the store, and I can substitute it in for x."}, {"video_title": "The beauty of algebra   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "So now all of a sudden, we can say that our discount is equal to 30% times x. Or if we wanted to write it as a decimal, we could write, if we wanted to write 30% as a decimal, we could write 0.30 times x. Now this is interesting. Now you give me the price of any product in the store, and I can substitute it in for x. And then I can essentially multiply 0.3 times that, and I would get the discount. So now we're starting to very slowly, we're starting to get into the abstraction of algebra. And we'll see that these will get much more nuanced and deep and frankly, more beautiful as we start studying more and more kind of algebraic ideas."}, {"video_title": "The beauty of algebra   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Now you give me the price of any product in the store, and I can substitute it in for x. And then I can essentially multiply 0.3 times that, and I would get the discount. So now we're starting to very slowly, we're starting to get into the abstraction of algebra. And we'll see that these will get much more nuanced and deep and frankly, more beautiful as we start studying more and more kind of algebraic ideas. But we aren't done here. We can abstract this even more. Over here, we've said we've generalized this for any product."}, {"video_title": "The beauty of algebra   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And we'll see that these will get much more nuanced and deep and frankly, more beautiful as we start studying more and more kind of algebraic ideas. But we aren't done here. We can abstract this even more. Over here, we've said we've generalized this for any product. We're not just saying for this $20 product. If there's a $10 product, we can put that $10 product in here for x, and then we would say 0.30 times 10. And the discount would be $3."}, {"video_title": "The beauty of algebra   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Over here, we've said we've generalized this for any product. We're not just saying for this $20 product. If there's a $10 product, we can put that $10 product in here for x, and then we would say 0.30 times 10. And the discount would be $3. It might be $100 project, then the discount would be $30. But let's generalize even more. Let's say, well, what is the discount for any given sale when the sale is a certain percentage?"}, {"video_title": "The beauty of algebra   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And the discount would be $3. It might be $100 project, then the discount would be $30. But let's generalize even more. Let's say, well, what is the discount for any given sale when the sale is a certain percentage? So now we can say that the discount, let me define a variable. So let's let m equal, I'll say p just so it makes sense. p is equal to the percentage off."}, {"video_title": "The beauty of algebra   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Let's say, well, what is the discount for any given sale when the sale is a certain percentage? So now we can say that the discount, let me define a variable. So let's let m equal, I'll say p just so it makes sense. p is equal to the percentage off. Now what can we do? Well, now we can say that the discount is equal to the percentage off. In these other examples, we were picking 30%."}, {"video_title": "The beauty of algebra   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "p is equal to the percentage off. Now what can we do? Well, now we can say that the discount is equal to the percentage off. In these other examples, we were picking 30%. But we can say now it's p. It's the percentage off. It's p. That's the percentage off times the product in question, times the price, the non-discount price of the product in question. Well, that was x."}, {"video_title": "The beauty of algebra   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "In these other examples, we were picking 30%. But we can say now it's p. It's the percentage off. It's p. That's the percentage off times the product in question, times the price, the non-discount price of the product in question. Well, that was x. The discount is equal to p times x. Now this is really interesting. Now we have a general way of calculating a discount for any given percentage off and any given product x."}, {"video_title": "The beauty of algebra   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Well, that was x. The discount is equal to p times x. Now this is really interesting. Now we have a general way of calculating a discount for any given percentage off and any given product x. And we didn't have to use these words and these letters. We could have said, let y equal the discount. Then we could have written the same underlying idea."}, {"video_title": "The beauty of algebra   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Now we have a general way of calculating a discount for any given percentage off and any given product x. And we didn't have to use these words and these letters. We could have said, let y equal the discount. Then we could have written the same underlying idea. Instead of writing discount, we could have written y is equal to the percentage off. Is equal to the percentage off p times the non-discount price of the product, times x. And you could have defined these letters any way you wanted."}, {"video_title": "The beauty of algebra   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Then we could have written the same underlying idea. Instead of writing discount, we could have written y is equal to the percentage off. Is equal to the percentage off p times the non-discount price of the product, times x. And you could have defined these letters any way you wanted. Instead of writing y there, you could have written a Greek letter. Or you could have written any symbol there, as long as you can keep track of that that symbol represents the actual dollar discount. But now things get really interesting."}, {"video_title": "The beauty of algebra   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And you could have defined these letters any way you wanted. Instead of writing y there, you could have written a Greek letter. Or you could have written any symbol there, as long as you can keep track of that that symbol represents the actual dollar discount. But now things get really interesting. Because we can use this type of a relationship, which is an equation. You're equating y to this right over here. That's why we call it an equation."}, {"video_title": "The beauty of algebra   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "But now things get really interesting. Because we can use this type of a relationship, which is an equation. You're equating y to this right over here. That's why we call it an equation. This can be used for things that are completely unrelated to the discount price at the store over here. So in physics, you'll see that force is equal to mass times acceleration. The letters are different, but these are fundamentally the same idea."}, {"video_title": "The beauty of algebra   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "That's why we call it an equation. This can be used for things that are completely unrelated to the discount price at the store over here. So in physics, you'll see that force is equal to mass times acceleration. The letters are different, but these are fundamentally the same idea. We could have let y is equal to force. And m is equal to, or mass is equal to p. So let me write p is equal to mass. And this wouldn't be an intuitive way to define it."}, {"video_title": "The beauty of algebra   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "The letters are different, but these are fundamentally the same idea. We could have let y is equal to force. And m is equal to, or mass is equal to p. So let me write p is equal to mass. And this wouldn't be an intuitive way to define it. But I want to show you that this is the same idea, the same relationship, but it's being applied to two completely different things. And we could say x is equal to acceleration. Well, then the famous force is equal to mass times acceleration can be rewritten."}, {"video_title": "The beauty of algebra   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And this wouldn't be an intuitive way to define it. But I want to show you that this is the same idea, the same relationship, but it's being applied to two completely different things. And we could say x is equal to acceleration. Well, then the famous force is equal to mass times acceleration can be rewritten. And it's really the same exact idea as y, which we've defined as force, can be equal to mass, which we're going to use the symbol p, which is equal to p times acceleration. We're just going to happen to use the letter x here, times x. Well, this is the exact same equation."}, {"video_title": "The beauty of algebra   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Well, then the famous force is equal to mass times acceleration can be rewritten. And it's really the same exact idea as y, which we've defined as force, can be equal to mass, which we're going to use the symbol p, which is equal to p times acceleration. We're just going to happen to use the letter x here, times x. Well, this is the exact same equation. We can see that we can take this equation and it can apply to things in economics, or it can apply to things in finance, or it can apply to things in computer science, or logic, or electrical engineering, or anything, accounting. There's an infinite number of applications of this one equation. And what's neat about mathematics, and what's neat about algebra in particular, is we can focus on this abstraction."}, {"video_title": "The beauty of algebra   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Well, this is the exact same equation. We can see that we can take this equation and it can apply to things in economics, or it can apply to things in finance, or it can apply to things in computer science, or logic, or electrical engineering, or anything, accounting. There's an infinite number of applications of this one equation. And what's neat about mathematics, and what's neat about algebra in particular, is we can focus on this abstraction. We can focus on the abstract here. And we can manipulate the abstract here. And what we discover from these ideas, from these manipulations, can then go and be reapplied to all of these other applications, to all of them."}, {"video_title": "The beauty of algebra   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And what's neat about mathematics, and what's neat about algebra in particular, is we can focus on this abstraction. We can focus on the abstract here. And we can manipulate the abstract here. And what we discover from these ideas, from these manipulations, can then go and be reapplied to all of these other applications, to all of them. And even neater, it's kind of telling us the true structure of the universe if you were to strip away all of these human definitions and all of these human applications. So for example, we could say, look, if y is equal to p times x, so literally if someone said, hey, this is y, and someone says on the other hand, they say, I have p times x, I could say, well, you have the same thing in both of your hands. And if you were to divide one of them by a number, and if you wanted them to still be equal, you would divide the other one by that number."}, {"video_title": "The beauty of algebra   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And what we discover from these ideas, from these manipulations, can then go and be reapplied to all of these other applications, to all of them. And even neater, it's kind of telling us the true structure of the universe if you were to strip away all of these human definitions and all of these human applications. So for example, we could say, look, if y is equal to p times x, so literally if someone said, hey, this is y, and someone says on the other hand, they say, I have p times x, I could say, well, you have the same thing in both of your hands. And if you were to divide one of them by a number, and if you wanted them to still be equal, you would divide the other one by that number. So for example, let's say we know that y is equal to p times x. What if you wanted to have them both be equal, and you say, well, what is y divided by x is going to be equal to? Well, y was equal to p times x, so y divided by x is going to be the same thing as p times x divided by x."}, {"video_title": "The beauty of algebra   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And if you were to divide one of them by a number, and if you wanted them to still be equal, you would divide the other one by that number. So for example, let's say we know that y is equal to p times x. What if you wanted to have them both be equal, and you say, well, what is y divided by x is going to be equal to? Well, y was equal to p times x, so y divided by x is going to be the same thing as p times x divided by x. But now this is interesting, because p times x divided by x, well, if you multiply by something and then divide by that something, it's just the same. You're going to get your original number. If you multiply by 5 and divide by 5, you're just going to start with p or whatever this number is."}, {"video_title": "The beauty of algebra   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Well, y was equal to p times x, so y divided by x is going to be the same thing as p times x divided by x. But now this is interesting, because p times x divided by x, well, if you multiply by something and then divide by that something, it's just the same. You're going to get your original number. If you multiply by 5 and divide by 5, you're just going to start with p or whatever this number is. So those would cancel out. But we were able to manipulate the abstraction here and get y over x is equal to p. Let me make that x green. y over x is equal to p. And now this has implications for every one of these ideas."}, {"video_title": "The beauty of algebra   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "If you multiply by 5 and divide by 5, you're just going to start with p or whatever this number is. So those would cancel out. But we were able to manipulate the abstraction here and get y over x is equal to p. Let me make that x green. y over x is equal to p. And now this has implications for every one of these ideas. One is telling us a fundamental truth about the universe, almost devoid of any of these applications. But now we can go and take them back to any place that we applied. And the really interesting thing is we're going to find new infinite number of applications."}, {"video_title": "The beauty of algebra   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "y over x is equal to p. And now this has implications for every one of these ideas. One is telling us a fundamental truth about the universe, almost devoid of any of these applications. But now we can go and take them back to any place that we applied. And the really interesting thing is we're going to find new infinite number of applications. And we don't even know, frankly, most of them. We're going to discover new ones for them in 1,000 years. And so hopefully this gives you a sense for why Galileo said what he said about really, mathematics is really the language with which we can understand the philosophy of the universe."}, {"video_title": "Ex 3 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So we're told that William is four times as old as Ben. 12 years ago, William was seven times as old as Ben. How old is Ben now? So once again, it's a good idea to try to do this on your own first, and then I'll work through it. So what's the unknown here? Well, the unknown here is how old is Ben now? So let's set a variable equal to that, and we do x or y, but since Ben starts with a b, let's use b for Ben."}, {"video_title": "Ex 3 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So once again, it's a good idea to try to do this on your own first, and then I'll work through it. So what's the unknown here? Well, the unknown here is how old is Ben now? So let's set a variable equal to that, and we do x or y, but since Ben starts with a b, let's use b for Ben. So let's let b equal Ben's current age, Ben's age now. So let's see how all of this other information relates to Ben's current age, and then maybe we can set up some equation and then solve for things. So I'll do it a little bit more structured in this one, and you could have done many of the problems we've been working on in this way."}, {"video_title": "Ex 3 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So let's set a variable equal to that, and we do x or y, but since Ben starts with a b, let's use b for Ben. So let's let b equal Ben's current age, Ben's age now. So let's see how all of this other information relates to Ben's current age, and then maybe we can set up some equation and then solve for things. So I'll do it a little bit more structured in this one, and you could have done many of the problems we've been working on in this way. So let's think about Ben. Let's think about Ben, and then let's think about William. I'll do William in blue here."}, {"video_title": "Ex 3 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So I'll do it a little bit more structured in this one, and you could have done many of the problems we've been working on in this way. So let's think about Ben. Let's think about Ben, and then let's think about William. I'll do William in blue here. So let's think about William. William. And then there's two points in time we're talking about."}, {"video_title": "Ex 3 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "I'll do William in blue here. So let's think about William. William. And then there's two points in time we're talking about. We're talking about now, today, and we're gonna talk about 12 years ago. So let's talk about, so over here, let's say today, or let's call that now. This will be our now column, and then this will be our 12 years ago."}, {"video_title": "Ex 3 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "And then there's two points in time we're talking about. We're talking about now, today, and we're gonna talk about 12 years ago. So let's talk about, so over here, let's say today, or let's call that now. This will be our now column, and then this will be our 12 years ago. 12 years ago. So let's see what we can fill in here. So what is Ben's age now?"}, {"video_title": "Ex 3 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "This will be our now column, and then this will be our 12 years ago. 12 years ago. So let's see what we can fill in here. So what is Ben's age now? Well, we just defined that as a variable b. That's the unknown. That's what we have to figure out."}, {"video_title": "Ex 3 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So what is Ben's age now? Well, we just defined that as a variable b. That's the unknown. That's what we have to figure out. So let's just stick that there. So that's going to be, that's just going to be b. Well, what's Ben's age 12 years ago?"}, {"video_title": "Ex 3 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "That's what we have to figure out. So let's just stick that there. So that's going to be, that's just going to be b. Well, what's Ben's age 12 years ago? And so we would maybe want to express it in terms of b. Well, if he's b now, 12 years ago, he was just b minus 12. Fair enough."}, {"video_title": "Ex 3 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Well, what's Ben's age 12 years ago? And so we would maybe want to express it in terms of b. Well, if he's b now, 12 years ago, he was just b minus 12. Fair enough. Now, what is William's age today? Well, this first sentence gave us the information. William is four times as old as Ben, and we can assume that they're talking about today."}, {"video_title": "Ex 3 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Fair enough. Now, what is William's age today? Well, this first sentence gave us the information. William is four times as old as Ben, and we can assume that they're talking about today. Is, is four times as old as Ben. So if Ben is b, William is going to be 4b. 4b."}, {"video_title": "Ex 3 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "William is four times as old as Ben, and we can assume that they're talking about today. Is, is four times as old as Ben. So if Ben is b, William is going to be 4b. 4b. And so how old was William 12 years ago? Well, if he's 4b right now, 12 years ago, he'll just be 12 less than that. So he's 4b now."}, {"video_title": "Ex 3 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "4b. And so how old was William 12 years ago? Well, if he's 4b right now, 12 years ago, he'll just be 12 less than that. So he's 4b now. 12 years ago, it was 4b minus 12. So that's kind of interesting. But we haven't quite yet made use of this second statement."}, {"video_title": "Ex 3 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So he's 4b now. 12 years ago, it was 4b minus 12. So that's kind of interesting. But we haven't quite yet made use of this second statement. 12 years ago, William, so this is William 12 years ago. 12 years ago, William was seven times as old as Ben. So 12 years ago, this number is going to be seven times this number."}, {"video_title": "Ex 3 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "But we haven't quite yet made use of this second statement. 12 years ago, William, so this is William 12 years ago. 12 years ago, William was seven times as old as Ben. So 12 years ago, this number is going to be seven times this number. Or another way to think about it, take this number and multiply it by seven. Seven, and you're going to get this number. 12 years ago, Ben was, Ben's age is 1 7th of William's age."}, {"video_title": "Ex 3 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So 12 years ago, this number is going to be seven times this number. Or another way to think about it, take this number and multiply it by seven. Seven, and you're going to get this number. 12 years ago, Ben was, Ben's age is 1 7th of William's age. Or William's age is seven times Ben's age. So let's see if we can set that up as an equation. So we can have seven times this, let me write this down."}, {"video_title": "Ex 3 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "12 years ago, Ben was, Ben's age is 1 7th of William's age. Or William's age is seven times Ben's age. So let's see if we can set that up as an equation. So we can have seven times this, let me write this down. Seven times Ben's age 12 years ago. Seven times Ben's age 12 years ago, b minus 12, is going to be equal to William's age. Is going to be equal to William's age."}, {"video_title": "Ex 3 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So we can have seven times this, let me write this down. Seven times Ben's age 12 years ago. Seven times Ben's age 12 years ago, b minus 12, is going to be equal to William's age. Is going to be equal to William's age. And it seems like we've done the hard part. We've set up the equation, now we just use a little bit of our algebraic tools to solve for b. So let's do that."}, {"video_title": "Ex 3 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Is going to be equal to William's age. And it seems like we've done the hard part. We've set up the equation, now we just use a little bit of our algebraic tools to solve for b. So let's do that. So the first thing we might want to do, we could distribute this seven. So seven times b, seven times essentially negative 12. So we have seven b minus seven times 12, let's see."}, {"video_title": "Ex 3 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So let's do that. So the first thing we might want to do, we could distribute this seven. So seven times b, seven times essentially negative 12. So we have seven b minus seven times 12, let's see. Seven, it's 84. 84 is going to be equal to four b minus 12. Is going to be equal to four b minus 12."}, {"video_title": "Ex 3 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So we have seven b minus seven times 12, let's see. Seven, it's 84. 84 is going to be equal to four b minus 12. Is going to be equal to four b minus 12. This whole expression is literally seven times Ben's age 12 years ago. Now what can we do to solve this? Well we can subtract four b from both sides."}, {"video_title": "Ex 3 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Is going to be equal to four b minus 12. This whole expression is literally seven times Ben's age 12 years ago. Now what can we do to solve this? Well we can subtract four b from both sides. So let's do that. Four b. I can do two steps at the same time. Well actually let's just keep it simple."}, {"video_title": "Ex 3 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Well we can subtract four b from both sides. So let's do that. Four b. I can do two steps at the same time. Well actually let's just keep it simple. So I'm going to subtract four b from both sides. That goes away. On the right hand side I have a negative 12."}, {"video_title": "Ex 3 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Well actually let's just keep it simple. So I'm going to subtract four b from both sides. That goes away. On the right hand side I have a negative 12. On the left hand side I'm left with seven b minus four b minus three b and then I still have a minus 84. Well I want to get rid of this negative 84, this minus 84 on the left hand side. So let's add 84 to both sides."}, {"video_title": "Ex 3 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "On the right hand side I have a negative 12. On the left hand side I'm left with seven b minus four b minus three b and then I still have a minus 84. Well I want to get rid of this negative 84, this minus 84 on the left hand side. So let's add 84 to both sides. Let's add 84 to both sides. On the left hand side I'm just left with three b. And on the right hand side I have negative 12 plus 84 or 84 minus 12 which is 72."}, {"video_title": "Ex 3 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So let's add 84 to both sides. Let's add 84 to both sides. On the left hand side I'm just left with three b. And on the right hand side I have negative 12 plus 84 or 84 minus 12 which is 72. 72. Now if I want to solve for b I just have to divide both sides of that equation by three. And so I am left with b is equal to, and now we have our drum roll, 72 divided by three."}, {"video_title": "Ex 3 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "And on the right hand side I have negative 12 plus 84 or 84 minus 12 which is 72. 72. Now if I want to solve for b I just have to divide both sides of that equation by three. And so I am left with b is equal to, and now we have our drum roll, 72 divided by three. And you might be able to do that in your head. It would be 24 I believe. And you could work it out on paper if you have trouble."}, {"video_title": "Ex 3 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "And so I am left with b is equal to, and now we have our drum roll, 72 divided by three. And you might be able to do that in your head. It would be 24 I believe. And you could work it out on paper if you have trouble. Let's just do it real quick. 72, three goes into seven two times. You get a two times three is six."}, {"video_title": "Ex 3 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "And you could work it out on paper if you have trouble. Let's just do it real quick. 72, three goes into seven two times. You get a two times three is six. Subtract. You get, bring down the two. Three goes into 12 four times."}, {"video_title": "Ex 3 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "You get a two times three is six. Subtract. You get, bring down the two. Three goes into 12 four times. So b is equal to 24. Going back to our question. What is Ben's age now?"}, {"video_title": "Ex 3 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Three goes into 12 four times. So b is equal to 24. Going back to our question. What is Ben's age now? It is 24. And let's verify that this is actually the case. So they're telling us that William is four times as old as Ben."}, {"video_title": "Ex 3 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "What is Ben's age now? It is 24. And let's verify that this is actually the case. So they're telling us that William is four times as old as Ben. So what is William's current age? Well four times 24 is 96. So William is a senior, we should call him Mr. William."}, {"video_title": "Ex 3 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So they're telling us that William is four times as old as Ben. So what is William's current age? Well four times 24 is 96. So William is a senior, we should call him Mr. William. So let's see, he is 96 years old. Maybe he's Ben's grandfather or great-grandfather. Then they say 12 years ago William, well 12 years ago William was 84 years old."}, {"video_title": "Ex 3 age word problem   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So William is a senior, we should call him Mr. William. So let's see, he is 96 years old. Maybe he's Ben's grandfather or great-grandfather. Then they say 12 years ago William, well 12 years ago William was 84 years old. So he was 84 years old. They say that's seven times as old as Ben. Well 12 years ago Ben, if he's 24 now, Ben was 12."}, {"video_title": "Checking ordered pair solutions to equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "Which of the ordered pairs is a solution of the following equation? Four x minus one is equal to three y plus five. Now when we look at an ordered pair and we want to figure out whether it's a solution, we just have to remind ourselves that in these ordered pairs, the convention, the standard is, is that the first coordinate is the x coordinate and the second coordinate is the y coordinate. So they're gonna, if this is a solution, if this ordered pair is a solution, that means that if x is equal to three and y is equal to two, that that would satisfy this equation up here. So let's try that out. So we have four times x. Well, we're saying x needs to be equal to three minus one is going to be equal to three times y."}, {"video_title": "Checking ordered pair solutions to equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "So they're gonna, if this is a solution, if this ordered pair is a solution, that means that if x is equal to three and y is equal to two, that that would satisfy this equation up here. So let's try that out. So we have four times x. Well, we're saying x needs to be equal to three minus one is going to be equal to three times y. Well, if this ordered pair is a solution, then y is going to be equal to two. So three times y, y is two, plus five. Notice all I did is wherever I saw the x, I substituted it with three, wherever I saw the y, I substituted it with two."}, {"video_title": "Checking ordered pair solutions to equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "Well, we're saying x needs to be equal to three minus one is going to be equal to three times y. Well, if this ordered pair is a solution, then y is going to be equal to two. So three times y, y is two, plus five. Notice all I did is wherever I saw the x, I substituted it with three, wherever I saw the y, I substituted it with two. Now let's see if this is true. Four times three is 12 minus one. Is this really the same thing as three times two, which is six plus five?"}, {"video_title": "Checking ordered pair solutions to equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "Notice all I did is wherever I saw the x, I substituted it with three, wherever I saw the y, I substituted it with two. Now let's see if this is true. Four times three is 12 minus one. Is this really the same thing as three times two, which is six plus five? See, 12 minus one is 11. Six plus five is also 11. This is true."}, {"video_title": "Checking ordered pair solutions to equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "Is this really the same thing as three times two, which is six plus five? See, 12 minus one is 11. Six plus five is also 11. This is true. 11 equals 11. This pair, three comma two, does satisfy this equation. Now let's see whether this one does."}, {"video_title": "Checking ordered pair solutions to equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "This is true. 11 equals 11. This pair, three comma two, does satisfy this equation. Now let's see whether this one does. Two comma three. So this is saying when x is equal to two, y would be equal to three for this equation. Let's see if that's true."}, {"video_title": "Checking ordered pair solutions to equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "Now let's see whether this one does. Two comma three. So this is saying when x is equal to two, y would be equal to three for this equation. Let's see if that's true. So four times x, we're now going to see if when x is two, y can be three. So four times x, four times two, minus one is equal to three times y. Now y we're testing to see if it can be three."}, {"video_title": "Checking ordered pair solutions to equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "Let's see if that's true. So four times x, we're now going to see if when x is two, y can be three. So four times x, four times two, minus one is equal to three times y. Now y we're testing to see if it can be three. Three times three plus five. Let's see if this is true. Four times two is eight minus one."}, {"video_title": "Checking ordered pair solutions to equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "Now y we're testing to see if it can be three. Three times three plus five. Let's see if this is true. Four times two is eight minus one. Is this equal to three times three? So that's nine plus five. So is seven equal to 14?"}, {"video_title": "Checking ordered pair solutions to equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "Four times two is eight minus one. Is this equal to three times three? So that's nine plus five. So is seven equal to 14? No, clearly seven is not equal to 14. So these things are not equal to each other. So this is not a solution."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "After you cross the Trolls Bridge and you save the prince or princess, you return them back to their father, the king, and he's so excited that you returned their son or daughter to him that he wants to throw a brunch in your honor. But he has a little bit of a conundrum in throwing the brunch. He wants to figure out how many cupcakes, how many cupcakes should he order? He doesn't want to waste any, but he wants to make sure that everyone has enough to eat. And you say, well, what's the problem here? And he says, well, I know adults eat a different number of cupcakes than children eat. And I know that in my kingdom, an adult will always eat the same amount and a child will always eat the same amount."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "He doesn't want to waste any, but he wants to make sure that everyone has enough to eat. And you say, well, what's the problem here? And he says, well, I know adults eat a different number of cupcakes than children eat. And I know that in my kingdom, an adult will always eat the same amount and a child will always eat the same amount. And so you say, king, well, what information can you give me? I might be able to help you out a little bit. You're feeling very confident after this Trolls situation."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "And I know that in my kingdom, an adult will always eat the same amount and a child will always eat the same amount. And so you say, king, well, what information can you give me? I might be able to help you out a little bit. You're feeling very confident after this Trolls situation. And he says, well, I know at the last party, we had 500 adults, 500 adults, and we had 200 children, 200 children. And combined, combined, they ate 2,900 cupcakes, 2,900 cupcakes. And you say, okay, that's interesting, but I think I'll need a little bit more information."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "You're feeling very confident after this Trolls situation. And he says, well, I know at the last party, we had 500 adults, 500 adults, and we had 200 children, 200 children. And combined, combined, they ate 2,900 cupcakes, 2,900 cupcakes. And you say, okay, that's interesting, but I think I'll need a little bit more information. Have you thrown parties before then? The king says, of course I have, I like to throw parties. Well, what happened at the party before that?"}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "And you say, okay, that's interesting, but I think I'll need a little bit more information. Have you thrown parties before then? The king says, of course I have, I like to throw parties. Well, what happened at the party before that? And he says, well, there we also had 500 adults, 500 adults, and we had 300 children, 300 children. And you say, well, how many cupcakes were eaten at that party? He says, well, we know, it was 3,100, 3,100 cupcakes."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "Well, what happened at the party before that? And he says, well, there we also had 500 adults, 500 adults, and we had 300 children, 300 children. And you say, well, how many cupcakes were eaten at that party? He says, well, we know, it was 3,100, 3,100 cupcakes. And so you get a tingling feeling that a little bit of algebra might apply over here. And you say, well, let me see, what do we need to figure out? We need to figure out how many, the number of cupcakes on average that an adult will eat."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "He says, well, we know, it was 3,100, 3,100 cupcakes. And so you get a tingling feeling that a little bit of algebra might apply over here. And you say, well, let me see, what do we need to figure out? We need to figure out how many, the number of cupcakes on average that an adult will eat. So number of cupcakes, but for an adult, and we also need to figure out the number for, number of cupcakes for a child. So these are the two things, these are the two things that we need to figure out, because then we can know how many adults and children are coming to the next brunch that are being held in your honor, and get the exact right number of cupcakes. So those are the things you're trying to figure out, and we don't know what those things are."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "We need to figure out how many, the number of cupcakes on average that an adult will eat. So number of cupcakes, but for an adult, and we also need to figure out the number for, number of cupcakes for a child. So these are the two things, these are the two things that we need to figure out, because then we can know how many adults and children are coming to the next brunch that are being held in your honor, and get the exact right number of cupcakes. So those are the things you're trying to figure out, and we don't know what those things are. Let's define some variables that represent those things. So let's let, well, let's do A for adults. Let's let A equal the number of cupcakes that on average each adult eats."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "So those are the things you're trying to figure out, and we don't know what those things are. Let's define some variables that represent those things. So let's let, well, let's do A for adults. Let's let A equal the number of cupcakes that on average each adult eats. And let's do C for children. So C is the number of cupcakes for a child on average. So given that information, let's see how we can represent what the king has told us algebraically."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "Let's let A equal the number of cupcakes that on average each adult eats. And let's do C for children. So C is the number of cupcakes for a child on average. So given that information, let's see how we can represent what the king has told us algebraically. So let's think about this orange information first. How could we represent this algebraically? Well, let's think about how many cupcakes the adults ate at that party."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "So given that information, let's see how we can represent what the king has told us algebraically. So let's think about this orange information first. How could we represent this algebraically? Well, let's think about how many cupcakes the adults ate at that party. You had 500 adults, and on average, each of them ate exactly A cupcakes, so the total number of cupcakes that the adults ate were 500 times A. How many did the children eat? Well, same logic."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "Well, let's think about how many cupcakes the adults ate at that party. You had 500 adults, and on average, each of them ate exactly A cupcakes, so the total number of cupcakes that the adults ate were 500 times A. How many did the children eat? Well, same logic. You had 200 children, and they each ate C cupcakes, so 200 times C is the total number of cupcakes that the children ate. Well, how much did they eat in total? Well, it's the total number that the adults ate plus the total number that the children ate, which is 2,900 cupcakes."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "Well, same logic. You had 200 children, and they each ate C cupcakes, so 200 times C is the total number of cupcakes that the children ate. Well, how much did they eat in total? Well, it's the total number that the adults ate plus the total number that the children ate, which is 2,900 cupcakes. 2,900. So let's do that, apply that same logic to the blue party right over here, this blue information. How can we represent this algebraically?"}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "Well, it's the total number that the adults ate plus the total number that the children ate, which is 2,900 cupcakes. 2,900. So let's do that, apply that same logic to the blue party right over here, this blue information. How can we represent this algebraically? Well, once again, how many total cupcakes did the adults eat? Well, you had 500 adults, and they each ate A cupcakes, which is an unknown right now. And then what about the children, where you had 300 children, and they each ate C cupcakes?"}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "How can we represent this algebraically? Well, once again, how many total cupcakes did the adults eat? Well, you had 500 adults, and they each ate A cupcakes, which is an unknown right now. And then what about the children, where you had 300 children, and they each ate C cupcakes? And so if you add up all the cupcakes that the adults ate plus all the cupcakes that the children ate, you get to 3,100 cupcakes. So this is starting to look interesting. I have two equations."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "And then what about the children, where you had 300 children, and they each ate C cupcakes? And so if you add up all the cupcakes that the adults ate plus all the cupcakes that the children ate, you get to 3,100 cupcakes. So this is starting to look interesting. I have two equations. I have a system of two equations with two unknowns, and you know from your experience with the troll that you should be able to solve this. You could solve it graphically like you did in the past, but now you feel that there could be another tool in your toolkit, which is really just an application of the algebra that you already know. So think a little bit about how you might do this."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "I have two equations. I have a system of two equations with two unknowns, and you know from your experience with the troll that you should be able to solve this. You could solve it graphically like you did in the past, but now you feel that there could be another tool in your toolkit, which is really just an application of the algebra that you already know. So think a little bit about how you might do this. So let's rewrite this first equation right over here. So we have 500A plus 200C is equal to 2,900 cupcakes. Now, it would be good if we could get rid of this 500A somehow."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "So think a little bit about how you might do this. So let's rewrite this first equation right over here. So we have 500A plus 200C is equal to 2,900 cupcakes. Now, it would be good if we could get rid of this 500A somehow. Well, you might say, well, let me just subtract 500A. So you might say, well, I just wanna subtract 500A, but if you subtracted 500A from the left-hand side, you'd also have to subtract 500A from the right-hand side, and so the A wouldn't just disappear. It would just end up on the right-hand side, and you'd still have one equation with two unknowns, which isn't too helpful."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "Now, it would be good if we could get rid of this 500A somehow. Well, you might say, well, let me just subtract 500A. So you might say, well, I just wanna subtract 500A, but if you subtracted 500A from the left-hand side, you'd also have to subtract 500A from the right-hand side, and so the A wouldn't just disappear. It would just end up on the right-hand side, and you'd still have one equation with two unknowns, which isn't too helpful. But you see something interesting. You're like, well, this is a 500A here. What if I subtracted a 500A and this 300C?"}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "It would just end up on the right-hand side, and you'd still have one equation with two unknowns, which isn't too helpful. But you see something interesting. You're like, well, this is a 500A here. What if I subtracted a 500A and this 300C? So if I subtracted the 500A and the 300C from the left-hand side, and you're like, well, why is that useful? You're gonna have to do the same thing on the right-hand side, and then you're gonna have an A and a C on the right-hand. And you just say, hold on, hold on one second here."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "What if I subtracted a 500A and this 300C? So if I subtracted the 500A and the 300C from the left-hand side, and you're like, well, why is that useful? You're gonna have to do the same thing on the right-hand side, and then you're gonna have an A and a C on the right-hand. And you just say, hold on, hold on one second here. Hold on, I guess you're talking to yourself. Hold on one second. I'm subtracting the left-hand side of this equation, but this left-hand side is the exact same thing as this right-hand side."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "And you just say, hold on, hold on one second here. Hold on, I guess you're talking to yourself. Hold on one second. I'm subtracting the left-hand side of this equation, but this left-hand side is the exact same thing as this right-hand side. So here I could subtract 500A and 300C, and I could do 500A and subtract 300C over here, but we know that subtracting 500A and 300C, that's the exact same thing as subtracting 3100. 500, let me make it clear. This is 500A, 500A minus 300C is the exact same thing as subtracting 500A plus 300C, and we know that 500A plus 300C is exactly 3100."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "I'm subtracting the left-hand side of this equation, but this left-hand side is the exact same thing as this right-hand side. So here I could subtract 500A and 300C, and I could do 500A and subtract 300C over here, but we know that subtracting 500A and 300C, that's the exact same thing as subtracting 3100. 500, let me make it clear. This is 500A, 500A minus 300C is the exact same thing as subtracting 500A plus 300C, and we know that 500A plus 300C is exactly 3100. This is 3100. This is what the second information gave us. So instead of subtracting 500A minus 300C, we can just subtract 3100."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "This is 500A, 500A minus 300C is the exact same thing as subtracting 500A plus 300C, and we know that 500A plus 300C is exactly 3100. This is 3100. This is what the second information gave us. So instead of subtracting 500A minus 300C, we can just subtract 3100. So let me do that. This is exciting. So let me clear that out."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "So instead of subtracting 500A minus 300C, we can just subtract 3100. So let me do that. This is exciting. So let me clear that out. So let's clear that out. And so here, instead of doing this, I can subtract the exact same value, which we know is 3100. Subtract 3100."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "So let me clear that out. So let's clear that out. And so here, instead of doing this, I can subtract the exact same value, which we know is 3100. Subtract 3100. So look at it this way. It looks like we're subtracting this bottom equation from the top equation, but we're really just subtracting the same thing from both sides. This is just very basic algebra here."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "Subtract 3100. So look at it this way. It looks like we're subtracting this bottom equation from the top equation, but we're really just subtracting the same thing from both sides. This is just very basic algebra here. But if we do that, let's see what happens. So on the left-hand side, 500A minus 500A, those cancel out. 200C minus 300C, that gives us negative 100C."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "This is just very basic algebra here. But if we do that, let's see what happens. So on the left-hand side, 500A minus 500A, those cancel out. 200C minus 300C, that gives us negative 100C. And on the right-hand side, 2900 minus 3100 is negative 200. Well, now we have one equation with one unknown, and we know how to solve this. We can divide both sides by negative 100."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "200C minus 300C, that gives us negative 100C. And on the right-hand side, 2900 minus 3100 is negative 200. Well, now we have one equation with one unknown, and we know how to solve this. We can divide both sides by negative 100. Divide both sides by negative 100. These cancel out. And then over here, you end up with a positive two."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "We can divide both sides by negative 100. Divide both sides by negative 100. These cancel out. And then over here, you end up with a positive two. So C is equal to positive two. So we've solved one of the unknowns. Each child, on average, drinks two cups."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "And then over here, you end up with a positive two. So C is equal to positive two. So we've solved one of the unknowns. Each child, on average, drinks two cups. So C is equal to two. So how can we figure out what A is? Well, now we can take this information and go back into either one of these and figure out what A has to be."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "Each child, on average, drinks two cups. So C is equal to two. So how can we figure out what A is? Well, now we can take this information and go back into either one of these and figure out what A has to be. So let's go back into the orange one right over here and figure out what A has to be. So we had 500A, 500A, plus 200C, but we know what C is. C is two."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "Well, now we can take this information and go back into either one of these and figure out what A has to be. So let's go back into the orange one right over here and figure out what A has to be. So we had 500A, 500A, plus 200C, but we know what C is. C is two. So 200 times two is equal to 2,900, is equal to 2,900. And now we just have to solve for A, one equation with one unknown. So we have 500A, 500A."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "C is two. So 200 times two is equal to 2,900, is equal to 2,900. And now we just have to solve for A, one equation with one unknown. So we have 500A, 500A. 200 times two is 400, plus 400 is equal to 2,900. We can subtract 400 from both sides of this equation. Let me do that."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "So we have 500A, 500A. 200 times two is 400, plus 400 is equal to 2,900. We can subtract 400 from both sides of this equation. Let me do that. Subtracting 400. And we are left with, this cancels out. And on the left-hand side, we have 500A."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "Let me do that. Subtracting 400. And we are left with, this cancels out. And on the left-hand side, we have 500A. This is very exciting. We're in the home stretch. On the right-hand side, you have 2,500."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "And on the left-hand side, we have 500A. This is very exciting. We're in the home stretch. On the right-hand side, you have 2,500. 500A equals 2,500. We can divide both sides by 500. And we are left with 2,500 divided by 500 is just five."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "On the right-hand side, you have 2,500. 500A equals 2,500. We can divide both sides by 500. And we are left with 2,500 divided by 500 is just five. So you have A is equal to five and you're done. You have solved the King's Conundrum. Each child on average drinks two cups of water."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "And we are left with 2,500 divided by 500 is just five. So you have A is equal to five and you're done. You have solved the King's Conundrum. Each child on average drinks two cups of water. Or sorry, not cups of water. I don't know where I got that from. Each child eats two cupcakes and each adult will eat five cupcakes."}, {"video_title": "Combining like terms, but more complicated   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Now we have a very, very, very hairy expression. And once again, I'm going to see if you can simplify this. And I'll give you a little time to do it. So this one is even crazier than the last few we've looked at. We've got y's and xy's and x squared's and x's and, well, more just xy's and y squared's and on and on and on. And there will be a temptation, because you see a y here and a y here, to say, oh, maybe I can add this negative 3y plus this 4xy somehow, since I see a y and a y. But the important thing to realize here is that a y is different than an xy."}, {"video_title": "Combining like terms, but more complicated   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "So this one is even crazier than the last few we've looked at. We've got y's and xy's and x squared's and x's and, well, more just xy's and y squared's and on and on and on. And there will be a temptation, because you see a y here and a y here, to say, oh, maybe I can add this negative 3y plus this 4xy somehow, since I see a y and a y. But the important thing to realize here is that a y is different than an xy. Think about if they were numbers. If y was 3 and an x was a 2, then a y would be a 3, while an xy would have been a 6. And a y is very different than a y squared."}, {"video_title": "Combining like terms, but more complicated   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "But the important thing to realize here is that a y is different than an xy. Think about if they were numbers. If y was 3 and an x was a 2, then a y would be a 3, while an xy would have been a 6. And a y is very different than a y squared. Once again, if the y took on the value 3, then the y squared would be the value 9. So even though you see the same letter here, they aren't the same. I guess you cannot add these two or subtract these two terms."}, {"video_title": "Combining like terms, but more complicated   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And a y is very different than a y squared. Once again, if the y took on the value 3, then the y squared would be the value 9. So even though you see the same letter here, they aren't the same. I guess you cannot add these two or subtract these two terms. A y is different than a y squared is different than an xy. Now with that said, let's see if we can, if there is anything that we can simplify. So first let's think about the y terms."}, {"video_title": "Combining like terms, but more complicated   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "I guess you cannot add these two or subtract these two terms. A y is different than a y squared is different than an xy. Now with that said, let's see if we can, if there is anything that we can simplify. So first let's think about the y terms. So you have a negative 3y there. Do we have any more y terms? Yes, we do."}, {"video_title": "Combining like terms, but more complicated   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "So first let's think about the y terms. So you have a negative 3y there. Do we have any more y terms? Yes, we do. We have this 2y right over there. So I'll just write it out. I'll just reorder it."}, {"video_title": "Combining like terms, but more complicated   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Yes, we do. We have this 2y right over there. So I'll just write it out. I'll just reorder it. So we have negative 3y plus 2y. Now let's think about, I'm just going in an arbitrary order, but since our next term is an xy term, let's think about all of the xy terms. So we have plus 4xy right over here."}, {"video_title": "Combining like terms, but more complicated   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "I'll just reorder it. So we have negative 3y plus 2y. Now let's think about, I'm just going in an arbitrary order, but since our next term is an xy term, let's think about all of the xy terms. So we have plus 4xy right over here. So let me just write it down. I'm just reordering the whole expression. Plus 4xy and then I have minus 4xy right over here."}, {"video_title": "Combining like terms, but more complicated   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "So we have plus 4xy right over here. So let me just write it down. I'm just reordering the whole expression. Plus 4xy and then I have minus 4xy right over here. So minus 4xy. Then let's go to the x squared terms. I have negative 2 times x squared or minus 2x squared."}, {"video_title": "Combining like terms, but more complicated   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Plus 4xy and then I have minus 4xy right over here. So minus 4xy. Then let's go to the x squared terms. I have negative 2 times x squared or minus 2x squared. So let's look at this. So I have minus 2x squared. Do I have any other x squareds?"}, {"video_title": "Combining like terms, but more complicated   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "I have negative 2 times x squared or minus 2x squared. So let's look at this. So I have minus 2x squared. Do I have any other x squareds? Yes, I do. I have this 3x squared right over there. So plus 3x squareds."}, {"video_title": "Combining like terms, but more complicated   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Do I have any other x squareds? Yes, I do. I have this 3x squared right over there. So plus 3x squareds. And then let's see. I have an x term right over here. And that actually looks like the only x term."}, {"video_title": "Combining like terms, but more complicated   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "So plus 3x squareds. And then let's see. I have an x term right over here. And that actually looks like the only x term. So that's plus 2x. And then I only have one y squared term. I'll circle that in orange."}, {"video_title": "Combining like terms, but more complicated   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And that actually looks like the only x term. So that's plus 2x. And then I only have one y squared term. I'll circle that in orange. So plus y squared. So all I have done is I've reordered the statement and I've color coded it based on the type of term we have. And now it should be a little bit simpler."}, {"video_title": "Combining like terms, but more complicated   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "I'll circle that in orange. So plus y squared. So all I have done is I've reordered the statement and I've color coded it based on the type of term we have. And now it should be a little bit simpler. So let's try it out. If I have negative 3 of something plus 2 of that something, what do I have? Or another way to say it, if I have 2 of something and I subtract 3 of that, what am I left with?"}, {"video_title": "Combining like terms, but more complicated   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And now it should be a little bit simpler. So let's try it out. If I have negative 3 of something plus 2 of that something, what do I have? Or another way to say it, if I have 2 of something and I subtract 3 of that, what am I left with? I'm left with negative 1 of that something. So I could write negative 1y or I could just write negative y. And another way to think about it, but I like to think about it intuitively more, is what's the coefficient here?"}, {"video_title": "Combining like terms, but more complicated   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Or another way to say it, if I have 2 of something and I subtract 3 of that, what am I left with? I'm left with negative 1 of that something. So I could write negative 1y or I could just write negative y. And another way to think about it, but I like to think about it intuitively more, is what's the coefficient here? It is negative 3. What's the coefficient here? It's 2."}, {"video_title": "Combining like terms, but more complicated   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And another way to think about it, but I like to think about it intuitively more, is what's the coefficient here? It is negative 3. What's the coefficient here? It's 2. We're obviously both dealing with y terms. Not xy terms, not y squared terms, just y. And so negative 3 plus 2 is negative 1."}, {"video_title": "Combining like terms, but more complicated   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "It's 2. We're obviously both dealing with y terms. Not xy terms, not y squared terms, just y. And so negative 3 plus 2 is negative 1. Or negative 1y is the same thing as negative y. So those simplify to this right over here. Now let's look at the xy terms."}, {"video_title": "Combining like terms, but more complicated   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And so negative 3 plus 2 is negative 1. Or negative 1y is the same thing as negative y. So those simplify to this right over here. Now let's look at the xy terms. If I have 4 of this, 4 xy's, and I were to take away 4 xy's, how many xy's am I left with? Well, I'm left with no xy's. Or you could say, add the coefficients, 4 plus negative 4 gives you 0 xy's."}, {"video_title": "Combining like terms, but more complicated   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Now let's look at the xy terms. If I have 4 of this, 4 xy's, and I were to take away 4 xy's, how many xy's am I left with? Well, I'm left with no xy's. Or you could say, add the coefficients, 4 plus negative 4 gives you 0 xy's. Either way, these two cancel out. If I have 4 of something and I take away those 4 of that something, I'm left with none of them. So I'm left with no xy's."}, {"video_title": "Combining like terms, but more complicated   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Or you could say, add the coefficients, 4 plus negative 4 gives you 0 xy's. Either way, these two cancel out. If I have 4 of something and I take away those 4 of that something, I'm left with none of them. So I'm left with no xy's. And then I have right over here, I could have written 0 xy, but that seems unnecessary. Then right over here, I have my x squared terms. Negative 2 plus 3 is 1."}, {"video_title": "Combining like terms, but more complicated   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "So I'm left with no xy's. And then I have right over here, I could have written 0 xy, but that seems unnecessary. Then right over here, I have my x squared terms. Negative 2 plus 3 is 1. Or another way of saying, if I have 3 x squared's and I were to take away 2 of those x squared's, I'm left with 1 x squared. So this right over here simplifies to 1 x squared. Or I could literally just write x squared."}, {"video_title": "Combining like terms, but more complicated   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Negative 2 plus 3 is 1. Or another way of saying, if I have 3 x squared's and I were to take away 2 of those x squared's, I'm left with 1 x squared. So this right over here simplifies to 1 x squared. Or I could literally just write x squared. 1 x squared is the same thing as x squared. So plus x squared. And then these, there's nothing really left to simplify."}, {"video_title": "Combining like terms, but more complicated   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Or I could literally just write x squared. 1 x squared is the same thing as x squared. So plus x squared. And then these, there's nothing really left to simplify. So plus 2x plus y squared. And we're done. And obviously, you might have gotten an answer in some other order, but the order in which I write these terms don't matter."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "What is the equation of the line? So let's just try to visualize this. So that is my x-axis. And you don't have to draw it to do this problem, but it always helps to visualize. That is my y-axis. And the first point is (-1, 6). So (-1, 1, 2, 3, 4, 5, 6)."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "And you don't have to draw it to do this problem, but it always helps to visualize. That is my y-axis. And the first point is (-1, 6). So (-1, 1, 2, 3, 4, 5, 6). So it's this point right over there. It's (-1, 6). And the other point is (-5, 4)."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "So (-1, 1, 2, 3, 4, 5, 6). So it's this point right over there. It's (-1, 6). And the other point is (-5, 4). So 1, 2, 3, 4, 5. And then we go down 4. So 1, 2, 3, 4."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "And the other point is (-5, 4). So 1, 2, 3, 4, 5. And then we go down 4. So 1, 2, 3, 4. So it's right over there. And so the line that connects them will look something like this. The line will draw a rough approximation."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "So 1, 2, 3, 4. So it's right over there. And so the line that connects them will look something like this. The line will draw a rough approximation. I could draw a straighter line than that. I'll draw a little dotted line maybe. Easier to do dotted lines."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "The line will draw a rough approximation. I could draw a straighter line than that. I'll draw a little dotted line maybe. Easier to do dotted lines. So the line will look something like that. So let's find its equation. So a good place to start is we could find its slope."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "Easier to do dotted lines. So the line will look something like that. So let's find its equation. So a good place to start is we could find its slope. Remember, we can find equation y is equal to mx plus b. This is a slope-intercept form where m is a slope and b is a y-intercept. We can first try to solve for m. We could find the slope of this line."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "So a good place to start is we could find its slope. Remember, we can find equation y is equal to mx plus b. This is a slope-intercept form where m is a slope and b is a y-intercept. We can first try to solve for m. We could find the slope of this line. So m, or the slope, is the change in y over the change in x. Or we could view it as the y values of our end point minus the y value of our starting point over the x values of our end point minus the x values of our starting point. Let me make that clear."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "We can first try to solve for m. We could find the slope of this line. So m, or the slope, is the change in y over the change in x. Or we could view it as the y values of our end point minus the y value of our starting point over the x values of our end point minus the x values of our starting point. Let me make that clear. So this is equal to change in y over change in x, which is the same thing as rise over run, which is the same thing as the y value of your ending point minus the y value of your starting point. This is the same exact thing as change in y. And that over the x value of your ending point minus the x value of your starting point."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "Let me make that clear. So this is equal to change in y over change in x, which is the same thing as rise over run, which is the same thing as the y value of your ending point minus the y value of your starting point. This is the same exact thing as change in y. And that over the x value of your ending point minus the x value of your starting point. This is the exact same thing as change in x. And you just have to pick one of these as a starting point and one as the ending point. So let's just make this over here our starting point and make that our ending point."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "And that over the x value of your ending point minus the x value of your starting point. This is the exact same thing as change in x. And you just have to pick one of these as a starting point and one as the ending point. So let's just make this over here our starting point and make that our ending point. So what is our change in y? So our change in y, we started at y is equal to 6. We started at y is equal to 6."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "So let's just make this over here our starting point and make that our ending point. So what is our change in y? So our change in y, we started at y is equal to 6. We started at y is equal to 6. And we go down all the way to y is equal to negative 4. So this is right here. That is our change in y."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "We started at y is equal to 6. And we go down all the way to y is equal to negative 4. So this is right here. That is our change in y. You could look at the graph. You say, well, if I start at 6 and I go to negative 4, I went down 10. Or if you just want to use this formula here, it'll give you the same thing."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "That is our change in y. You could look at the graph. You say, well, if I start at 6 and I go to negative 4, I went down 10. Or if you just want to use this formula here, it'll give you the same thing. We finished at negative 4. We finished at negative 4. And from that, we want to subtract 6."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "Or if you just want to use this formula here, it'll give you the same thing. We finished at negative 4. We finished at negative 4. And from that, we want to subtract 6. This right here is y2. This is our ending y. And this is our beginning y."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "And from that, we want to subtract 6. This right here is y2. This is our ending y. And this is our beginning y. This is y1. So y2, negative 4, minus y1. Negative 6."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "And this is our beginning y. This is y1. So y2, negative 4, minus y1. Negative 6. So negative 4 minus 6. That is equal to negative 10. And all that's doing is telling us the change in y."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "Negative 6. So negative 4 minus 6. That is equal to negative 10. And all that's doing is telling us the change in y. To go from this point to that point, we had to go down. Our rise was negative. We had to go down 10."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "And all that's doing is telling us the change in y. To go from this point to that point, we had to go down. Our rise was negative. We had to go down 10. That's where the negative 10 comes from. Now we just have to find our change in x. So we can look at this graph over here."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "We had to go down 10. That's where the negative 10 comes from. Now we just have to find our change in x. So we can look at this graph over here. We started at x is equal to negative 1. And we go all the way to x is equal to 5. So we start at x is equal to negative 1."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "So we can look at this graph over here. We started at x is equal to negative 1. And we go all the way to x is equal to 5. So we start at x is equal to negative 1. And we go all the way to x is equal to 5. So it takes us 1 to get to 0, and then 5 more. So our change in x is 6."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "So we start at x is equal to negative 1. And we go all the way to x is equal to 5. So it takes us 1 to get to 0, and then 5 more. So our change in x is 6. You can look at it visually there. Or you could use this formula. Same exact idea."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "So our change in x is 6. You can look at it visually there. Or you could use this formula. Same exact idea. Our ending x value is 5. And our starting x value is negative 1. 5 minus negative 1."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "Same exact idea. Our ending x value is 5. And our starting x value is negative 1. 5 minus negative 1. 5 minus negative 1 is the same thing as 5 plus 1. So it is 6. So our slope here is negative 10 over 6, which is the exact same thing as negative 5 thirds."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "5 minus negative 1. 5 minus negative 1 is the same thing as 5 plus 1. So it is 6. So our slope here is negative 10 over 6, which is the exact same thing as negative 5 thirds. As negative 5 over 3. Divide the numerator and denominator by 2. So we now know our equation will be y is equal to negative 5 thirds, that's our slope, x plus b."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "So our slope here is negative 10 over 6, which is the exact same thing as negative 5 thirds. As negative 5 over 3. Divide the numerator and denominator by 2. So we now know our equation will be y is equal to negative 5 thirds, that's our slope, x plus b. So we still need to solve for our y-intercept to get our equation. And to do that, we can use the information that we know. Or we have several points of information."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "So we now know our equation will be y is equal to negative 5 thirds, that's our slope, x plus b. So we still need to solve for our y-intercept to get our equation. And to do that, we can use the information that we know. Or we have several points of information. But we can use the fact that the line goes through the point negative 1, 6. We could use the other point as well. But we know that when x is equal to negative 1, so y is equal to 6."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "Or we have several points of information. But we can use the fact that the line goes through the point negative 1, 6. We could use the other point as well. But we know that when x is equal to negative 1, so y is equal to 6. So y is equal to 6 when x is equal to negative 1. So negative 5 thirds times x. When x is equal to negative 1, y is equal to 6."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "But we know that when x is equal to negative 1, so y is equal to 6. So y is equal to 6 when x is equal to negative 1. So negative 5 thirds times x. When x is equal to negative 1, y is equal to 6. So we literally just substitute this x and y value back into this. And now we can solve for b. So let's see."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "When x is equal to negative 1, y is equal to 6. So we literally just substitute this x and y value back into this. And now we can solve for b. So let's see. This negative 1 times negative 5 thirds. So we get 6 is equal to positive 5 thirds plus b. And now we can subtract 5 thirds from both sides of this equation."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "So let's see. This negative 1 times negative 5 thirds. So we get 6 is equal to positive 5 thirds plus b. And now we can subtract 5 thirds from both sides of this equation. So we have subtract the left-hand side from the left-hand side. And subtract from the right-hand side. And then we get, what's 6 minus 5 thirds?"}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "And now we can subtract 5 thirds from both sides of this equation. So we have subtract the left-hand side from the left-hand side. And subtract from the right-hand side. And then we get, what's 6 minus 5 thirds? So that's going to be, let me do it over here. We could take a common denominator. So 6 is the same thing as, let me just do it over here."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "And then we get, what's 6 minus 5 thirds? So that's going to be, let me do it over here. We could take a common denominator. So 6 is the same thing as, let me just do it over here. So 6 minus 5 over 3 is the same thing as 6 is 18 over 3 minus 5 over 3. That's 6 is 18 over 3. And this is just 13 over 3."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "So 6 is the same thing as, let me just do it over here. So 6 minus 5 over 3 is the same thing as 6 is 18 over 3 minus 5 over 3. That's 6 is 18 over 3. And this is just 13 over 3. So this is 13 over 3. And then, of course, these cancel out. So we get b is equal to 13 thirds."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "And this is just 13 over 3. So this is 13 over 3. And then, of course, these cancel out. So we get b is equal to 13 thirds. So we're done. We know the slope and we know the y-intercept. The equation of our line is y is equal to negative 5 thirds x plus our y-intercept, which is 13 over 3."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "So we get b is equal to 13 thirds. So we're done. We know the slope and we know the y-intercept. The equation of our line is y is equal to negative 5 thirds x plus our y-intercept, which is 13 over 3. And we could write these as mixed numbers if it's easier to visualize. 13 over 3 is 4 and 1 third. So this y-intercept right over here, that's 0 comma 13 over 3 or 0 comma 4 and 1 third."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "The equation of our line is y is equal to negative 5 thirds x plus our y-intercept, which is 13 over 3. And we could write these as mixed numbers if it's easier to visualize. 13 over 3 is 4 and 1 third. So this y-intercept right over here, that's 0 comma 13 over 3 or 0 comma 4 and 1 third. And even with my very roughly drawn diagram, it does look like this. And this slope, negative 5 thirds, that's the same thing as negative 1 and 2 thirds. And you can see here, the slope is downward sloping."}, {"video_title": "Slope-intercept equation from two solutions example   Algebra I   Khan Academy.mp3", "Sentence": "So this y-intercept right over here, that's 0 comma 13 over 3 or 0 comma 4 and 1 third. And even with my very roughly drawn diagram, it does look like this. And this slope, negative 5 thirds, that's the same thing as negative 1 and 2 thirds. And you can see here, the slope is downward sloping. It's negative. And it's a little bit steeper than a slope of 1. It's not quite a negative 2."}, {"video_title": "Subtracting decimals example 1   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "Let's try to subtract 9.57 minus 8.09. So try to pause this video and figure this out first before we work through it together. All right, well let's just rewrite it. Let's rewrite it. And when I rewrite it, I like to line up the decimals. This one is a little intuitive. We have 8.09, just like that."}, {"video_title": "Subtracting decimals example 1   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "Let's rewrite it. And when I rewrite it, I like to line up the decimals. This one is a little intuitive. We have 8.09, just like that. And now we're ready to subtract. And we want to subtract 9 hundredths from 7 hundredths. Well, we don't have enough hundredths up here, so let's move over here."}, {"video_title": "Subtracting decimals example 1   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "We have 8.09, just like that. And now we're ready to subtract. And we want to subtract 9 hundredths from 7 hundredths. Well, we don't have enough hundredths up here, so let's move over here. Let's see if we can do some regrouping so we always have a higher number on top. So over here, we want to subtract 0 tenths from 5 tenths. We have enough tenths over here, so let's regroup."}, {"video_title": "Subtracting decimals example 1   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "Well, we don't have enough hundredths up here, so let's move over here. Let's see if we can do some regrouping so we always have a higher number on top. So over here, we want to subtract 0 tenths from 5 tenths. We have enough tenths over here, so let's regroup. So instead of 5 tenths, I'm going to have 4 tenths. And then I'm going to give that other tenth, which is the same thing as 10 hundredths over here. So this becomes 17 hundredths."}, {"video_title": "Subtracting decimals example 1   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "We have enough tenths over here, so let's regroup. So instead of 5 tenths, I'm going to have 4 tenths. And then I'm going to give that other tenth, which is the same thing as 10 hundredths over here. So this becomes 17 hundredths. 17 minus 9 is 8. 4 minus 0 is 4. And then I have 9 minus 8 is 1."}, {"video_title": "Dividing decimals with hundredths   Arithmetic operations   5th grade   Khan Academy.mp3", "Sentence": "And try pausing the video and solving it on your own before I work through it. So there's a couple of ways you could think about it. We could just write it as 30.24 divided by 0.42. But what do you do now? Well, the important realization is when you're doing a division problem like this, you will get the same answer as long as you multiply or divide both numbers by the same thing. And to understand that, we could rewrite this division as 30.42 over 0.42. We could write it really as a fraction."}, {"video_title": "Dividing decimals with hundredths   Arithmetic operations   5th grade   Khan Academy.mp3", "Sentence": "But what do you do now? Well, the important realization is when you're doing a division problem like this, you will get the same answer as long as you multiply or divide both numbers by the same thing. And to understand that, we could rewrite this division as 30.42 over 0.42. We could write it really as a fraction. And we know that when we have a fraction like this, we're not changing the value of the fraction. If we multiply the numerator and the denominator by the same quantity. And so what could we multiply this denominator by to make it a whole number?"}, {"video_title": "Dividing decimals with hundredths   Arithmetic operations   5th grade   Khan Academy.mp3", "Sentence": "We could write it really as a fraction. And we know that when we have a fraction like this, we're not changing the value of the fraction. If we multiply the numerator and the denominator by the same quantity. And so what could we multiply this denominator by to make it a whole number? Well, we could multiply it by 10 and then another 10. So we could multiply it by 100. So let's do that."}, {"video_title": "Dividing decimals with hundredths   Arithmetic operations   5th grade   Khan Academy.mp3", "Sentence": "And so what could we multiply this denominator by to make it a whole number? Well, we could multiply it by 10 and then another 10. So we could multiply it by 100. So let's do that. If we multiply the denominator by 100, in order to not change the value of this, we also need to multiply the numerator by 100. We're essentially multiplying by 100 over 100, which is just 1. So we're not changing the value of this fraction or of this division problem."}, {"video_title": "Dividing decimals with hundredths   Arithmetic operations   5th grade   Khan Academy.mp3", "Sentence": "So let's do that. If we multiply the denominator by 100, in order to not change the value of this, we also need to multiply the numerator by 100. We're essentially multiplying by 100 over 100, which is just 1. So we're not changing the value of this fraction or of this division problem. So this is going to be 30.42 times 100. Move the decimal two places to the right. Gets you 3,042."}, {"video_title": "Dividing decimals with hundredths   Arithmetic operations   5th grade   Khan Academy.mp3", "Sentence": "So we're not changing the value of this fraction or of this division problem. So this is going to be 30.42 times 100. Move the decimal two places to the right. Gets you 3,042. The decimal is now there if you care about it. And 0.42 times 100. Once again, move the decimal one, two places to the right."}, {"video_title": "Dividing decimals with hundredths   Arithmetic operations   5th grade   Khan Academy.mp3", "Sentence": "Gets you 3,042. The decimal is now there if you care about it. And 0.42 times 100. Once again, move the decimal one, two places to the right. It is now 42. So this is going to be the exact same thing as 3,042 divided by 42. So once again, we can move the decimal here two to the right."}, {"video_title": "Dividing decimals with hundredths   Arithmetic operations   5th grade   Khan Academy.mp3", "Sentence": "Once again, move the decimal one, two places to the right. It is now 42. So this is going to be the exact same thing as 3,042 divided by 42. So once again, we can move the decimal here two to the right. And if we move that two to the right, then we can move this two to the right. Or we need to move this two to the right. And so this is where now the decimal place is."}, {"video_title": "Dividing decimals with hundredths   Arithmetic operations   5th grade   Khan Academy.mp3", "Sentence": "So once again, we can move the decimal here two to the right. And if we move that two to the right, then we can move this two to the right. Or we need to move this two to the right. And so this is where now the decimal place is. You could view this as 3,024. Let me clear that. 3,024 divided by 42."}, {"video_title": "Dividing decimals with hundredths   Arithmetic operations   5th grade   Khan Academy.mp3", "Sentence": "And so this is where now the decimal place is. You could view this as 3,024. Let me clear that. 3,024 divided by 42. Let me clear that. And we already know how to tackle that already, so let's do it step by step. How many times does 42 go into 3?"}, {"video_title": "Dividing decimals with hundredths   Arithmetic operations   5th grade   Khan Academy.mp3", "Sentence": "3,024 divided by 42. Let me clear that. And we already know how to tackle that already, so let's do it step by step. How many times does 42 go into 3? Well, it doesn't go at all, so we can move on to 30. How many times does 42 go into 30? Well, it doesn't go into 30, so we can move on to 302."}, {"video_title": "Dividing decimals with hundredths   Arithmetic operations   5th grade   Khan Academy.mp3", "Sentence": "How many times does 42 go into 3? Well, it doesn't go at all, so we can move on to 30. How many times does 42 go into 30? Well, it doesn't go into 30, so we can move on to 302. How many times does 42 go into 302? And like always, this is a bit of an art when you're dividing by a two-digit or a multi-digit number, I should say. So let's think about it a little bit."}, {"video_title": "Dividing decimals with hundredths   Arithmetic operations   5th grade   Khan Academy.mp3", "Sentence": "Well, it doesn't go into 30, so we can move on to 302. How many times does 42 go into 302? And like always, this is a bit of an art when you're dividing by a two-digit or a multi-digit number, I should say. So let's think about it a little bit. So this is roughly 40. This is roughly 300. So how many times does 40 go into 300?"}, {"video_title": "Dividing decimals with hundredths   Arithmetic operations   5th grade   Khan Academy.mp3", "Sentence": "So let's think about it a little bit. So this is roughly 40. This is roughly 300. So how many times does 40 go into 300? Well, how many times does 4 go into 30? It looks like it's about 7 times, so I'm going to try out a 7, see if it works out. 7 times 2 is 14."}, {"video_title": "Dividing decimals with hundredths   Arithmetic operations   5th grade   Khan Academy.mp3", "Sentence": "So how many times does 40 go into 300? Well, how many times does 4 go into 30? It looks like it's about 7 times, so I'm going to try out a 7, see if it works out. 7 times 2 is 14. 7 times 4 is 28, plus 1 is 29. And now I can subtract to do a little bit of regrouping here. So let's see."}, {"video_title": "Dividing decimals with hundredths   Arithmetic operations   5th grade   Khan Academy.mp3", "Sentence": "7 times 2 is 14. 7 times 4 is 28, plus 1 is 29. And now I can subtract to do a little bit of regrouping here. So let's see. If I regroup, I take 100 from the 300. That becomes a 200. Then our 0 tens, now I have 10 tens."}, {"video_title": "Dividing decimals with hundredths   Arithmetic operations   5th grade   Khan Academy.mp3", "Sentence": "So let's see. If I regroup, I take 100 from the 300. That becomes a 200. Then our 0 tens, now I have 10 tens. But I'm going to need one of those 10 tens, so that's going to be 9 tens. And I'm going to give it over here. So this is going to be a 12."}, {"video_title": "Dividing decimals with hundredths   Arithmetic operations   5th grade   Khan Academy.mp3", "Sentence": "Then our 0 tens, now I have 10 tens. But I'm going to need one of those 10 tens, so that's going to be 9 tens. And I'm going to give it over here. So this is going to be a 12. 12 minus 4 is 8. 9 minus 9 is 0. 2 minus 2 is 0."}, {"video_title": "Dividing decimals with hundredths   Arithmetic operations   5th grade   Khan Academy.mp3", "Sentence": "So this is going to be a 12. 12 minus 4 is 8. 9 minus 9 is 0. 2 minus 2 is 0. So what I got left over is less than 42, so I know that 7 is the right number. I want to go as many times as possible into 302 without going over. So now let's bring down the next digit."}, {"video_title": "Dividing decimals with hundredths   Arithmetic operations   5th grade   Khan Academy.mp3", "Sentence": "2 minus 2 is 0. So what I got left over is less than 42, so I know that 7 is the right number. I want to go as many times as possible into 302 without going over. So now let's bring down the next digit. Let's bring down this 4 over here. How many times does 42 go into 84? Well, that jumps out at you."}, {"video_title": "Dividing decimals with hundredths   Arithmetic operations   5th grade   Khan Academy.mp3", "Sentence": "So now let's bring down the next digit. Let's bring down this 4 over here. How many times does 42 go into 84? Well, that jumps out at you. Or hopefully it jumps. It starts to. It goes 2 times."}, {"video_title": "Dividing decimals with hundredths   Arithmetic operations   5th grade   Khan Academy.mp3", "Sentence": "Well, that jumps out at you. Or hopefully it jumps. It starts to. It goes 2 times. 2 times 2 is 4. 2 times 4 is 8. You subtract, and we have no remainder."}, {"video_title": "Dividing decimals with hundredths   Arithmetic operations   5th grade   Khan Academy.mp3", "Sentence": "It goes 2 times. 2 times 2 is 4. 2 times 4 is 8. You subtract, and we have no remainder. So 3,042 divided by 42 is the same thing as 30.42 divided by 0.42. And it's going to be equal to 72. It's going to be equal to 72."}, {"video_title": "Dividing decimals with hundredths   Arithmetic operations   5th grade   Khan Academy.mp3", "Sentence": "You subtract, and we have no remainder. So 3,042 divided by 42 is the same thing as 30.42 divided by 0.42. And it's going to be equal to 72. It's going to be equal to 72. Actually, I didn't have to copy and paste that. I'll just write this. This is equal to 72."}, {"video_title": "Dividing decimals with hundredths   Arithmetic operations   5th grade   Khan Academy.mp3", "Sentence": "It's going to be equal to 72. Actually, I didn't have to copy and paste that. I'll just write this. This is equal to 72. This is equal to 72. 72. Just like that."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "So for example, the equation y is equal to two x minus three. This is a linear equation. Now why do we call it a linear equation? Well if you were to take the set of all of the xy pairs that satisfy this equation, and if you were to graph them on the coordinate plane, you would actually get a line. That's why it's called a linear equation. Let's actually feel good about that statement. Let's see if, let's plot some of the xy pairs that satisfy this equation, and then feel good that it does indeed generate a line."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "Well if you were to take the set of all of the xy pairs that satisfy this equation, and if you were to graph them on the coordinate plane, you would actually get a line. That's why it's called a linear equation. Let's actually feel good about that statement. Let's see if, let's plot some of the xy pairs that satisfy this equation, and then feel good that it does indeed generate a line. So I'm just gonna pick some x values that make it easy to calculate the corresponding y values. So if x is equal to zero, y is gonna be two times zero minus three, which is negative three. And on our coordinate plane here, that's, we're gonna move zero in the x direction, zero in the horizontal direction, and we're gonna go down three in the vertical direction, in the y direction."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "Let's see if, let's plot some of the xy pairs that satisfy this equation, and then feel good that it does indeed generate a line. So I'm just gonna pick some x values that make it easy to calculate the corresponding y values. So if x is equal to zero, y is gonna be two times zero minus three, which is negative three. And on our coordinate plane here, that's, we're gonna move zero in the x direction, zero in the horizontal direction, and we're gonna go down three in the vertical direction, in the y direction. So that's that point there. If x is equal to one, what is y equal to? Well two times one is two minus three is negative one."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "And on our coordinate plane here, that's, we're gonna move zero in the x direction, zero in the horizontal direction, and we're gonna go down three in the vertical direction, in the y direction. So that's that point there. If x is equal to one, what is y equal to? Well two times one is two minus three is negative one. So we move positive one in the x direction, and negative one, or down one, in the y direction. Now let's see, if x is equal to two, what is y? Two times two is four, minus three is one."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "Well two times one is two minus three is negative one. So we move positive one in the x direction, and negative one, or down one, in the y direction. Now let's see, if x is equal to two, what is y? Two times two is four, minus three is one. When x is equal to two, y is equal to one. And hopefully you're seeing now, that if I were to keep going, and I encourage you to, if you want, pause the video, try x equals three, or x equals negative one, and keep going, you will see that this is going to generate a line. And in fact, let me connect these dots, and you will see, you will see the line that I'm talking about."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "Two times two is four, minus three is one. When x is equal to two, y is equal to one. And hopefully you're seeing now, that if I were to keep going, and I encourage you to, if you want, pause the video, try x equals three, or x equals negative one, and keep going, you will see that this is going to generate a line. And in fact, let me connect these dots, and you will see, you will see the line that I'm talking about. So, let me see if I can draw, I'm gonna use the line tool here. Try to connect the dots as neatly as I can. There you go."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "And in fact, let me connect these dots, and you will see, you will see the line that I'm talking about. So, let me see if I can draw, I'm gonna use the line tool here. Try to connect the dots as neatly as I can. There you go. This line that I have just drawn, this is the graph, this is the graph of y is equal to two x minus three. So if you were to graph all of the xy pairs that satisfy this equation, you are going to get this line. And you might be saying, hey wait, wait, hold on Sal, you just tried some particular points, why don't I just get a bunch of points, how do I actually get a line?"}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "There you go. This line that I have just drawn, this is the graph, this is the graph of y is equal to two x minus three. So if you were to graph all of the xy pairs that satisfy this equation, you are going to get this line. And you might be saying, hey wait, wait, hold on Sal, you just tried some particular points, why don't I just get a bunch of points, how do I actually get a line? Well I just tried, over here, I just tried integer values of x, but you could try any value in between here, all of these, it's actually a pretty neat concept. Any value of x that you input into this, you find the corresponding value for y, it will sit on this line. So for example, for example, if we were to take x is equal to, actually let's say x is equal to negative.5."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "And you might be saying, hey wait, wait, hold on Sal, you just tried some particular points, why don't I just get a bunch of points, how do I actually get a line? Well I just tried, over here, I just tried integer values of x, but you could try any value in between here, all of these, it's actually a pretty neat concept. Any value of x that you input into this, you find the corresponding value for y, it will sit on this line. So for example, for example, if we were to take x is equal to, actually let's say x is equal to negative.5. So if x is equal to negative.5, if we look at the line, when x is equal to negative.5, it looks like, it looks like y is equal to negative four. That looks like that sits on the line. Well let's verify that."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "So for example, for example, if we were to take x is equal to, actually let's say x is equal to negative.5. So if x is equal to negative.5, if we look at the line, when x is equal to negative.5, it looks like, it looks like y is equal to negative four. That looks like that sits on the line. Well let's verify that. If x is equal to negative, I'll write that as negative 1 1 2, then what is y equal to? Let's see, two times negative 1 1 2, I'll try it out, two times negative, two times negative 1 1 2 minus three. Well this is two times negative 1 1 2 is negative one minus three is indeed negative four."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "Well let's verify that. If x is equal to negative, I'll write that as negative 1 1 2, then what is y equal to? Let's see, two times negative 1 1 2, I'll try it out, two times negative, two times negative 1 1 2 minus three. Well this is two times negative 1 1 2 is negative one minus three is indeed negative four. It is indeed negative four. So you can literally take any, for any x value that you put here and the corresponding y value, it is going to sit on the line. This point right over here represents a solution to this linear equation."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "Well this is two times negative 1 1 2 is negative one minus three is indeed negative four. It is indeed negative four. So you can literally take any, for any x value that you put here and the corresponding y value, it is going to sit on the line. This point right over here represents a solution to this linear equation. Let me do this in a color you can see. So this point represents a solution to a linear equation. This point represents a solution to a linear equation."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "This point right over here represents a solution to this linear equation. Let me do this in a color you can see. So this point represents a solution to a linear equation. This point represents a solution to a linear equation. This point is not a solution to a linear equation. So if x is equal to five, then y is not going to be equal to three. If x is going to be equal to five, you go to the line to see what the solution to the linear equation is."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "This point represents a solution to a linear equation. This point is not a solution to a linear equation. So if x is equal to five, then y is not going to be equal to three. If x is going to be equal to five, you go to the line to see what the solution to the linear equation is. If x is five, this shows us that y is going to be seven. And it's indeed, that's indeed the case. Two times five is 10 minus three is seven."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "If x is going to be equal to five, you go to the line to see what the solution to the linear equation is. If x is five, this shows us that y is going to be seven. And it's indeed, that's indeed the case. Two times five is 10 minus three is seven. The point, the point five comma seven is on, or it satisfies this linear equation. So if you take all of the xy pairs that satisfy it, you get a line. That is why it's called a linear equation."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "Two times five is 10 minus three is seven. The point, the point five comma seven is on, or it satisfies this linear equation. So if you take all of the xy pairs that satisfy it, you get a line. That is why it's called a linear equation. Now this isn't the only way that we could write a linear equation. You could write a linear equation like, let me do this in a new color. You could write a linear equation like this."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "That is why it's called a linear equation. Now this isn't the only way that we could write a linear equation. You could write a linear equation like, let me do this in a new color. You could write a linear equation like this. Four x minus three y is equal to 12. This also is a linear equation. And we can see that if we were to graph the xy pairs that satisfy this, we would once again get a line, x and y."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "You could write a linear equation like this. Four x minus three y is equal to 12. This also is a linear equation. And we can see that if we were to graph the xy pairs that satisfy this, we would once again get a line, x and y. If x is equal to zero, then this goes away. You have negative three y is equal to 12. See if negative three y equals 12, then y would be equal to negative four."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "And we can see that if we were to graph the xy pairs that satisfy this, we would once again get a line, x and y. If x is equal to zero, then this goes away. You have negative three y is equal to 12. See if negative three y equals 12, then y would be equal to negative four. Negative zero comma negative four. You can verify that. Four times zero minus three times negative four, well that's going to be equal to positive 12."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "See if negative three y equals 12, then y would be equal to negative four. Negative zero comma negative four. You can verify that. Four times zero minus three times negative four, well that's going to be equal to positive 12. And let's see, if y were to equal zero, if y were to equal zero, then this is going to be four times x is equal to 12. Well then x is equal to three. And so you have the point zero comma negative four."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "Four times zero minus three times negative four, well that's going to be equal to positive 12. And let's see, if y were to equal zero, if y were to equal zero, then this is going to be four times x is equal to 12. Well then x is equal to three. And so you have the point zero comma negative four. Zero comma negative four on this line. And you have the point three comma zero on this line. Three comma zero."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "And so you have the point zero comma negative four. Zero comma negative four on this line. And you have the point three comma zero on this line. Three comma zero. Did I do that right? Yep. So zero comma negative four, and then three comma zero."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "Three comma zero. Did I do that right? Yep. So zero comma negative four, and then three comma zero. These are both going to be on this line. Three comma zero. Is also on this line."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "So zero comma negative four, and then three comma zero. These are both going to be on this line. Three comma zero. Is also on this line. So this line is going to look something like, I'll just try to hand draw it, something like that. So once again, all of the xy pairs that satisfy this, if you were to plot them out, it forms a line. Now what are some examples of, maybe you're saying, well isn't any equation a linear equation?"}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "Is also on this line. So this line is going to look something like, I'll just try to hand draw it, something like that. So once again, all of the xy pairs that satisfy this, if you were to plot them out, it forms a line. Now what are some examples of, maybe you're saying, well isn't any equation a linear equation? And the simple answer is no. Not any equation is a linear equation. I'll give you some examples of nonlinear equations."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "Now what are some examples of, maybe you're saying, well isn't any equation a linear equation? And the simple answer is no. Not any equation is a linear equation. I'll give you some examples of nonlinear equations. So nonlinear, whoops, let me write a little bit neater than that. Nonlinear equations. Well, those could include something like y is equal to x squared."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "I'll give you some examples of nonlinear equations. So nonlinear, whoops, let me write a little bit neater than that. Nonlinear equations. Well, those could include something like y is equal to x squared. If you graph this, you'll see that this is going to be a curve. It could be something like x times y is equal to 12. This is also not going to be a line."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "Well, those could include something like y is equal to x squared. If you graph this, you'll see that this is going to be a curve. It could be something like x times y is equal to 12. This is also not going to be a line. Or it could be something like five over x plus y is equal to 10. This also is not going to be a line. So now, and at some point, I encourage you to try to graph these things."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "This is also not going to be a line. Or it could be something like five over x plus y is equal to 10. This also is not going to be a line. So now, and at some point, I encourage you to try to graph these things. These are actually quite interesting. But given that we've now seen examples of linear equations and nonlinear equations, let's see if we can come up with a definition for linear equations. One way to think about it is it's an equation that if you were to graph all the x and y pairs that satisfy this equation, you'll get a line."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "So now, and at some point, I encourage you to try to graph these things. These are actually quite interesting. But given that we've now seen examples of linear equations and nonlinear equations, let's see if we can come up with a definition for linear equations. One way to think about it is it's an equation that if you were to graph all the x and y pairs that satisfy this equation, you'll get a line. And that's actually literally where the word linear equation comes from. But another way to think about it is it's going to be an equation where every term is either going to be a constant. So for example, 12 is a constant."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "One way to think about it is it's an equation that if you were to graph all the x and y pairs that satisfy this equation, you'll get a line. And that's actually literally where the word linear equation comes from. But another way to think about it is it's going to be an equation where every term is either going to be a constant. So for example, 12 is a constant. It's not going to change based on the value of some variable. 12 is 12. Or negative three is negative three."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "So for example, 12 is a constant. It's not going to change based on the value of some variable. 12 is 12. Or negative three is negative three. So every term is either going to be a constant or it's going to be a constant times a variable raised to the first power. So this is a constant two times x to the first power. This is the variable y raised to the first power."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "Or negative three is negative three. So every term is either going to be a constant or it's going to be a constant times a variable raised to the first power. So this is a constant two times x to the first power. This is the variable y raised to the first power. You could say this is just one y. We're not dividing by x or y. We're not multiplying, or we don't have a term that has x to the second power or x to the third power or y to the fifth power."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "This is the variable y raised to the first power. You could say this is just one y. We're not dividing by x or y. We're not multiplying, or we don't have a term that has x to the second power or x to the third power or y to the fifth power. We just have y to the first power. We have x to the first power. We're not multiplying x and y together like we did over here."}, {"video_title": "Factoring polynomials common factor area model   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "And we can see the area right over here, and they broke it up. This green area is 12x to the fourth, this purple area is 6x to the third, this blue area is 15x squared. You add them all together, you get this entire rectangle, which would be the combined areas, 12x to the fourth plus 6x to the third plus 15x squared. The length of the rectangle in meters, so this is the length right over here that we're talking about, we're talking about this distance. The length of the rectangle in meters is equal to the greatest common monomial factor of 12x to the fourth, 6x to the third, and 15x squared. What is the length and width of the rectangle? I encourage you to pause the video and try to work through it on your own."}, {"video_title": "Factoring polynomials common factor area model   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "The length of the rectangle in meters, so this is the length right over here that we're talking about, we're talking about this distance. The length of the rectangle in meters is equal to the greatest common monomial factor of 12x to the fourth, 6x to the third, and 15x squared. What is the length and width of the rectangle? I encourage you to pause the video and try to work through it on your own. Well, the key realization here is that the length times the width, the length times the width, is going to be equal to this area. If the length is the greatest common monomial factor of these terms, of 12x to the fourth, 6x to the third, and 15x squared, well then we can factor that out, and then what we have left over is going to be the width. So let's figure out what is the greatest common monomial factor of these three terms."}, {"video_title": "Factoring polynomials common factor area model   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "I encourage you to pause the video and try to work through it on your own. Well, the key realization here is that the length times the width, the length times the width, is going to be equal to this area. If the length is the greatest common monomial factor of these terms, of 12x to the fourth, 6x to the third, and 15x squared, well then we can factor that out, and then what we have left over is going to be the width. So let's figure out what is the greatest common monomial factor of these three terms. The first thing we can look at is let's look at the coefficients. Let's figure out what's the greatest common factor of 12, 6, and 15. And there's a couple of ways you could do it."}, {"video_title": "Factoring polynomials common factor area model   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "So let's figure out what is the greatest common monomial factor of these three terms. The first thing we can look at is let's look at the coefficients. Let's figure out what's the greatest common factor of 12, 6, and 15. And there's a couple of ways you could do it. You could do it by looking at a prime factorization. You could say, all right, well 12 is two times six, which is two times three. That's the prime factorization of 12."}, {"video_title": "Factoring polynomials common factor area model   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "And there's a couple of ways you could do it. You could do it by looking at a prime factorization. You could say, all right, well 12 is two times six, which is two times three. That's the prime factorization of 12. Prime factorization of six is just two times three. Prime factorization of 15 is three times five. And so the greatest common factor, the largest factor that's divisible into all of them, so let's see, we can throw a three in there."}, {"video_title": "Factoring polynomials common factor area model   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "That's the prime factorization of 12. Prime factorization of six is just two times three. Prime factorization of 15 is three times five. And so the greatest common factor, the largest factor that's divisible into all of them, so let's see, we can throw a three in there. Three is divisible into all of them. And that's it, because we can't say a three and a two. A three and a two would be divisible into 12 and six, but there's no two that's divisible into 15."}, {"video_title": "Factoring polynomials common factor area model   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "And so the greatest common factor, the largest factor that's divisible into all of them, so let's see, we can throw a three in there. Three is divisible into all of them. And that's it, because we can't say a three and a two. A three and a two would be divisible into 12 and six, but there's no two that's divisible into 15. We can't say a three and a five, because five isn't divisible into 12 or six. So the greatest common factor is going to be three. Another way we could have done this is we could have said, what are the non-prime factors of each of these numbers?"}, {"video_title": "Factoring polynomials common factor area model   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "A three and a two would be divisible into 12 and six, but there's no two that's divisible into 15. We can't say a three and a five, because five isn't divisible into 12 or six. So the greatest common factor is going to be three. Another way we could have done this is we could have said, what are the non-prime factors of each of these numbers? 12, you could have said, okay, I can get 12 by saying one times 12, or two times six, or three times four. Six, you could have said, let's see, that could be one times six, or two times three. So those are the factors of six."}, {"video_title": "Factoring polynomials common factor area model   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "Another way we could have done this is we could have said, what are the non-prime factors of each of these numbers? 12, you could have said, okay, I can get 12 by saying one times 12, or two times six, or three times four. Six, you could have said, let's see, that could be one times six, or two times three. So those are the factors of six. And then 15, you could have said, well, one times 15, or three times five. And so you say the greatest common factor, well, three is the largest number that I've listed here that is common to all three of these factors. So once again, the greatest common factor of 12, six, and 15 is three."}, {"video_title": "Factoring polynomials common factor area model   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "So those are the factors of six. And then 15, you could have said, well, one times 15, or three times five. And so you say the greatest common factor, well, three is the largest number that I've listed here that is common to all three of these factors. So once again, the greatest common factor of 12, six, and 15 is three. So when we're looking at the greatest common monomial factor, the coefficient is going to be three. And then we look at these powers of x. We have x to the fourth, we have x to the, let me do this in a different color."}, {"video_title": "Factoring polynomials common factor area model   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "So once again, the greatest common factor of 12, six, and 15 is three. So when we're looking at the greatest common monomial factor, the coefficient is going to be three. And then we look at these powers of x. We have x to the fourth, we have x to the, let me do this in a different color. We have x to the fourth, x to the third, and x squared. Well, what's the largest power of x that's divisible into all of those? Well, it's going to be x squared."}, {"video_title": "Factoring polynomials common factor area model   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "We have x to the fourth, we have x to the, let me do this in a different color. We have x to the fourth, x to the third, and x squared. Well, what's the largest power of x that's divisible into all of those? Well, it's going to be x squared. X squared is divisible into x to the fourth and x to the third, and of course, x squared itself. So the greatest common monomial factor is three x squared. This length right over here, this is three x squared."}, {"video_title": "Factoring polynomials common factor area model   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "Well, it's going to be x squared. X squared is divisible into x to the fourth and x to the third, and of course, x squared itself. So the greatest common monomial factor is three x squared. This length right over here, this is three x squared. So if this is three x squared, we can then figure out what the width is. So what's, if we were to divide 12 x to the fourth by three x squared, what do we get? Well, 12 divided by three is four, and x to the fourth divided by x squared is x squared."}, {"video_title": "Factoring polynomials common factor area model   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "This length right over here, this is three x squared. So if this is three x squared, we can then figure out what the width is. So what's, if we were to divide 12 x to the fourth by three x squared, what do we get? Well, 12 divided by three is four, and x to the fourth divided by x squared is x squared. Notice, three x squared times four x squared is 12 x to the fourth. And then we move over to this purple section. If we take six x to the third divided by three x squared, six divided by three is two, and then x to the third divided by x squared is just going to be x."}, {"video_title": "Factoring polynomials common factor area model   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "Well, 12 divided by three is four, and x to the fourth divided by x squared is x squared. Notice, three x squared times four x squared is 12 x to the fourth. And then we move over to this purple section. If we take six x to the third divided by three x squared, six divided by three is two, and then x to the third divided by x squared is just going to be x. And then last but not least, we have 15 divided by three is going to be five. X squared divided by x squared is just one, so it's just going to be five. So the width is going to be four x squared plus two x plus five."}, {"video_title": "Factoring polynomials common factor area model   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "If we take six x to the third divided by three x squared, six divided by three is two, and then x to the third divided by x squared is just going to be x. And then last but not least, we have 15 divided by three is going to be five. X squared divided by x squared is just one, so it's just going to be five. So the width is going to be four x squared plus two x plus five. So once again, the length, we figured that out, it was the greatest common monomial factor of these terms. It's three x squared, and the width is four x squared plus two x plus five. And one way to think about it is we just factored, we just factored this expression over here."}, {"video_title": "Factoring polynomials common factor area model   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "So the width is going to be four x squared plus two x plus five. So once again, the length, we figured that out, it was the greatest common monomial factor of these terms. It's three x squared, and the width is four x squared plus two x plus five. And one way to think about it is we just factored, we just factored this expression over here. We could write, we could write that, excuse me, I wanna see the original thing. We could write that three x squared times four x squared plus two x plus five, which is the entire width, well that's going to be equal to the area. That's going to be equal to our original expression, 12x to the fourth power plus six x to the third plus 15x squared."}, {"video_title": "Similarity example where same side plays different roles   Similarity   Geometry   Khan Academy.mp3", "Sentence": "Maybe we can establish similarity between some of the triangles. There's actually three different triangles that I can see here. This triangle, this triangle, and this larger triangle. If we can establish some similarity here, maybe we can use ratios between sides somehow to figure out what BC is. When you look at it, you have a right angle right over here. In triangle BDC, you have one right angle. In triangle ABC, you have another right angle."}, {"video_title": "Similarity example where same side plays different roles   Similarity   Geometry   Khan Academy.mp3", "Sentence": "If we can establish some similarity here, maybe we can use ratios between sides somehow to figure out what BC is. When you look at it, you have a right angle right over here. In triangle BDC, you have one right angle. In triangle ABC, you have another right angle. If we can show that they have another angle or another corresponding set of angles that are congruent to each other, then we can show that they're similar. Actually, both of those triangles, both BDC and ABC, both share this angle right over here. If they share that angle, then they definitely share two angles."}, {"video_title": "Similarity example where same side plays different roles   Similarity   Geometry   Khan Academy.mp3", "Sentence": "In triangle ABC, you have another right angle. If we can show that they have another angle or another corresponding set of angles that are congruent to each other, then we can show that they're similar. Actually, both of those triangles, both BDC and ABC, both share this angle right over here. If they share that angle, then they definitely share two angles. They both share that angle right over there. Let me do that in a different color just to make it different than those right angles. They both share that angle there."}, {"video_title": "Similarity example where same side plays different roles   Similarity   Geometry   Khan Academy.mp3", "Sentence": "If they share that angle, then they definitely share two angles. They both share that angle right over there. Let me do that in a different color just to make it different than those right angles. They both share that angle there. We know that two triangles that have at least two of their angles, that have at least two congruent angles, they are going to be similar triangles. We know that triangle, I'll write this, triangle ABC, we went from the unlabeled angle to the yellow right angle to the orange angle. Let me write it this way."}, {"video_title": "Similarity example where same side plays different roles   Similarity   Geometry   Khan Academy.mp3", "Sentence": "They both share that angle there. We know that two triangles that have at least two of their angles, that have at least two congruent angles, they are going to be similar triangles. We know that triangle, I'll write this, triangle ABC, we went from the unlabeled angle to the yellow right angle to the orange angle. Let me write it this way. We went from the unlabeled angle right over here to the orange angle, or sorry, to the yellow angle, I'm having trouble with colors, to the orange angle, ABC. We want to do this very carefully here because the same points or the same vertices might not play the same role in both triangles. We want to make sure we're getting the similarity right."}, {"video_title": "Similarity example where same side plays different roles   Similarity   Geometry   Khan Academy.mp3", "Sentence": "Let me write it this way. We went from the unlabeled angle right over here to the orange angle, or sorry, to the yellow angle, I'm having trouble with colors, to the orange angle, ABC. We want to do this very carefully here because the same points or the same vertices might not play the same role in both triangles. We want to make sure we're getting the similarity right. White vertex to the 90 degree angle vertex to the orange vertex, that is going to be similar to triangle. Which is the one that is neither right angle, so we're looking at the smaller triangle right over here, which is the one that is neither a right angle or the orange angle? It's going to be vertex B. Vertex B had the right angle when you think about the larger triangle, but we haven't thought about just that little angle right over there."}, {"video_title": "Similarity example where same side plays different roles   Similarity   Geometry   Khan Academy.mp3", "Sentence": "We want to make sure we're getting the similarity right. White vertex to the 90 degree angle vertex to the orange vertex, that is going to be similar to triangle. Which is the one that is neither right angle, so we're looking at the smaller triangle right over here, which is the one that is neither a right angle or the orange angle? It's going to be vertex B. Vertex B had the right angle when you think about the larger triangle, but we haven't thought about just that little angle right over there. We're going to start at vertex B, then we're going to go to the right angle. The right angle is vertex D, and then we go to vertex C, which is in orange. We have shown that they are similar."}, {"video_title": "Similarity example where same side plays different roles   Similarity   Geometry   Khan Academy.mp3", "Sentence": "It's going to be vertex B. Vertex B had the right angle when you think about the larger triangle, but we haven't thought about just that little angle right over there. We're going to start at vertex B, then we're going to go to the right angle. The right angle is vertex D, and then we go to vertex C, which is in orange. We have shown that they are similar. Now that we know that they are similar, we can attempt to take ratios between the sides. Let's think about it. We know what the length of AC is."}, {"video_title": "Similarity example where same side plays different roles   Similarity   Geometry   Khan Academy.mp3", "Sentence": "We have shown that they are similar. Now that we know that they are similar, we can attempt to take ratios between the sides. Let's think about it. We know what the length of AC is. AC is going to be equal to 8, 6 plus 2. We know that AC, what's the corresponding side on this triangle right over here? You can literally look at the letters."}, {"video_title": "Similarity example where same side plays different roles   Similarity   Geometry   Khan Academy.mp3", "Sentence": "We know what the length of AC is. AC is going to be equal to 8, 6 plus 2. We know that AC, what's the corresponding side on this triangle right over here? You can literally look at the letters. A and C is going to correspond to BC. The first and the third, first and the third. AC is going to correspond to BC."}, {"video_title": "Similarity example where same side plays different roles   Similarity   Geometry   Khan Academy.mp3", "Sentence": "You can literally look at the letters. A and C is going to correspond to BC. The first and the third, first and the third. AC is going to correspond to BC. This is interesting because we're already involving BC. What is going to correspond to, and then if we look at BC on the larger triangle, so if we look at BC on the larger triangle, BC is going to correspond to what on the smaller triangle? It's going to correspond to DC."}, {"video_title": "Similarity example where same side plays different roles   Similarity   Geometry   Khan Academy.mp3", "Sentence": "AC is going to correspond to BC. This is interesting because we're already involving BC. What is going to correspond to, and then if we look at BC on the larger triangle, so if we look at BC on the larger triangle, BC is going to correspond to what on the smaller triangle? It's going to correspond to DC. It's good because we know what AC is, and we know what DC is, and so we can solve for BC. I want to take one more step to show you what we just did here because BC is playing two different roles. On this first statement right over here, we're thinking of BC."}, {"video_title": "Similarity example where same side plays different roles   Similarity   Geometry   Khan Academy.mp3", "Sentence": "It's going to correspond to DC. It's good because we know what AC is, and we know what DC is, and so we can solve for BC. I want to take one more step to show you what we just did here because BC is playing two different roles. On this first statement right over here, we're thinking of BC. BC on our smaller triangle corresponds to AC on our larger triangle. Then in the second statement, BC on our larger triangle corresponds to DC on our smaller triangle. In both of these cases, these are our larger triangles, and then this is from the smaller triangle right over here, corresponding sides."}, {"video_title": "Similarity example where same side plays different roles   Similarity   Geometry   Khan Academy.mp3", "Sentence": "On this first statement right over here, we're thinking of BC. BC on our smaller triangle corresponds to AC on our larger triangle. Then in the second statement, BC on our larger triangle corresponds to DC on our smaller triangle. In both of these cases, these are our larger triangles, and then this is from the smaller triangle right over here, corresponding sides. This is a cool problem because BC plays two different roles in both triangles. Now we have enough information to solve for BC. We know that AC is equal to 9."}, {"video_title": "Similarity example where same side plays different roles   Similarity   Geometry   Khan Academy.mp3", "Sentence": "In both of these cases, these are our larger triangles, and then this is from the smaller triangle right over here, corresponding sides. This is a cool problem because BC plays two different roles in both triangles. Now we have enough information to solve for BC. We know that AC is equal to 9. We know that AC is equal to 8. 6 plus 2 is 8. We know that DC is equal to 2."}, {"video_title": "Similarity example where same side plays different roles   Similarity   Geometry   Khan Academy.mp3", "Sentence": "We know that AC is equal to 9. We know that AC is equal to 8. 6 plus 2 is 8. We know that DC is equal to 2. That's given. Now we can cross multiply. 8 times 2 is 16, is equal to BC times BC, is equal to BC squared."}, {"video_title": "Similarity example where same side plays different roles   Similarity   Geometry   Khan Academy.mp3", "Sentence": "We know that DC is equal to 2. That's given. Now we can cross multiply. 8 times 2 is 16, is equal to BC times BC, is equal to BC squared. BC is going to be equal to the principal root of 16, which is 4. BC is equal to 4. BC is equal to 4, and we're done."}, {"video_title": "Similarity example where same side plays different roles   Similarity   Geometry   Khan Academy.mp3", "Sentence": "8 times 2 is 16, is equal to BC times BC, is equal to BC squared. BC is going to be equal to the principal root of 16, which is 4. BC is equal to 4. BC is equal to 4, and we're done. The hardest part about this problem is just realizing that BC plays two different roles and just keeping your head straight on those two different roles. Just to make it clear, let me actually draw these two triangles separately. If I drew ABC separately, it would look like this."}, {"video_title": "Similarity example where same side plays different roles   Similarity   Geometry   Khan Academy.mp3", "Sentence": "BC is equal to 4, and we're done. The hardest part about this problem is just realizing that BC plays two different roles and just keeping your head straight on those two different roles. Just to make it clear, let me actually draw these two triangles separately. If I drew ABC separately, it would look like this. It would look like this. This is my triangle ABC. Then this is a right angle."}, {"video_title": "Similarity example where same side plays different roles   Similarity   Geometry   Khan Academy.mp3", "Sentence": "If I drew ABC separately, it would look like this. It would look like this. This is my triangle ABC. Then this is a right angle. This is our orange angle. We know that the length of this side right over here is 8, and we know that the length of this side when we figure it out through this problem is 4. Then if we wanted to draw BDC, we would draw it like this."}, {"video_title": "Similarity example where same side plays different roles   Similarity   Geometry   Khan Academy.mp3", "Sentence": "Then this is a right angle. This is our orange angle. We know that the length of this side right over here is 8, and we know that the length of this side when we figure it out through this problem is 4. Then if we wanted to draw BDC, we would draw it like this. BDC looks like this. This is BDC. That's a little bit easier to visualize because this is our right angle."}, {"video_title": "Similarity example where same side plays different roles   Similarity   Geometry   Khan Academy.mp3", "Sentence": "Then if we wanted to draw BDC, we would draw it like this. BDC looks like this. This is BDC. That's a little bit easier to visualize because this is our right angle. This is our orange angle. This is 4, and this right over here is 2. I did it this way to show you that you kind of have to flip this triangle over and rotate it just to have kind of a similar orientation."}, {"video_title": "Similarity example where same side plays different roles   Similarity   Geometry   Khan Academy.mp3", "Sentence": "That's a little bit easier to visualize because this is our right angle. This is our orange angle. This is 4, and this right over here is 2. I did it this way to show you that you kind of have to flip this triangle over and rotate it just to have kind of a similar orientation. Then it might make it look a little bit clearer. If you found this part confusing, I encourage you to try to flip and rotate BDC in such a way that it seems to look a lot like ABC. Then this ratio should hopefully make a lot more sense."}, {"video_title": "How to evaluate expressions with two variables   6th grade   Khan Academy.mp3", "Sentence": "And I want to evaluate this expression when a is equal to 7 and b is equal to 2. And I encourage you to pause this and try this on your own. Well, wherever we see the a, we would just replace it with the 7. And wherever we see the b, we'd replace it with the 2. So when a equals 7 and b equals 2, this expression will be 7 plus 2, which, of course, is equal to 9. So this expression would be equal to 9 in this circumstance. Let's do a slightly more complicated one."}, {"video_title": "How to evaluate expressions with two variables   6th grade   Khan Academy.mp3", "Sentence": "And wherever we see the b, we'd replace it with the 2. So when a equals 7 and b equals 2, this expression will be 7 plus 2, which, of course, is equal to 9. So this expression would be equal to 9 in this circumstance. Let's do a slightly more complicated one. Let's say we have the expression x times y minus y plus x. Actually, let's make it plus 3x, or another way of saying it, plus 3 times x. So let's evaluate this when x is equal to 3 and y is equal to 2."}, {"video_title": "How to evaluate expressions with two variables   6th grade   Khan Academy.mp3", "Sentence": "Let's do a slightly more complicated one. Let's say we have the expression x times y minus y plus x. Actually, let's make it plus 3x, or another way of saying it, plus 3 times x. So let's evaluate this when x is equal to 3 and y is equal to 2. And once again, I encourage you to pause this video and try this on your own. Well, everywhere we see an x, let's replace it with a 3. Every place we see a y, let's replace it with a 2."}, {"video_title": "How to evaluate expressions with two variables   6th grade   Khan Academy.mp3", "Sentence": "So let's evaluate this when x is equal to 3 and y is equal to 2. And once again, I encourage you to pause this video and try this on your own. Well, everywhere we see an x, let's replace it with a 3. Every place we see a y, let's replace it with a 2. So this is going to be equal to 3 times y. And y is 2 in this case. 3 times 2 minus 2 plus this 3 times x."}, {"video_title": "How to evaluate expressions with two variables   6th grade   Khan Academy.mp3", "Sentence": "Every place we see a y, let's replace it with a 2. So this is going to be equal to 3 times y. And y is 2 in this case. 3 times 2 minus 2 plus this 3 times x. But x is also now equal to 3. So what is this going to be equal to? Well, this is going to be equal to 3 times 2 is 6."}, {"video_title": "How to evaluate expressions with two variables   6th grade   Khan Academy.mp3", "Sentence": "3 times 2 minus 2 plus this 3 times x. But x is also now equal to 3. So what is this going to be equal to? Well, this is going to be equal to 3 times 2 is 6. This 3 times 3 is 9. So it simplifies to 6 minus 2, which is 4, plus 9, which is equal to 13. So in this case, it is equal to 13."}, {"video_title": "Constructing and solving a two-step inequality   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "However, both farms made up for some of this shortfall by purchasing equal quantities of apples from farms in neighboring states. What can you say about the number of apples available at each farm? Does one farm have more than the other, or do they have the same amount? How do I know? Let's define some variables here. Let's let m be equal to number of apples at Maple Farms. And who's the other guy?"}, {"video_title": "Constructing and solving a two-step inequality   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "How do I know? Let's define some variables here. Let's let m be equal to number of apples at Maple Farms. And who's the other guy? River Orchards. So let's r be equal to the number of apples at River Orchards. So this first sentence, they say, let me do this in a different color."}, {"video_title": "Constructing and solving a two-step inequality   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "And who's the other guy? River Orchards. So let's r be equal to the number of apples at River Orchards. So this first sentence, they say, let me do this in a different color. They say, for the past few years, Old Maple Farms has grown about one thousand more apples than their chief rival in the region, River Orchards. So we could say, m is approximately river plus one thousand. Or, since we don't know the exact amount, it says it's about a thousand more."}, {"video_title": "Constructing and solving a two-step inequality   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So this first sentence, they say, let me do this in a different color. They say, for the past few years, Old Maple Farms has grown about one thousand more apples than their chief rival in the region, River Orchards. So we could say, m is approximately river plus one thousand. Or, since we don't know the exact amount, it says it's about a thousand more. So we don't know it's exactly a thousand more. We can just say that in a normal year, Old Maple Farms, which we denote by m, has a larger amount of apples than River Orchard. So in a normal year, m is greater than r. It has about a thousand more apples than Old Maple Farms."}, {"video_title": "Constructing and solving a two-step inequality   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Or, since we don't know the exact amount, it says it's about a thousand more. So we don't know it's exactly a thousand more. We can just say that in a normal year, Old Maple Farms, which we denote by m, has a larger amount of apples than River Orchard. So in a normal year, m is greater than r. It has about a thousand more apples than Old Maple Farms. Now, they say, due to cold weather this year, so let's talk about this year now, the harvest at both farms were down about a third. So this isn't a normal year. Let's talk about what's going to happen this year."}, {"video_title": "Constructing and solving a two-step inequality   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So in a normal year, m is greater than r. It has about a thousand more apples than Old Maple Farms. Now, they say, due to cold weather this year, so let's talk about this year now, the harvest at both farms were down about a third. So this isn't a normal year. Let's talk about what's going to happen this year. In this year, each of these characters are going to be down by a third. Now, if I go down by a third, that's the same thing as being two thirds of what I was before. Let me do an example."}, {"video_title": "Constructing and solving a two-step inequality   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Let's talk about what's going to happen this year. In this year, each of these characters are going to be down by a third. Now, if I go down by a third, that's the same thing as being two thirds of what I was before. Let me do an example. If I'm at x and I take away one third x, I'm left with two thirds x. So going down by a third is the same thing as multiplying the quantity by two thirds. So if we multiply each of these quantities by two thirds, we can still hold this inequality, we're doing the same thing to both sides of this inequality, and we're multiplying by a positive number."}, {"video_title": "Constructing and solving a two-step inequality   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Let me do an example. If I'm at x and I take away one third x, I'm left with two thirds x. So going down by a third is the same thing as multiplying the quantity by two thirds. So if we multiply each of these quantities by two thirds, we can still hold this inequality, we're doing the same thing to both sides of this inequality, and we're multiplying by a positive number. If we were multiplying by a negative number, we would have to swap the inequality. So we can multiply both sides of this by two thirds. So two thirds of m is still going to be greater than two thirds of r. And you could even draw that in a number line if you like."}, {"video_title": "Constructing and solving a two-step inequality   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So if we multiply each of these quantities by two thirds, we can still hold this inequality, we're doing the same thing to both sides of this inequality, and we're multiplying by a positive number. If we were multiplying by a negative number, we would have to swap the inequality. So we can multiply both sides of this by two thirds. So two thirds of m is still going to be greater than two thirds of r. And you could even draw that in a number line if you like. Let's do this in a number line. This all might be a little intuitive for you, and if it is, I apologize, but if it's not, it never hurts. So that's zero on our number line."}, {"video_title": "Constructing and solving a two-step inequality   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So two thirds of m is still going to be greater than two thirds of r. And you could even draw that in a number line if you like. Let's do this in a number line. This all might be a little intuitive for you, and if it is, I apologize, but if it's not, it never hurts. So that's zero on our number line. So in a normal year, m has a thousand more than r. So in a normal year, m might be over here, and maybe r is over here. Let's say r is over there. Now, if we take two thirds of m, that's going to stick us someplace around, oh, I don't know, two thirds is right about there."}, {"video_title": "Constructing and solving a two-step inequality   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So that's zero on our number line. So in a normal year, m has a thousand more than r. So in a normal year, m might be over here, and maybe r is over here. Let's say r is over there. Now, if we take two thirds of m, that's going to stick us someplace around, oh, I don't know, two thirds is right about there. So that is m, this is, let me write this, this is two thirds m. And what's two thirds of r going to be? Well, if you take two thirds of this, you get to right about there. That is two thirds r. So you can see two thirds r is still less than two thirds m. Two thirds r is still less than two thirds m, or two thirds m is greater than two thirds r. Now, they say both farms made up for some of the shortfall by purchasing equal quantities of apples from farms in neighboring states."}, {"video_title": "Constructing and solving a two-step inequality   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Now, if we take two thirds of m, that's going to stick us someplace around, oh, I don't know, two thirds is right about there. So that is m, this is, let me write this, this is two thirds m. And what's two thirds of r going to be? Well, if you take two thirds of this, you get to right about there. That is two thirds r. So you can see two thirds r is still less than two thirds m. Two thirds r is still less than two thirds m, or two thirds m is greater than two thirds r. Now, they say both farms made up for some of the shortfall by purchasing equal quantities of apples from farms in neighboring states. So let's let a be equal to the quantity of apples both purchased. So they're telling us that they both purchased the same amount. So we could add a to both sides of this equation, it will not change the inequality."}, {"video_title": "Constructing and solving a two-step inequality   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "That is two thirds r. So you can see two thirds r is still less than two thirds m. Two thirds r is still less than two thirds m, or two thirds m is greater than two thirds r. Now, they say both farms made up for some of the shortfall by purchasing equal quantities of apples from farms in neighboring states. So let's let a be equal to the quantity of apples both purchased. So they're telling us that they both purchased the same amount. So we could add a to both sides of this equation, it will not change the inequality. As long as you add or subtract the same value to both sides, it will not change the inequality. So if you add a to both sides, you have a plus two thirds m is greater than two thirds r plus a. This is the amount that old maple farms has after purchasing the apples, and this is the amount that, what's it called, river orchards has."}, {"video_title": "Constructing and solving a two-step inequality   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So we could add a to both sides of this equation, it will not change the inequality. As long as you add or subtract the same value to both sides, it will not change the inequality. So if you add a to both sides, you have a plus two thirds m is greater than two thirds r plus a. This is the amount that old maple farms has after purchasing the apples, and this is the amount that, what's it called, river orchards has. So after everything is said and done, old maple farms still has more apples. And you can see that here. Maple farms, normal year, this year they only had two thirds of the production, but then they purchased a apple."}, {"video_title": "Constructing and solving a two-step inequality   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "This is the amount that old maple farms has after purchasing the apples, and this is the amount that, what's it called, river orchards has. So after everything is said and done, old maple farms still has more apples. And you can see that here. Maple farms, normal year, this year they only had two thirds of the production, but then they purchased a apple. So let's say a is about, let's say that a is, I don't know, that many apples. So they got back to their normal amount. So let's say they got back to their normal amount."}, {"video_title": "Constructing and solving a two-step inequality   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Maple farms, normal year, this year they only had two thirds of the production, but then they purchased a apple. So let's say a is about, let's say that a is, I don't know, that many apples. So they got back to their normal amount. So let's say they got back to their normal amount. So that's how many apples they purchased, so he got back to m. Now if r, if river orchards also purchased a apples, that same distance, a, if you go along here, gets you to right about over there. So once again, this is, let me do it a little bit different because I don't want it overlapping. So let me do it like this."}, {"video_title": "Constructing and solving a two-step inequality   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So let's say they got back to their normal amount. So that's how many apples they purchased, so he got back to m. Now if r, if river orchards also purchased a apples, that same distance, a, if you go along here, gets you to right about over there. So once again, this is, let me do it a little bit different because I don't want it overlapping. So let me do it like this. So let's say this guy, m, I keep forgetting their names, old maple farms purchases a apples. Gets them that far, so that's a apples. But river orchards also purchases a apples."}, {"video_title": "Constructing and solving a two-step inequality   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So let me do it like this. So let's say this guy, m, I keep forgetting their names, old maple farms purchases a apples. Gets them that far, so that's a apples. But river orchards also purchases a apples. So let's add that same amount. I'm just going to copy and paste it so it's the exact same amount. Copy and paste."}, {"video_title": "Constructing and solving a two-step inequality   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "But river orchards also purchases a apples. So let's add that same amount. I'm just going to copy and paste it so it's the exact same amount. Copy and paste. So old river orchards also purchases a. So it also purchases that same amount. So when all is said and done, river orchards is going to have this many apples in the year that they had less production but they went and purchased it."}, {"video_title": "Constructing and solving a two-step inequality   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Copy and paste. So old river orchards also purchases a. So it also purchases that same amount. So when all is said and done, river orchards is going to have this many apples in the year that they had less production but they went and purchased it. So this right here is, this value right here is 2 thirds r plus a. That's what river orchards has and then old maple farms has this value right here which is 2 thirds m plus a. Everything said and done, old maple farms still has more apples."}, {"video_title": "How to solve one-step equations   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "All it's saying is something plus 7 is equal to 10, and you might be able to figure it out in your head, but if you want to do it a little bit more systematically, you're like, well, all I want on the left-hand side is an x. Well, if all I want on the left-hand side is an x, I'd want to get rid of the 7. I want to subtract 7 from the left-hand side, but if I want to maintain an equality here, whatever I do to the left-hand side, I also have to do to the right-hand side, going back to our scales. That's so that we can keep our scale balanced. So we can say that the left is still equal to the right, and so what we're going to be left with is x, and then the 7s cancel out, is equal to 10 minus 7 is equal to 3. So the unknown is 3, and you can verify it. 3 plus 7 is indeed equal to 10."}, {"video_title": "How to solve one-step equations   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "That's so that we can keep our scale balanced. So we can say that the left is still equal to the right, and so what we're going to be left with is x, and then the 7s cancel out, is equal to 10 minus 7 is equal to 3. So the unknown is 3, and you can verify it. 3 plus 7 is indeed equal to 10. Let's try one more. Let's say we have a minus 5 is equal to negative 2. So this is a little bit more interesting since we have all of these negative numbers here, but we can use the exact same logic."}, {"video_title": "How to solve one-step equations   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "3 plus 7 is indeed equal to 10. Let's try one more. Let's say we have a minus 5 is equal to negative 2. So this is a little bit more interesting since we have all of these negative numbers here, but we can use the exact same logic. We just want an a over here on the left-hand side, so we have to get rid of this negative 5 somehow. Well, the best way of getting rid of a negative 5 is to add 5 to it, so I'll do that. So I will add 5 to the left-hand side, but if I want the left-hand side to stay equal to the right-hand side, whatever I do to the left, I have to do to the right, so I'm going to have to add 5 on the right-hand side as well."}, {"video_title": "How to solve one-step equations   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So this is a little bit more interesting since we have all of these negative numbers here, but we can use the exact same logic. We just want an a over here on the left-hand side, so we have to get rid of this negative 5 somehow. Well, the best way of getting rid of a negative 5 is to add 5 to it, so I'll do that. So I will add 5 to the left-hand side, but if I want the left-hand side to stay equal to the right-hand side, whatever I do to the left, I have to do to the right, so I'm going to have to add 5 on the right-hand side as well. So on the left-hand side, I'm left with a, and then the negative 5 and the positive 5 cancel out, and on the right-hand side, and they're going to stay equal because I did the same thing to both sides, we have negative 2 plus 5, which is equal to 3. So a is equal to 3. Once again, you can verify it."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "But in these, it won't be a one-step elimination. We're going to have to massage the equations a little bit in order to prepare them for elimination. So let's say that we have an equation 5x minus 10y is equal to 15. And we have another equation. 3x minus 2y is equal to 3. And I said we want to do this using elimination. Once again, we could use substitution."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "And we have another equation. 3x minus 2y is equal to 3. And I said we want to do this using elimination. Once again, we could use substitution. We could graph both of these lines and figure out where they intersect. But we're going to use elimination. But the first thing you might say, hey, Sal, with elimination, you are subtracting the left-hand side of one equation from another or adding the two and then adding the two right-hand sides."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "Once again, we could use substitution. We could graph both of these lines and figure out where they intersect. But we're going to use elimination. But the first thing you might say, hey, Sal, with elimination, you are subtracting the left-hand side of one equation from another or adding the two and then adding the two right-hand sides. And I could do that because it's essentially adding the same thing to both sides of the equation. But here, it's not obvious that that would be of any help. If we added these two left-hand sides, you would get 8x minus 12y."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "But the first thing you might say, hey, Sal, with elimination, you are subtracting the left-hand side of one equation from another or adding the two and then adding the two right-hand sides. And I could do that because it's essentially adding the same thing to both sides of the equation. But here, it's not obvious that that would be of any help. If we added these two left-hand sides, you would get 8x minus 12y. That wouldn't eliminate any variables. And on the right-hand side, you would just be left with a number. And if you subtracted, that wouldn't eliminate any variables."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "If we added these two left-hand sides, you would get 8x minus 12y. That wouldn't eliminate any variables. And on the right-hand side, you would just be left with a number. And if you subtracted, that wouldn't eliminate any variables. So how is elimination going to help here? And the answer is we can multiply both of these equations in such a way that maybe we can get one of these terms to cancel out with one of the others. So let's say we want, and you could really pick which term you want to cancel out."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "And if you subtracted, that wouldn't eliminate any variables. So how is elimination going to help here? And the answer is we can multiply both of these equations in such a way that maybe we can get one of these terms to cancel out with one of the others. So let's say we want, and you could really pick which term you want to cancel out. Let's say we want to cancel out the y terms. So I'll just rewrite this 5x minus 10y here. 5x minus 10y is equal to 15."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "So let's say we want, and you could really pick which term you want to cancel out. Let's say we want to cancel out the y terms. So I'll just rewrite this 5x minus 10y here. 5x minus 10y is equal to 15. Now, is there anything that I can multiply this green equation by so that this negative 2y term becomes a term that will cancel out with the negative 10y? So I essentially want to make this negative 2y into a positive 10y, right? Because if this was a positive 10y, it'll cancel out when I add the left-hand side to this equation."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "5x minus 10y is equal to 15. Now, is there anything that I can multiply this green equation by so that this negative 2y term becomes a term that will cancel out with the negative 10y? So I essentially want to make this negative 2y into a positive 10y, right? Because if this was a positive 10y, it'll cancel out when I add the left-hand side to this equation. So what could I multiply this equation by? Well, if I multiply it by negative 5, negative 5 times negative 2 right here would be positive 10. So let's do that."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "Because if this was a positive 10y, it'll cancel out when I add the left-hand side to this equation. So what could I multiply this equation by? Well, if I multiply it by negative 5, negative 5 times negative 2 right here would be positive 10. So let's do that. Let's multiply this equation times 5, times negative 5. So you multiply the left-hand side by negative 5 and multiply the right-hand side by negative 5. And what do you get?"}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "So let's do that. Let's multiply this equation times 5, times negative 5. So you multiply the left-hand side by negative 5 and multiply the right-hand side by negative 5. And what do you get? Remember, we're not fundamentally changing the equation. We're not changing the information in the equation. We're doing the same thing to both sides of it."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "And what do you get? Remember, we're not fundamentally changing the equation. We're not changing the information in the equation. We're doing the same thing to both sides of it. So the left-hand side of the equation becomes negative 5 times 3x is negative 15x. And the negative 5 times negative 2y is plus 10y is equal to 3 times negative 5 is negative 15. And now we're ready to do our elimination."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "We're doing the same thing to both sides of it. So the left-hand side of the equation becomes negative 5 times 3x is negative 15x. And the negative 5 times negative 2y is plus 10y is equal to 3 times negative 5 is negative 15. And now we're ready to do our elimination. If we add this to the left-hand side of the yellow equation and we add the negative 15 to the right-hand side of the yellow equation, we are adding the same thing to both sides of the equation because this is equal to that. So let's do that. So 5x minus 15y, we have this little negative sign there."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "And now we're ready to do our elimination. If we add this to the left-hand side of the yellow equation and we add the negative 15 to the right-hand side of the yellow equation, we are adding the same thing to both sides of the equation because this is equal to that. So let's do that. So 5x minus 15y, we have this little negative sign there. We don't want to lose that. That's negative 10x. The y's cancel out."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "So 5x minus 15y, we have this little negative sign there. We don't want to lose that. That's negative 10x. The y's cancel out. Negative 10y plus 10y, that's 0y. That was the whole point behind multiplying this by negative 5. Is going to be equal to 15 minus 15 is 0."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "The y's cancel out. Negative 10y plus 10y, that's 0y. That was the whole point behind multiplying this by negative 5. Is going to be equal to 15 minus 15 is 0. So negative 10x is equal to 0. Divide both sides by negative 10. And you get x is equal to 0."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "Is going to be equal to 15 minus 15 is 0. So negative 10x is equal to 0. Divide both sides by negative 10. And you get x is equal to 0. And now we can substitute back into either of these equations to figure out what y must be equal to. Let's substitute into the top equation. So we get 5 times 0 minus 10y is equal to 15."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "And you get x is equal to 0. And now we can substitute back into either of these equations to figure out what y must be equal to. Let's substitute into the top equation. So we get 5 times 0 minus 10y is equal to 15. Or negative 10y is equal to 15. Let me write that. Negative 10y is equal to 15."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "So we get 5 times 0 minus 10y is equal to 15. Or negative 10y is equal to 15. Let me write that. Negative 10y is equal to 15. Divide both sides by negative 10. And we are left with y is equal to 15 over 10 is negative 3 over 2. So if you were to graph it, the point of intersection would be the point 0, negative 3 halves."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "Negative 10y is equal to 15. Divide both sides by negative 10. And we are left with y is equal to 15 over 10 is negative 3 over 2. So if you were to graph it, the point of intersection would be the point 0, negative 3 halves. And you can verify that it also satisfies this equation. The original equation over here was 3x minus 2y is equal to 3. 3 times 0 is equal to 3."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "So if you were to graph it, the point of intersection would be the point 0, negative 3 halves. And you can verify that it also satisfies this equation. The original equation over here was 3x minus 2y is equal to 3. 3 times 0 is equal to 3. 3 times 0, which is 0, minus 2 times negative 3 halves is 0. This is positive 3. These cancel out."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "3 times 0 is equal to 3. 3 times 0, which is 0, minus 2 times negative 3 halves is 0. This is positive 3. These cancel out. These become positive. Plus positive 3 is equal to 3. So this does indeed satisfy both equations."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "These cancel out. These become positive. Plus positive 3 is equal to 3. So this does indeed satisfy both equations. Let's do another one of these where we have to multiply and to massage the equations. And then we can eliminate one of the variables. Let's say we have 5x plus 7y is equal to 15."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "So this does indeed satisfy both equations. Let's do another one of these where we have to multiply and to massage the equations. And then we can eliminate one of the variables. Let's say we have 5x plus 7y is equal to 15. And we have 7x minus 3y is equal to 5. Now once again, if you just added or subtracted both the left-hand sides, you're not eliminating any variables. These aren't in any way kind of have the same coefficient or the negative of their coefficient."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "Let's say we have 5x plus 7y is equal to 15. And we have 7x minus 3y is equal to 5. Now once again, if you just added or subtracted both the left-hand sides, you're not eliminating any variables. These aren't in any way kind of have the same coefficient or the negative of their coefficient. So let's pick a variable to eliminate. Let's say we want to eliminate the x's this time. And you could literally pick on one of the variables or another."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "These aren't in any way kind of have the same coefficient or the negative of their coefficient. So let's pick a variable to eliminate. Let's say we want to eliminate the x's this time. And you could literally pick on one of the variables or another. Doesn't matter. You could say, oh, let's eliminate the y's first. But I'm going to choose to eliminate the x's first."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "And you could literally pick on one of the variables or another. Doesn't matter. You could say, oh, let's eliminate the y's first. But I'm going to choose to eliminate the x's first. And so what I need to do is massage one or both of these equations in a way that these guys have the same coefficients or their coefficients are the negatives of each other. So that when I add the left-hand sides, they're going to eliminate each other. Now, there's nothing obvious."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "But I'm going to choose to eliminate the x's first. And so what I need to do is massage one or both of these equations in a way that these guys have the same coefficients or their coefficients are the negatives of each other. So that when I add the left-hand sides, they're going to eliminate each other. Now, there's nothing obvious. I could multiply this by a fraction to make it equal to negative 5. Or I could multiply this by a fraction to make it equal to negative 7. But even a more fun thing to do is I can try to get both of them to be their least common multiple."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "Now, there's nothing obvious. I could multiply this by a fraction to make it equal to negative 5. Or I could multiply this by a fraction to make it equal to negative 7. But even a more fun thing to do is I can try to get both of them to be their least common multiple. I could get both of these to 35. And the way I can do it is by multiplying by each other. So I can multiply this top equation by 7."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "But even a more fun thing to do is I can try to get both of them to be their least common multiple. I could get both of these to 35. And the way I can do it is by multiplying by each other. So I can multiply this top equation by 7. And I'm picking 7 so that this becomes a 35. And I could multiply this bottom equation by negative 5. And the reason why I'm doing that is so that this becomes a negative 35."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "So I can multiply this top equation by 7. And I'm picking 7 so that this becomes a 35. And I could multiply this bottom equation by negative 5. And the reason why I'm doing that is so that this becomes a negative 35. Remember, my point is I want to eliminate the x's. So if I make this a 35 and if I make this a negative 35, then I'm going to be all set. I can add the two, the left-hand and the right-hand sides of the equations."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "And the reason why I'm doing that is so that this becomes a negative 35. Remember, my point is I want to eliminate the x's. So if I make this a 35 and if I make this a negative 35, then I'm going to be all set. I can add the two, the left-hand and the right-hand sides of the equations. So this top equation, when you multiply by 7, it becomes. Let me scroll up a little bit. When you multiply by 7, it becomes 35x plus 49y is equal to 70 plus 35 is equal to 105."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "I can add the two, the left-hand and the right-hand sides of the equations. So this top equation, when you multiply by 7, it becomes. Let me scroll up a little bit. When you multiply by 7, it becomes 35x plus 49y is equal to 70 plus 35 is equal to 105. 15 at 70 plus 35 is equal to 105. That's what the top equation becomes. This bottom equation becomes negative 5 times 7x is negative 35x."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "When you multiply by 7, it becomes 35x plus 49y is equal to 70 plus 35 is equal to 105. 15 at 70 plus 35 is equal to 105. That's what the top equation becomes. This bottom equation becomes negative 5 times 7x is negative 35x. Negative 5 times negative 3y is plus 15y. The negatives cancel out. And then 5, this isn't a minus 5."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "This bottom equation becomes negative 5 times 7x is negative 35x. Negative 5 times negative 3y is plus 15y. The negatives cancel out. And then 5, this isn't a minus 5. This is times negative 5. 5 times negative 5 is equal to negative 25. Now we can start with this top equation and add the same thing to both sides, where that same thing is negative 25, which is also equal to this expression."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "And then 5, this isn't a minus 5. This is times negative 5. 5 times negative 5 is equal to negative 25. Now we can start with this top equation and add the same thing to both sides, where that same thing is negative 25, which is also equal to this expression. So let's add the left-hand sides and the right-hand sides because we're really adding the same thing to both sides of the equation. So the left-hand side, the x's cancel out. 35x minus 35x."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "Now we can start with this top equation and add the same thing to both sides, where that same thing is negative 25, which is also equal to this expression. So let's add the left-hand sides and the right-hand sides because we're really adding the same thing to both sides of the equation. So the left-hand side, the x's cancel out. 35x minus 35x. That was the whole point. They cancel out. And on the y's, you get 49y plus 15y."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "35x minus 35x. That was the whole point. They cancel out. And on the y's, you get 49y plus 15y. That is 64y is equal to 105 minus 25 is equal to 80. Divide both sides by 64. And you get y is equal to 80 over 64."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "And on the y's, you get 49y plus 15y. That is 64y is equal to 105 minus 25 is equal to 80. Divide both sides by 64. And you get y is equal to 80 over 64. And let's see, if you divide the numerator and the denominator by 8, actually, if I do 16, 16 would be better. Well, let's do 8 first just because we know our 8 times tables. So that becomes 10 over 8."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "And you get y is equal to 80 over 64. And let's see, if you divide the numerator and the denominator by 8, actually, if I do 16, 16 would be better. Well, let's do 8 first just because we know our 8 times tables. So that becomes 10 over 8. And you can divide this by 2, and you get 5 over 4. If you divided just straight up by 16, you would have gone straight to 5 over 4. So y is equal to 5 over 4."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "So that becomes 10 over 8. And you can divide this by 2, and you get 5 over 4. If you divided just straight up by 16, you would have gone straight to 5 over 4. So y is equal to 5 over 4. Let's figure out what x is. So we can substitute either into one of these equations or into one of the original equations. Let's substitute into the second of the original equations where we had 7x minus 3y is equal to 5."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "So y is equal to 5 over 4. Let's figure out what x is. So we can substitute either into one of these equations or into one of the original equations. Let's substitute into the second of the original equations where we had 7x minus 3y is equal to 5. That was the original version of the second equation that we later transformed into this. So we get 7x minus 3 times y times 5 over 4 is equal to 5. Or 7x minus 15 over 4 is equal to 5."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "Let's substitute into the second of the original equations where we had 7x minus 3y is equal to 5. That was the original version of the second equation that we later transformed into this. So we get 7x minus 3 times y times 5 over 4 is equal to 5. Or 7x minus 15 over 4 is equal to 5. Let's add 15 over 4. Oh, sorry, I didn't do that right. This would be 7x minus 3 times 4."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "Or 7x minus 15 over 4 is equal to 5. Let's add 15 over 4. Oh, sorry, I didn't do that right. This would be 7x minus 3 times 4. Oh, sorry, that was right. What am I doing? 3 times 15 over 4 is equal to 5."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "This would be 7x minus 3 times 4. Oh, sorry, that was right. What am I doing? 3 times 15 over 4 is equal to 5. Let's add 15 over 4 to both sides. So plus 15 over 4. Plus 15 over 4."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "3 times 15 over 4 is equal to 5. Let's add 15 over 4 to both sides. So plus 15 over 4. Plus 15 over 4. And what do we get? The left-hand side just becomes a 7x. These guys cancel out."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "Plus 15 over 4. And what do we get? The left-hand side just becomes a 7x. These guys cancel out. And that's going to be equal to 5 is the same thing as 20 over 4. 20 over 4 plus 15 over 4. Or we get that, let me scroll down a little bit, 7x is equal to 35 over 4."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "These guys cancel out. And that's going to be equal to 5 is the same thing as 20 over 4. 20 over 4 plus 15 over 4. Or we get that, let me scroll down a little bit, 7x is equal to 35 over 4. We can multiply both sides by 1 7th. Or we could divide both sides by 7. Same thing."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "Or we get that, let me scroll down a little bit, 7x is equal to 35 over 4. We can multiply both sides by 1 7th. Or we could divide both sides by 7. Same thing. Let's multiply both sides by 1 7th. The same thing as dividing by 7. So these cancel out."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "Same thing. Let's multiply both sides by 1 7th. The same thing as dividing by 7. So these cancel out. And you're left with x is equal to. Here, if you divide 35 by 7, you get 5. You divide 7 by 7, you get 1."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "So these cancel out. And you're left with x is equal to. Here, if you divide 35 by 7, you get 5. You divide 7 by 7, you get 1. So x is equal to 5 4ths as well. So the point of intersection of this right here is both x and y are going to be equal to 5 4ths. So if you looked at it as a graph, it would be 5 4ths comma 5 4ths."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "You divide 7 by 7, you get 1. So x is equal to 5 4ths as well. So the point of intersection of this right here is both x and y are going to be equal to 5 4ths. So if you looked at it as a graph, it would be 5 4ths comma 5 4ths. Now let's verify that this satisfies the top equation. If you take 5 times 5 over 4 plus 7 times 5 over 4, what do you get? It should be equal to 15."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "So if you looked at it as a graph, it would be 5 4ths comma 5 4ths. Now let's verify that this satisfies the top equation. If you take 5 times 5 over 4 plus 7 times 5 over 4, what do you get? It should be equal to 15. So this is equal to 25 over 4 plus 35 over 4, which is equal to 60 over 4, which is indeed equal to 15. So it does definitely satisfy that top equation. And you can check out this bottom equation for yourself."}, {"video_title": "Linear equation using segment   Geometry   8th grade   Khan Academy.mp3", "Sentence": "And they tell us that jk is equal to 7x plus 9. So this distance right over here is equal to 7x plus 9. Then they tell us that jl is equal to 114. So jl is the entire length of the segment. So this entire thing is equal to 114. And then they tell us that kl is equal to 9x plus 9. So this right over here is equal to 9x plus 9."}, {"video_title": "Linear equation using segment   Geometry   8th grade   Khan Academy.mp3", "Sentence": "So jl is the entire length of the segment. So this entire thing is equal to 114. And then they tell us that kl is equal to 9x plus 9. So this right over here is equal to 9x plus 9. And they say find kl. So they essentially want us to figure out, what does 9x plus 9 equal? And to figure that out, we have to figure out what x equals."}, {"video_title": "Linear equation using segment   Geometry   8th grade   Khan Academy.mp3", "Sentence": "So this right over here is equal to 9x plus 9. And they say find kl. So they essentially want us to figure out, what does 9x plus 9 equal? And to figure that out, we have to figure out what x equals. And lucky for us, they've given us all the information we need. They tell us that the entire segment is 114 long. And they don't give us any units."}, {"video_title": "Linear equation using segment   Geometry   8th grade   Khan Academy.mp3", "Sentence": "And to figure that out, we have to figure out what x equals. And lucky for us, they've given us all the information we need. They tell us that the entire segment is 114 long. And they don't give us any units. It's just 114. It has the length 114. And we know that this segment, jk plus kl, added together is going to be equal to the length of the entire thing."}, {"video_title": "Linear equation using segment   Geometry   8th grade   Khan Academy.mp3", "Sentence": "And they don't give us any units. It's just 114. It has the length 114. And we know that this segment, jk plus kl, added together is going to be equal to the length of the entire thing. So we could say that 7x plus 9. Actually, let me write it this way. We could write that jk, so the length of segment jk, plus the length of segment kl is going to be equal to 114."}, {"video_title": "Linear equation using segment   Geometry   8th grade   Khan Academy.mp3", "Sentence": "And we know that this segment, jk plus kl, added together is going to be equal to the length of the entire thing. So we could say that 7x plus 9. Actually, let me write it this way. We could write that jk, so the length of segment jk, plus the length of segment kl is going to be equal to 114. 114. And we know that the length of segment jk is 7x plus 9. We know that the length of segment kl is 9x plus 9."}, {"video_title": "Linear equation using segment   Geometry   8th grade   Khan Academy.mp3", "Sentence": "We could write that jk, so the length of segment jk, plus the length of segment kl is going to be equal to 114. 114. And we know that the length of segment jk is 7x plus 9. We know that the length of segment kl is 9x plus 9. So plus 9x plus 9. And this is going to be equal to 114. And now we just have to break out a little bit of our algebraic skills."}, {"video_title": "Linear equation using segment   Geometry   8th grade   Khan Academy.mp3", "Sentence": "We know that the length of segment kl is 9x plus 9. So plus 9x plus 9. And this is going to be equal to 114. And now we just have to break out a little bit of our algebraic skills. So the first thing we might want to do is, let's see, we have two terms that have x's in them. If I have 7x's and I were to add that to another 9x's, that means I'm going to have 16x's. And then if I have, so this is just plain old 9, and I add it to another 9, that's going to give me 18."}, {"video_title": "Linear equation using segment   Geometry   8th grade   Khan Academy.mp3", "Sentence": "And now we just have to break out a little bit of our algebraic skills. So the first thing we might want to do is, let's see, we have two terms that have x's in them. If I have 7x's and I were to add that to another 9x's, that means I'm going to have 16x's. And then if I have, so this is just plain old 9, and I add it to another 9, that's going to give me 18. And that's going to be equal to 114. Now I just subtract 18 from both sides. Let me do that explicitly."}, {"video_title": "Linear equation using segment   Geometry   8th grade   Khan Academy.mp3", "Sentence": "And then if I have, so this is just plain old 9, and I add it to another 9, that's going to give me 18. And that's going to be equal to 114. Now I just subtract 18 from both sides. Let me do that explicitly. So I subtract 18 from both sides. On the left-hand side, I have 16x. And on the right-hand side, I'm going to have 114 minus 18."}, {"video_title": "Linear equation using segment   Geometry   8th grade   Khan Academy.mp3", "Sentence": "Let me do that explicitly. So I subtract 18 from both sides. On the left-hand side, I have 16x. And on the right-hand side, I'm going to have 114 minus 18. Well, if I were to subtract 14, that would get me to 100. And I'm going to subtract 4 more than that. So that's going to get me to 96."}, {"video_title": "Linear equation using segment   Geometry   8th grade   Khan Academy.mp3", "Sentence": "And on the right-hand side, I'm going to have 114 minus 18. Well, if I were to subtract 14, that would get me to 100. And I'm going to subtract 4 more than that. So that's going to get me to 96. And now we just divide both sides by 16. And let's see, this looks like 96 divided by 16. We could do it explicitly, but it looks like it's going to be 6."}, {"video_title": "Linear equation using segment   Geometry   8th grade   Khan Academy.mp3", "Sentence": "So that's going to get me to 96. And now we just divide both sides by 16. And let's see, this looks like 96 divided by 16. We could do it explicitly, but it looks like it's going to be 6. 6 times 10 is 60. 6 times 6 is 36. 60 plus 36 is 96."}, {"video_title": "Linear equation using segment   Geometry   8th grade   Khan Academy.mp3", "Sentence": "We could do it explicitly, but it looks like it's going to be 6. 6 times 10 is 60. 6 times 6 is 36. 60 plus 36 is 96. So we get x is equal to 6. Now we're not done yet. We're not looking just for x."}, {"video_title": "Linear equation using segment   Geometry   8th grade   Khan Academy.mp3", "Sentence": "60 plus 36 is 96. So we get x is equal to 6. Now we're not done yet. We're not looking just for x. We're looking for the length of KL. KL is 9x plus 9. Let me write that down."}, {"video_title": "Linear equation using segment   Geometry   8th grade   Khan Academy.mp3", "Sentence": "We're not looking just for x. We're looking for the length of KL. KL is 9x plus 9. Let me write that down. So KL is equal to 9x plus 9. They told us that right over there is equal to 9x plus 9. We just figured out that x is equal to 6."}, {"video_title": "Linear equation using segment   Geometry   8th grade   Khan Academy.mp3", "Sentence": "Let me write that down. So KL is equal to 9x plus 9. They told us that right over there is equal to 9x plus 9. We just figured out that x is equal to 6. So this is equal to 9 times 6 plus 9. And this is equal to 9 times 6 is 54 plus 9 is equal to 63. So KL is equal to 63."}, {"video_title": "Example 3 Multiplying a monomial by a polynomial   Algebra I   Khan Academy.mp3", "Sentence": "So we're going to have to distribute this negative 4x squared over every term in the expression. So first we can start with negative 4x squared times 3x squared. So we can write that, we're going to have negative 4x squared times 3x squared. And to that we're going to add negative 4x squared times 25x. And to that we're going to add negative 4x squared times negative 7. So let's just simplify this a little bit. Now we can obviously swap the order."}, {"video_title": "Example 3 Multiplying a monomial by a polynomial   Algebra I   Khan Academy.mp3", "Sentence": "And to that we're going to add negative 4x squared times 25x. And to that we're going to add negative 4x squared times negative 7. So let's just simplify this a little bit. Now we can obviously swap the order. We're just multiplying negative 4 times x squared times 3 times x squared. And actually I'll do out every step. Eventually you can do some of this in your head."}, {"video_title": "Example 3 Multiplying a monomial by a polynomial   Algebra I   Khan Academy.mp3", "Sentence": "Now we can obviously swap the order. We're just multiplying negative 4 times x squared times 3 times x squared. And actually I'll do out every step. Eventually you can do some of this in your head. This is the exact same thing as negative 4 times 3 times x squared times x squared. And what is that equal to? Well negative 4 times 3 is negative 12."}, {"video_title": "Example 3 Multiplying a monomial by a polynomial   Algebra I   Khan Academy.mp3", "Sentence": "Eventually you can do some of this in your head. This is the exact same thing as negative 4 times 3 times x squared times x squared. And what is that equal to? Well negative 4 times 3 is negative 12. And x squared times x squared, same base, we're taking the product. That's going to be x to the fourth. So this right here is negative 12x to the fourth."}, {"video_title": "Example 3 Multiplying a monomial by a polynomial   Algebra I   Khan Academy.mp3", "Sentence": "Well negative 4 times 3 is negative 12. And x squared times x squared, same base, we're taking the product. That's going to be x to the fourth. So this right here is negative 12x to the fourth. Now let's think about this term over here. This is the same thing as, of course we have this plus out here. And this part right here is the exact same thing as 25 times negative 4 times x squared times x."}, {"video_title": "Example 3 Multiplying a monomial by a polynomial   Algebra I   Khan Academy.mp3", "Sentence": "So this right here is negative 12x to the fourth. Now let's think about this term over here. This is the same thing as, of course we have this plus out here. And this part right here is the exact same thing as 25 times negative 4 times x squared times x. So let's just multiply the numbers out here. These were the coefficients. 25 times negative 4 is negative 100."}, {"video_title": "Example 3 Multiplying a monomial by a polynomial   Algebra I   Khan Academy.mp3", "Sentence": "And this part right here is the exact same thing as 25 times negative 4 times x squared times x. So let's just multiply the numbers out here. These were the coefficients. 25 times negative 4 is negative 100. So it will be plus negative 100 or we can just say it's minus 100. And then we have x squared times x or x squared times x to the first power. Same base, we can add the exponents."}, {"video_title": "Example 3 Multiplying a monomial by a polynomial   Algebra I   Khan Academy.mp3", "Sentence": "25 times negative 4 is negative 100. So it will be plus negative 100 or we can just say it's minus 100. And then we have x squared times x or x squared times x to the first power. Same base, we can add the exponents. 2 plus 1 is 3. So this is negative 100x to the third power. And then let's look at this last term over here."}, {"video_title": "Example 3 Multiplying a monomial by a polynomial   Algebra I   Khan Academy.mp3", "Sentence": "Same base, we can add the exponents. 2 plus 1 is 3. So this is negative 100x to the third power. And then let's look at this last term over here. We have negative 4x squared. So this is going to be plus. That's this plus right over here."}, {"video_title": "Example 3 Multiplying a monomial by a polynomial   Algebra I   Khan Academy.mp3", "Sentence": "And then let's look at this last term over here. We have negative 4x squared. So this is going to be plus. That's this plus right over here. We have negative 4. We can multiply that times negative 7. And then multiply that times x squared."}, {"video_title": "Example 3 Multiplying a monomial by a polynomial   Algebra I   Khan Academy.mp3", "Sentence": "That's this plus right over here. We have negative 4. We can multiply that times negative 7. And then multiply that times x squared. I'm just changing the order in which we multiply it. So negative 4 times negative 7 is positive 28. And then I'm going to multiply that times the x squared."}, {"video_title": "Example 3 Multiplying a monomial by a polynomial   Algebra I   Khan Academy.mp3", "Sentence": "And then multiply that times x squared. I'm just changing the order in which we multiply it. So negative 4 times negative 7 is positive 28. And then I'm going to multiply that times the x squared. There's no simplification to do. No like terms. These are different powers of x."}, {"video_title": "Dividing positive and negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "And what you'll see is that it's actually a very similar methodology, that if both are positive, you will get a positive answer. If one is negative, but not both, you're going to get a negative answer. And if both are negative, they will cancel out and you will get a positive answer. But let's apply it, and I encourage you to pause this video and try these out yourself and then see if you get the same answer that I'm going to get. So 8 divided by negative 2. So if I just had 8 divided by 2, that would be a positive 4. But since exactly one of these two numbers are negative, this one right over here, the answer is going to be negative."}, {"video_title": "Dividing positive and negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "But let's apply it, and I encourage you to pause this video and try these out yourself and then see if you get the same answer that I'm going to get. So 8 divided by negative 2. So if I just had 8 divided by 2, that would be a positive 4. But since exactly one of these two numbers are negative, this one right over here, the answer is going to be negative. So 8 divided by negative 2 is negative 4. Now negative 16 divided by positive 4, we have to be very careful here. If I just had positive 16 divided by positive 4, that would just be 4."}, {"video_title": "Dividing positive and negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "But since exactly one of these two numbers are negative, this one right over here, the answer is going to be negative. So 8 divided by negative 2 is negative 4. Now negative 16 divided by positive 4, we have to be very careful here. If I just had positive 16 divided by positive 4, that would just be 4. But because one of these two numbers is negative, and exactly one of these two numbers is negative, then I'm going to get a negative answer. Now I have negative 30 divided by negative 5. If I just had 30 divided by 5, that would be positive 6."}, {"video_title": "Dividing positive and negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "If I just had positive 16 divided by positive 4, that would just be 4. But because one of these two numbers is negative, and exactly one of these two numbers is negative, then I'm going to get a negative answer. Now I have negative 30 divided by negative 5. If I just had 30 divided by 5, that would be positive 6. And because I have a negative divided by a negative, the negatives cancel out, and so my answer will still be positive 6. And I could even write a positive out here. I don't have to."}, {"video_title": "Dividing positive and negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "If I just had 30 divided by 5, that would be positive 6. And because I have a negative divided by a negative, the negatives cancel out, and so my answer will still be positive 6. And I could even write a positive out here. I don't have to. But this is a positive 6. A negative divided by a negative, just like a negative times a negative, you're going to get a positive answer. 18 divided by 2."}, {"video_title": "Dividing positive and negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "I don't have to. But this is a positive 6. A negative divided by a negative, just like a negative times a negative, you're going to get a positive answer. 18 divided by 2. And this is a little bit of a trick question. This is what you knew how to do before we even talked about negative numbers. This is a positive divided by a positive, which is going to be a positive."}, {"video_title": "Dividing positive and negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "18 divided by 2. And this is a little bit of a trick question. This is what you knew how to do before we even talked about negative numbers. This is a positive divided by a positive, which is going to be a positive. So that is going to be equal to positive 9. Now we start doing some interesting things. Here's this kind of a compound problem."}, {"video_title": "Dividing positive and negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "This is a positive divided by a positive, which is going to be a positive. So that is going to be equal to positive 9. Now we start doing some interesting things. Here's this kind of a compound problem. You have some multiplication and some division going on. And so first, right over here, the way this is written, we're going to want to multiply the numerator out. And if you're not familiar with this little dot symbol, it's just another way of writing multiplication."}, {"video_title": "Dividing positive and negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "Here's this kind of a compound problem. You have some multiplication and some division going on. And so first, right over here, the way this is written, we're going to want to multiply the numerator out. And if you're not familiar with this little dot symbol, it's just another way of writing multiplication. I could have written this little x thing over here. And what you're going to see is in algebra, the dot becomes much more common because the x gets used for other things. The times symbol, people don't want to confuse it with the letter x, which gets used a lot in algebra."}, {"video_title": "Dividing positive and negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "And if you're not familiar with this little dot symbol, it's just another way of writing multiplication. I could have written this little x thing over here. And what you're going to see is in algebra, the dot becomes much more common because the x gets used for other things. The times symbol, people don't want to confuse it with the letter x, which gets used a lot in algebra. And so that's why they use the dot very often. So this just says negative 7 times 3 in the numerator. And then we're going to take that product and divide it by negative 1."}, {"video_title": "Dividing positive and negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "The times symbol, people don't want to confuse it with the letter x, which gets used a lot in algebra. And so that's why they use the dot very often. So this just says negative 7 times 3 in the numerator. And then we're going to take that product and divide it by negative 1. So in the numerator, negative 7 times 3. Positive 7 times 3 would be 21. But since exactly one of these two are negative, this is going to be negative 21."}, {"video_title": "Dividing positive and negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "And then we're going to take that product and divide it by negative 1. So in the numerator, negative 7 times 3. Positive 7 times 3 would be 21. But since exactly one of these two are negative, this is going to be negative 21. And it's going to be negative 21 over negative 1. And so negative 21 divided by negative 1, negative divided by a negative is going to be a positive. So this is just going to be positive 21."}, {"video_title": "Dividing positive and negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "But since exactly one of these two are negative, this is going to be negative 21. And it's going to be negative 21 over negative 1. And so negative 21 divided by negative 1, negative divided by a negative is going to be a positive. So this is just going to be positive 21. Let me write all these things down. So if I were to take a positive divided by a negative, that's going to give me a negative. If I have a negative divided by a positive, that's also going to give me a negative."}, {"video_title": "Dividing positive and negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "So this is just going to be positive 21. Let me write all these things down. So if I were to take a positive divided by a negative, that's going to give me a negative. If I have a negative divided by a positive, that's also going to give me a negative. If I have a negative divided by a negative, that's going to give me a positive. And if I have, obviously, a positive divided by a positive, that's also going to give me a positive. Now let's do this last one over here."}, {"video_title": "Dividing positive and negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "If I have a negative divided by a positive, that's also going to give me a negative. If I have a negative divided by a negative, that's going to give me a positive. And if I have, obviously, a positive divided by a positive, that's also going to give me a positive. Now let's do this last one over here. This actually is all multiplication, but it's interesting because we're multiplying three things, which we haven't done yet. And we could just go from left to right over here, and we could first think about negative 2 times negative 7. Negative 2 times negative 7, they are both negative."}, {"video_title": "Dividing positive and negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "Now let's do this last one over here. This actually is all multiplication, but it's interesting because we're multiplying three things, which we haven't done yet. And we could just go from left to right over here, and we could first think about negative 2 times negative 7. Negative 2 times negative 7, they are both negative. The negatives cancel out. So this will give us, this part right over here, will give us positive 14. And so we're going to multiply the positive 14 times this negative 1."}, {"video_title": "Dividing positive and negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "Negative 2 times negative 7, they are both negative. The negatives cancel out. So this will give us, this part right over here, will give us positive 14. And so we're going to multiply the positive 14 times this negative 1. Now we have a positive times a negative. Exactly one of them is negative. So this is going to give me a negative answer."}, {"video_title": "Dividing positive and negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "And so we're going to multiply the positive 14 times this negative 1. Now we have a positive times a negative. Exactly one of them is negative. So this is going to give me a negative answer. It's going to give me negative 14. Now let me give you a couple more, I guess we could call these trick problems. What would happen if I had 0 divided by negative 5?"}, {"video_title": "Dividing positive and negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "So this is going to give me a negative answer. It's going to give me negative 14. Now let me give you a couple more, I guess we could call these trick problems. What would happen if I had 0 divided by negative 5? Well, this is 0 negative 5. So 0 divided by anything that's non-zero is just going to be equal to 0. What if we were to do it the other way around?"}, {"video_title": "Dividing positive and negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "What would happen if I had 0 divided by negative 5? Well, this is 0 negative 5. So 0 divided by anything that's non-zero is just going to be equal to 0. What if we were to do it the other way around? What happens if we said negative 5 divided by 0? Well, we don't know what happens when you divide things by 0. We haven't defined that."}, {"video_title": "Dividing positive and negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "What if we were to do it the other way around? What happens if we said negative 5 divided by 0? Well, we don't know what happens when you divide things by 0. We haven't defined that. There's arguments for multiple ways to conceptualize this. So we traditionally just say that this is undefined. We haven't defined what happens when something is divided by 0."}, {"video_title": "Slope from two ordered pairs example 2   Algebra I   Khan Academy.mp3", "Sentence": "Let me just do a quick graph of these just so we can visualize what they look like. So let me draw a quick graph over here. So our first point is 7, negative 1. So 1, 2, 3, 4, 5, 6, 7. This is the x-axis. 7, negative 1. So 7, negative 1 is right over there."}, {"video_title": "Slope from two ordered pairs example 2   Algebra I   Khan Academy.mp3", "Sentence": "So 1, 2, 3, 4, 5, 6, 7. This is the x-axis. 7, negative 1. So 7, negative 1 is right over there. 7, negative 1. This of course is the y-axis. And then the next point is negative 3, negative 1."}, {"video_title": "Slope from two ordered pairs example 2   Algebra I   Khan Academy.mp3", "Sentence": "So 7, negative 1 is right over there. 7, negative 1. This of course is the y-axis. And then the next point is negative 3, negative 1. So we go back 3 in the horizontal direction. Negative 3. But the y-coordinate is still negative 1."}, {"video_title": "Slope from two ordered pairs example 2   Algebra I   Khan Academy.mp3", "Sentence": "And then the next point is negative 3, negative 1. So we go back 3 in the horizontal direction. Negative 3. But the y-coordinate is still negative 1. It's still negative 1. So the line that connects these two points will look like this. It will look like that."}, {"video_title": "Slope from two ordered pairs example 2   Algebra I   Khan Academy.mp3", "Sentence": "But the y-coordinate is still negative 1. It's still negative 1. So the line that connects these two points will look like this. It will look like that. Now, they're asking us to find the slope of the line that goes through the ordered pairs. Find the slope of this line. And just to give a little bit of intuition here, slope is a measure of a line's inclination."}, {"video_title": "Slope from two ordered pairs example 2   Algebra I   Khan Academy.mp3", "Sentence": "It will look like that. Now, they're asking us to find the slope of the line that goes through the ordered pairs. Find the slope of this line. And just to give a little bit of intuition here, slope is a measure of a line's inclination. And the way that it's defined, slope is defined as rise over run, or change in y over change in x, or sometimes you'll see it defined as the variable m, and then they'll define change in y as just being the second y-coordinate minus the first y-coordinate, and then the change in x as the second x-coordinate minus the first x-coordinate. These are all different variations in slope, but hopefully you appreciate that these are measuring inclination. If I rise a ton when I run a little bit, if I move a little bit in the x-direction and I rise a bunch, then I have a very steep line."}, {"video_title": "Slope from two ordered pairs example 2   Algebra I   Khan Academy.mp3", "Sentence": "And just to give a little bit of intuition here, slope is a measure of a line's inclination. And the way that it's defined, slope is defined as rise over run, or change in y over change in x, or sometimes you'll see it defined as the variable m, and then they'll define change in y as just being the second y-coordinate minus the first y-coordinate, and then the change in x as the second x-coordinate minus the first x-coordinate. These are all different variations in slope, but hopefully you appreciate that these are measuring inclination. If I rise a ton when I run a little bit, if I move a little bit in the x-direction and I rise a bunch, then I have a very steep line. I have a very steep upward sloping line. If I don't change at all when I run a bit, then I have a very low slope, and that's actually what's happening here. I'm going from, you could either view this as the starting point or view this as the starting point, but let's view this as the starting point."}, {"video_title": "Slope from two ordered pairs example 2   Algebra I   Khan Academy.mp3", "Sentence": "If I rise a ton when I run a little bit, if I move a little bit in the x-direction and I rise a bunch, then I have a very steep line. I have a very steep upward sloping line. If I don't change at all when I run a bit, then I have a very low slope, and that's actually what's happening here. I'm going from, you could either view this as the starting point or view this as the starting point, but let's view this as the starting point. So this negative 3 comma 1. If I go from negative 3 comma negative 1 to 7 comma negative 1, I'm running a good bit. I'm going from negative 3, my x value is negative 3 here, and it goes all the way to 7."}, {"video_title": "Slope from two ordered pairs example 2   Algebra I   Khan Academy.mp3", "Sentence": "I'm going from, you could either view this as the starting point or view this as the starting point, but let's view this as the starting point. So this negative 3 comma 1. If I go from negative 3 comma negative 1 to 7 comma negative 1, I'm running a good bit. I'm going from negative 3, my x value is negative 3 here, and it goes all the way to 7. So my change in x here is 10. To go from negative 3 to 7, I change my x value by 10. But what's my change in y?"}, {"video_title": "Slope from two ordered pairs example 2   Algebra I   Khan Academy.mp3", "Sentence": "I'm going from negative 3, my x value is negative 3 here, and it goes all the way to 7. So my change in x here is 10. To go from negative 3 to 7, I change my x value by 10. But what's my change in y? Well, my y value here is negative 1, and my y value over here is still negative 1. So my change in y is 0. My change in y is going to be 0."}, {"video_title": "Slope from two ordered pairs example 2   Algebra I   Khan Academy.mp3", "Sentence": "But what's my change in y? Well, my y value here is negative 1, and my y value over here is still negative 1. So my change in y is 0. My change in y is going to be 0. My y value does not change no matter how much I change my x value. So the slope here is going to be, when we run 10, what was our rise? How much did we change in y?"}, {"video_title": "Slope from two ordered pairs example 2   Algebra I   Khan Academy.mp3", "Sentence": "My change in y is going to be 0. My y value does not change no matter how much I change my x value. So the slope here is going to be, when we run 10, what was our rise? How much did we change in y? Well, we didn't rise at all. We didn't go up or down. So the slope here is 0."}, {"video_title": "Slope from two ordered pairs example 2   Algebra I   Khan Academy.mp3", "Sentence": "How much did we change in y? Well, we didn't rise at all. We didn't go up or down. So the slope here is 0. Or another way to think about it is this line has no inclination. It's a completely flat, it's a completely horizontal line. So this should make sense."}, {"video_title": "Slope from two ordered pairs example 2   Algebra I   Khan Academy.mp3", "Sentence": "So the slope here is 0. Or another way to think about it is this line has no inclination. It's a completely flat, it's a completely horizontal line. So this should make sense. This is a 0. The slope here is 0. And just to make sure that this gels with all of these other formulas that you might know, but I want to make it very clear, these are all just telling you rise over run, or change in y over change in x, a way to measure inclination."}, {"video_title": "Slope from two ordered pairs example 2   Algebra I   Khan Academy.mp3", "Sentence": "So this should make sense. This is a 0. The slope here is 0. And just to make sure that this gels with all of these other formulas that you might know, but I want to make it very clear, these are all just telling you rise over run, or change in y over change in x, a way to measure inclination. But let's just apply them just so hopefully it all makes sense to you. So we could also say slope is change in y over change in x. If we take this to be our start, and if we take this to be our end point, then we would call this over here x1, and then this is over here, this is y1, and then we would call this x2, and we would call this y2."}, {"video_title": "Slope from two ordered pairs example 2   Algebra I   Khan Academy.mp3", "Sentence": "And just to make sure that this gels with all of these other formulas that you might know, but I want to make it very clear, these are all just telling you rise over run, or change in y over change in x, a way to measure inclination. But let's just apply them just so hopefully it all makes sense to you. So we could also say slope is change in y over change in x. If we take this to be our start, and if we take this to be our end point, then we would call this over here x1, and then this is over here, this is y1, and then we would call this x2, and we would call this y2. If this is our start point and that is our end point. And so the slope here, the change in y, y2 minus y1, so it's negative 1 minus negative 1. All of that over x2, negative 3 minus x1, minus 7."}, {"video_title": "Slope from two ordered pairs example 2   Algebra I   Khan Academy.mp3", "Sentence": "If we take this to be our start, and if we take this to be our end point, then we would call this over here x1, and then this is over here, this is y1, and then we would call this x2, and we would call this y2. If this is our start point and that is our end point. And so the slope here, the change in y, y2 minus y1, so it's negative 1 minus negative 1. All of that over x2, negative 3 minus x1, minus 7. So the numerator, negative 1 minus negative 1, that's the same thing as negative 1 plus 1. And our denominator is negative 3 minus 7, which is negative 10. So once again, negative 1 plus 1 is 0 over negative 10."}, {"video_title": "Slope from two ordered pairs example 2   Algebra I   Khan Academy.mp3", "Sentence": "All of that over x2, negative 3 minus x1, minus 7. So the numerator, negative 1 minus negative 1, that's the same thing as negative 1 plus 1. And our denominator is negative 3 minus 7, which is negative 10. So once again, negative 1 plus 1 is 0 over negative 10. And this is still going to be 0. And the only reason why we have a negative 10 here and a positive 10 there is because we swapped the starting and the ending points. In this example right over here, we took this as the start point and made this coordinate over here as the end point."}, {"video_title": "Slope from two ordered pairs example 2   Algebra I   Khan Academy.mp3", "Sentence": "So once again, negative 1 plus 1 is 0 over negative 10. And this is still going to be 0. And the only reason why we have a negative 10 here and a positive 10 there is because we swapped the starting and the ending points. In this example right over here, we took this as the start point and made this coordinate over here as the end point. Over here we swapped them around. 7, negative 1 was our start point, and negative 3, negative 1 is our end point. So if we start over here, our change in x is going to be negative 10, but our change in y is still going to be 0."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Let's graph ourselves some inequalities. So let's say I had the inequality y is less than or equal to 4x plus 3. And on our xy-coordinate plane, we want to show all of the x and y points that satisfy this condition right here. So a good starting point might be to break up this less than or equal to, because we know how to graph y is equal to 4x plus 3. So this thing is the same thing as y could be less than 4x plus 3, or y could be equal to 4x plus 3. That's what less than or equal means. It could be less than or equal."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So a good starting point might be to break up this less than or equal to, because we know how to graph y is equal to 4x plus 3. So this thing is the same thing as y could be less than 4x plus 3, or y could be equal to 4x plus 3. That's what less than or equal means. It could be less than or equal. And the reason why I did that on this first example problem is because we know how to graph that. So let's graph that. I'll try to draw it a little bit neater than that."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "It could be less than or equal. And the reason why I did that on this first example problem is because we know how to graph that. So let's graph that. I'll try to draw it a little bit neater than that. So that is my vertical axis, my y-axis. This is my x-axis right there. That is the x-axis."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "I'll try to draw it a little bit neater than that. So that is my vertical axis, my y-axis. This is my x-axis right there. That is the x-axis. And then we know the y-intercept. The y-intercept is 3, so the point 0, 3, 1, 2, 3, is on the line, and we know we have a slope of 4, which means if we go 1 in the x direction, we're going to go up 4 in the y. So 1, 2, 3, 4."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "That is the x-axis. And then we know the y-intercept. The y-intercept is 3, so the point 0, 3, 1, 2, 3, is on the line, and we know we have a slope of 4, which means if we go 1 in the x direction, we're going to go up 4 in the y. So 1, 2, 3, 4. So it's going to be right here. And that's enough to draw a line, but we could even go back in the x direction. If we go 1 back in the x direction, we're going to go down 4."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So 1, 2, 3, 4. So it's going to be right here. And that's enough to draw a line, but we could even go back in the x direction. If we go 1 back in the x direction, we're going to go down 4. 1, 2, 3, 4. So that's also going to be a point on the line. So my best attempt at drawing this line is going to look something like something."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "If we go 1 back in the x direction, we're going to go down 4. 1, 2, 3, 4. So that's also going to be a point on the line. So my best attempt at drawing this line is going to look something like something. This is the hardest part. It's going to look something like that. That is a line."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So my best attempt at drawing this line is going to look something like something. This is the hardest part. It's going to look something like that. That is a line. It should be straight. I think you get the idea. That right there is the graph of y is equal to 4x plus 3."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "That is a line. It should be straight. I think you get the idea. That right there is the graph of y is equal to 4x plus 3. So let's think about what it means to be less than. So all of these points satisfy this inequality, but we have more. This is just these points over here."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "That right there is the graph of y is equal to 4x plus 3. So let's think about what it means to be less than. So all of these points satisfy this inequality, but we have more. This is just these points over here. What about all of these where y is less than 4x plus 3? So let's think about what this means. When, let's pick up some values for x."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "This is just these points over here. What about all of these where y is less than 4x plus 3? So let's think about what this means. When, let's pick up some values for x. When x is equal to 0, what does this say? When x is equal to 0, then that means y is going to be less than 0 plus 3. y is less than 3. When x is equal to negative 1, what is this telling us?"}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "When, let's pick up some values for x. When x is equal to 0, what does this say? When x is equal to 0, then that means y is going to be less than 0 plus 3. y is less than 3. When x is equal to negative 1, what is this telling us? 4 times negative 1 is negative 4, plus 3 is negative 1. y would be less than negative 1. When x is equal to 1, what is this telling us? 4 times 1 is 4, plus 3 is 7."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "When x is equal to negative 1, what is this telling us? 4 times negative 1 is negative 4, plus 3 is negative 1. y would be less than negative 1. When x is equal to 1, what is this telling us? 4 times 1 is 4, plus 3 is 7. So y is going to be less than 7. So let's at least try to plot these. So when x is equal to, let's plot this one first."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "4 times 1 is 4, plus 3 is 7. So y is going to be less than 7. So let's at least try to plot these. So when x is equal to, let's plot this one first. When x is equal to 0, y is less than 3. So when x is equal to 0, y is less than 3. So it's all of these points here that I'm shading in in green, satisfy that right there."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So when x is equal to, let's plot this one first. When x is equal to 0, y is less than 3. So when x is equal to 0, y is less than 3. So it's all of these points here that I'm shading in in green, satisfy that right there. If I were to look at this one over here, when x is negative 1, y is less than negative 1. So y has to be all of these points down here. When x is equal to 1, y is less than 7."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So it's all of these points here that I'm shading in in green, satisfy that right there. If I were to look at this one over here, when x is negative 1, y is less than negative 1. So y has to be all of these points down here. When x is equal to 1, y is less than 7. So it's all of these points down here. And in general, you take any point x. Let's say you take this point x right there."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "When x is equal to 1, y is less than 7. So it's all of these points down here. And in general, you take any point x. Let's say you take this point x right there. If you evaluate 4x plus 3, you're going to get the point on the line. That is that x times 4 plus 3. Now, the y's that satisfy it, it could be equal to that point on the line, or it could be less than."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Let's say you take this point x right there. If you evaluate 4x plus 3, you're going to get the point on the line. That is that x times 4 plus 3. Now, the y's that satisfy it, it could be equal to that point on the line, or it could be less than. So it's going to go below the line. So if you were to do this for all the possible x's, you would not only get all the points on this line, which we've drawn, you would get all the points below the line. So now we have graphed this inequality."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Now, the y's that satisfy it, it could be equal to that point on the line, or it could be less than. So it's going to go below the line. So if you were to do this for all the possible x's, you would not only get all the points on this line, which we've drawn, you would get all the points below the line. So now we have graphed this inequality. It's essentially this line, 4x plus 3, with all of the area below it shaded. Now, if this was just a less than, not less than or equal sign, we would not include the actual line. And the convention to do that is to actually make the line a dashed line."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So now we have graphed this inequality. It's essentially this line, 4x plus 3, with all of the area below it shaded. Now, if this was just a less than, not less than or equal sign, we would not include the actual line. And the convention to do that is to actually make the line a dashed line. This is the situation if we were dealing with just less than 4x plus 3, because in that situation, this wouldn't apply, and we would just have that. So the line itself wouldn't have satisfied it, just the area below it. Let's do one like that."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "And the convention to do that is to actually make the line a dashed line. This is the situation if we were dealing with just less than 4x plus 3, because in that situation, this wouldn't apply, and we would just have that. So the line itself wouldn't have satisfied it, just the area below it. Let's do one like that. So let's say we have y is greater than negative x over 2 minus 6. So a good way to start, the way I like to start these problems, is to just graph this equation right here. So let me just graph, just for fun, y is equal to, this is the same thing as negative 1 half minus 6."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Let's do one like that. So let's say we have y is greater than negative x over 2 minus 6. So a good way to start, the way I like to start these problems, is to just graph this equation right here. So let me just graph, just for fun, y is equal to, this is the same thing as negative 1 half minus 6. So if we were to graph it, that is my vertical axis, that is my horizontal axis, and our y-intercept is negative 6. So 1, 2, 3, 4, 5, 6. So that's my y-intercept."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So let me just graph, just for fun, y is equal to, this is the same thing as negative 1 half minus 6. So if we were to graph it, that is my vertical axis, that is my horizontal axis, and our y-intercept is negative 6. So 1, 2, 3, 4, 5, 6. So that's my y-intercept. And my slope is negative 1 half. Well, that should be an x there. Negative 1 half x minus 6."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So that's my y-intercept. And my slope is negative 1 half. Well, that should be an x there. Negative 1 half x minus 6. So my slope is negative 1 half, which means when I go 2 to the right, I go down 1. So if I go 2 to the right, I'm going to go down 1. So 2 to the right, down 1."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Negative 1 half x minus 6. So my slope is negative 1 half, which means when I go 2 to the right, I go down 1. So if I go 2 to the right, I'm going to go down 1. So 2 to the right, down 1. If I go 2 to the left, if I go negative 2, I'm going to go up 1. So negative 2, up 1. So my line is going to look like this."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So 2 to the right, down 1. If I go 2 to the left, if I go negative 2, I'm going to go up 1. So negative 2, up 1. So my line is going to look like this. My line is going to look like that. That's my best attempt at drawing the line. So that's the line of y is equal to negative 1 half x minus 6."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So my line is going to look like this. My line is going to look like that. That's my best attempt at drawing the line. So that's the line of y is equal to negative 1 half x minus 6. Now, our inequality is not greater than or equal. It's just greater than negative x over 2 minus 6, or greater than negative 1 half x minus 6. So using the same logic as before, for any x, let's say that's a particular x we want to pick."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So that's the line of y is equal to negative 1 half x minus 6. Now, our inequality is not greater than or equal. It's just greater than negative x over 2 minus 6, or greater than negative 1 half x minus 6. So using the same logic as before, for any x, let's say that's a particular x we want to pick. If you evaluate negative x over 2 minus 6, you're going to get that point right there. You're going to get the point on the line. But the y's that satisfy this inequality are the y's greater than that."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So using the same logic as before, for any x, let's say that's a particular x we want to pick. If you evaluate negative x over 2 minus 6, you're going to get that point right there. You're going to get the point on the line. But the y's that satisfy this inequality are the y's greater than that. So it's going to be not that point. In fact, you would draw an open circle there, because you can't include the point of negative 1 half x minus 6. But it's going to be all the y's greater than that."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "But the y's that satisfy this inequality are the y's greater than that. So it's going to be not that point. In fact, you would draw an open circle there, because you can't include the point of negative 1 half x minus 6. But it's going to be all the y's greater than that. So it's going to be all the y's greater than that. And that would be true for any x. You take this x, you evaluate negative 1 half or negative x over 2 minus 6, you're going to get this point over here."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "But it's going to be all the y's greater than that. So it's going to be all the y's greater than that. And that would be true for any x. You take this x, you evaluate negative 1 half or negative x over 2 minus 6, you're going to get this point over here. But the y's that satisfy it are all the y's above that. So all of the y's that satisfy this equation, or all of the coordinates that satisfy the equation, is this entire area above the line. And we're not going to include the line."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "You take this x, you evaluate negative 1 half or negative x over 2 minus 6, you're going to get this point over here. But the y's that satisfy it are all the y's above that. So all of the y's that satisfy this equation, or all of the coordinates that satisfy the equation, is this entire area above the line. And we're not going to include the line. So the convention is to make this line into a dashed line. And let me try my best to turn it into a dashed line. I'll just erase sections of the line."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "And we're not going to include the line. So the convention is to make this line into a dashed line. And let me try my best to turn it into a dashed line. I'll just erase sections of the line. And hopefully it will look dashed to you. So I'm turning that solid line into a dashed line to show that it's just a boundary, but it's not included in the coordinates that satisfy our inequality. The coordinates that satisfy our inequality are all of this yellow stuff that I'm shading above the line."}, {"video_title": "Example 2 Factoring quadratics by grouping   Algebra I   Khan Academy (2).mp3", "Sentence": "And whenever you have an expression like this, where you have a non-1 coefficient on the y squared, or on the second degree term, it could have been an x squared, the best way to do this is by grouping. And to factor by grouping, we need to look for 2 numbers whose product is equal to 4 times negative 15. So we're looking for 2 numbers whose product, let's call those a and b, is going to be equal to 4 times negative 15. 4 times negative 15, or negative 60. And the sum of those 2 numbers, a plus b, needs to be equal to this 4 right there. Needs to be equal to 4. So let's think about all the factors of negative 60 or 60."}, {"video_title": "Example 2 Factoring quadratics by grouping   Algebra I   Khan Academy (2).mp3", "Sentence": "4 times negative 15, or negative 60. And the sum of those 2 numbers, a plus b, needs to be equal to this 4 right there. Needs to be equal to 4. So let's think about all the factors of negative 60 or 60. And we're looking for ones that are essentially 4 apart, because the numbers are going to be of different signs, because their product is negative. So when you take 2 numbers of different signs and you sum them, you kind of view it as the difference of their absolute values. If that confuses you, don't worry about it."}, {"video_title": "Example 2 Factoring quadratics by grouping   Algebra I   Khan Academy (2).mp3", "Sentence": "So let's think about all the factors of negative 60 or 60. And we're looking for ones that are essentially 4 apart, because the numbers are going to be of different signs, because their product is negative. So when you take 2 numbers of different signs and you sum them, you kind of view it as the difference of their absolute values. If that confuses you, don't worry about it. But this tells you that the numbers, since they're going to be of different size, their absolute values are going to be roughly 4 apart. So we could try out things like 5 and 12. 5 and negative 12, because 1 has to be negative."}, {"video_title": "Example 2 Factoring quadratics by grouping   Algebra I   Khan Academy (2).mp3", "Sentence": "If that confuses you, don't worry about it. But this tells you that the numbers, since they're going to be of different size, their absolute values are going to be roughly 4 apart. So we could try out things like 5 and 12. 5 and negative 12, because 1 has to be negative. If you add these 2, you get negative 7. If you did negative 5 and 12, you'd get positive 7. They're just still too far apart."}, {"video_title": "Example 2 Factoring quadratics by grouping   Algebra I   Khan Academy (2).mp3", "Sentence": "5 and negative 12, because 1 has to be negative. If you add these 2, you get negative 7. If you did negative 5 and 12, you'd get positive 7. They're just still too far apart. What if we tried 6 and negative 10? Then you'd get a negative 4 if you added these 2. We want a positive 4, so let's do negative 6 and 10."}, {"video_title": "Example 2 Factoring quadratics by grouping   Algebra I   Khan Academy (2).mp3", "Sentence": "They're just still too far apart. What if we tried 6 and negative 10? Then you'd get a negative 4 if you added these 2. We want a positive 4, so let's do negative 6 and 10. Negative 6 plus 10 is positive 4. So those will be our 2 numbers, negative 6 and positive 10. Now, what we want to do is we want to break up this middle term here."}, {"video_title": "Example 2 Factoring quadratics by grouping   Algebra I   Khan Academy (2).mp3", "Sentence": "We want a positive 4, so let's do negative 6 and 10. Negative 6 plus 10 is positive 4. So those will be our 2 numbers, negative 6 and positive 10. Now, what we want to do is we want to break up this middle term here. The whole point of figuring out the negative 6 and the 10 is to break up the 4y into a negative 6y and a 10y. So let's do that. So this 4y can be rewritten as negative 6y plus 10y."}, {"video_title": "Example 2 Factoring quadratics by grouping   Algebra I   Khan Academy (2).mp3", "Sentence": "Now, what we want to do is we want to break up this middle term here. The whole point of figuring out the negative 6 and the 10 is to break up the 4y into a negative 6y and a 10y. So let's do that. So this 4y can be rewritten as negative 6y plus 10y. Because if you add those, you get 4y. And on the other sides of it, you have your 4y squared, and then you have your minus 15. All I did is expand this into these 2 numbers as being the coefficients on the y."}, {"video_title": "Example 2 Factoring quadratics by grouping   Algebra I   Khan Academy (2).mp3", "Sentence": "So this 4y can be rewritten as negative 6y plus 10y. Because if you add those, you get 4y. And on the other sides of it, you have your 4y squared, and then you have your minus 15. All I did is expand this into these 2 numbers as being the coefficients on the y. If you add these, you get the 4y again. Now, this is where the grouping comes in. You group the terms."}, {"video_title": "Example 2 Factoring quadratics by grouping   Algebra I   Khan Academy (2).mp3", "Sentence": "All I did is expand this into these 2 numbers as being the coefficients on the y. If you add these, you get the 4y again. Now, this is where the grouping comes in. You group the terms. So let's see. Let me do it in a different color. So if I take these 2 guys, what can I factor out of those 2 guys?"}, {"video_title": "Example 2 Factoring quadratics by grouping   Algebra I   Khan Academy (2).mp3", "Sentence": "You group the terms. So let's see. Let me do it in a different color. So if I take these 2 guys, what can I factor out of those 2 guys? Well, there's a common factor. It looks like there's a common factor of 2y. So if we factor out 2y, we get 2y times 4y squared divided by 2y is 2y."}, {"video_title": "Example 2 Factoring quadratics by grouping   Algebra I   Khan Academy (2).mp3", "Sentence": "So if I take these 2 guys, what can I factor out of those 2 guys? Well, there's a common factor. It looks like there's a common factor of 2y. So if we factor out 2y, we get 2y times 4y squared divided by 2y is 2y. And then negative 6y divided by 2y is negative 3. Negative 3. So this group gets factored into 2y times 2y minus 3."}, {"video_title": "Example 2 Factoring quadratics by grouping   Algebra I   Khan Academy (2).mp3", "Sentence": "So if we factor out 2y, we get 2y times 4y squared divided by 2y is 2y. And then negative 6y divided by 2y is negative 3. Negative 3. So this group gets factored into 2y times 2y minus 3. Now, let's look at this other group right here. This was the whole point about breaking it up like this. And in other videos, I've explained why this works."}, {"video_title": "Example 2 Factoring quadratics by grouping   Algebra I   Khan Academy (2).mp3", "Sentence": "So this group gets factored into 2y times 2y minus 3. Now, let's look at this other group right here. This was the whole point about breaking it up like this. And in other videos, I've explained why this works. Now, here, the greatest common factor is a 5. So we can factor out a 5. So this is equal to plus 5 times 10y divided by 5 is 2y."}, {"video_title": "Example 2 Factoring quadratics by grouping   Algebra I   Khan Academy (2).mp3", "Sentence": "And in other videos, I've explained why this works. Now, here, the greatest common factor is a 5. So we can factor out a 5. So this is equal to plus 5 times 10y divided by 5 is 2y. Negative 15 divided by 5 is 3. And so we have 2y times 2y minus 3 plus 5 times 2y minus 3. So now you have 2 terms, and 2y minus 3 is a common factor to both."}, {"video_title": "Example 2 Factoring quadratics by grouping   Algebra I   Khan Academy (2).mp3", "Sentence": "So this is equal to plus 5 times 10y divided by 5 is 2y. Negative 15 divided by 5 is 3. And so we have 2y times 2y minus 3 plus 5 times 2y minus 3. So now you have 2 terms, and 2y minus 3 is a common factor to both. So let's factor out a 2y minus 3. So this is equal to 2y minus 3 times 2y, times that 2y, plus that 5. There's no magic happening here."}, {"video_title": "Example 2 Factoring quadratics by grouping   Algebra I   Khan Academy (2).mp3", "Sentence": "So now you have 2 terms, and 2y minus 3 is a common factor to both. So let's factor out a 2y minus 3. So this is equal to 2y minus 3 times 2y, times that 2y, plus that 5. There's no magic happening here. All I did is undistribute the 2y minus 3. I factored it out of both of these guys and took it out of the parentheses. If I distributed it in, you'd get back to this expression."}, {"video_title": "Example 2 Factoring quadratics by grouping   Algebra I   Khan Academy (2).mp3", "Sentence": "There's no magic happening here. All I did is undistribute the 2y minus 3. I factored it out of both of these guys and took it out of the parentheses. If I distributed it in, you'd get back to this expression. But we're done. We factored it. We factored it into 2 binomial expressions."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "And just to start ourselves out, let's look at some examples of linear equations. So for example, the equation y is equal to two x minus three. This is a linear equation. Now why do we call it a linear equation? Well if you were to take the set of all of the xy pairs that satisfy this equation, and if you were to graph them on the coordinate plane, you would actually get a line. That's why it's called a linear equation. Let's actually feel good about that statement."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "Now why do we call it a linear equation? Well if you were to take the set of all of the xy pairs that satisfy this equation, and if you were to graph them on the coordinate plane, you would actually get a line. That's why it's called a linear equation. Let's actually feel good about that statement. Let's see if, let's plot some of the xy pairs that satisfy this equation, and then feel good that it does indeed generate a line. So I'm just gonna pick some x values that make it easy to calculate the corresponding y values. So if x is equal to zero, y is gonna be two times zero minus three, which is negative three."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "Let's actually feel good about that statement. Let's see if, let's plot some of the xy pairs that satisfy this equation, and then feel good that it does indeed generate a line. So I'm just gonna pick some x values that make it easy to calculate the corresponding y values. So if x is equal to zero, y is gonna be two times zero minus three, which is negative three. And on our coordinate plane here, that's, we're gonna move zero in the x direction, zero in the horizontal direction, and we're gonna go down three in the vertical direction, in the y direction. So that's that point there. If x is equal to one, what is y equal to?"}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "So if x is equal to zero, y is gonna be two times zero minus three, which is negative three. And on our coordinate plane here, that's, we're gonna move zero in the x direction, zero in the horizontal direction, and we're gonna go down three in the vertical direction, in the y direction. So that's that point there. If x is equal to one, what is y equal to? Well two times one is two minus three is negative one. So we move positive one in the x direction, and negative one, or down one, in the y direction. Now let's see, if x is equal to two, what is y?"}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "If x is equal to one, what is y equal to? Well two times one is two minus three is negative one. So we move positive one in the x direction, and negative one, or down one, in the y direction. Now let's see, if x is equal to two, what is y? Two times two is four, minus three is one. When x is equal to two, y is equal to one. And hopefully you're seeing now, that if I were to keep going, and I encourage you to, if you want, pause the video, try x equals three, or x equals negative one, and keep going, you will see that this is going to generate a line."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "Now let's see, if x is equal to two, what is y? Two times two is four, minus three is one. When x is equal to two, y is equal to one. And hopefully you're seeing now, that if I were to keep going, and I encourage you to, if you want, pause the video, try x equals three, or x equals negative one, and keep going, you will see that this is going to generate a line. And in fact, let me connect these dots, and you will see, you will see the line that I'm talking about. So, let me see if I can draw, I'm gonna use the line tool here. Try to connect the dots as neatly as I can."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "And hopefully you're seeing now, that if I were to keep going, and I encourage you to, if you want, pause the video, try x equals three, or x equals negative one, and keep going, you will see that this is going to generate a line. And in fact, let me connect these dots, and you will see, you will see the line that I'm talking about. So, let me see if I can draw, I'm gonna use the line tool here. Try to connect the dots as neatly as I can. There you go. This line that I have just drawn, this is the graph, this is the graph of y is equal to two x minus three. So if you were to graph all of the xy pairs that satisfy this equation, you are going to get this line."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "Try to connect the dots as neatly as I can. There you go. This line that I have just drawn, this is the graph, this is the graph of y is equal to two x minus three. So if you were to graph all of the xy pairs that satisfy this equation, you are going to get this line. And you might be saying, hey wait, wait, hold on Sal, you just tried some particular points, why don't I just get a bunch of points, how do I actually get a line? Well I just tried, over here, I just tried integer values of x, but you could try any value in between here, all of these, it's actually a pretty neat concept. Any value of x that you input into this, you find the corresponding value for y, it will sit on this line."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "So if you were to graph all of the xy pairs that satisfy this equation, you are going to get this line. And you might be saying, hey wait, wait, hold on Sal, you just tried some particular points, why don't I just get a bunch of points, how do I actually get a line? Well I just tried, over here, I just tried integer values of x, but you could try any value in between here, all of these, it's actually a pretty neat concept. Any value of x that you input into this, you find the corresponding value for y, it will sit on this line. So for example, for example, if we were to take x is equal to, actually let's say x is equal to negative.5. So if x is equal to negative.5, if we look at the line, when x is equal to negative.5, it looks like, it looks like y is equal to negative four. That looks like that sits on the line."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "Any value of x that you input into this, you find the corresponding value for y, it will sit on this line. So for example, for example, if we were to take x is equal to, actually let's say x is equal to negative.5. So if x is equal to negative.5, if we look at the line, when x is equal to negative.5, it looks like, it looks like y is equal to negative four. That looks like that sits on the line. Well let's verify that. If x is equal to negative, I'll write that as negative 1 1 2, then what is y equal to? Let's see, two times negative 1 1 2, I'll try it out, two times negative, two times negative 1 1 2 minus three."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "That looks like that sits on the line. Well let's verify that. If x is equal to negative, I'll write that as negative 1 1 2, then what is y equal to? Let's see, two times negative 1 1 2, I'll try it out, two times negative, two times negative 1 1 2 minus three. Well this is two times negative 1 1 2 is negative one minus three is indeed negative four. It is indeed negative four. So you can literally take any, for any x value that you put here and the corresponding y value, it is going to sit on the line."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "Let's see, two times negative 1 1 2, I'll try it out, two times negative, two times negative 1 1 2 minus three. Well this is two times negative 1 1 2 is negative one minus three is indeed negative four. It is indeed negative four. So you can literally take any, for any x value that you put here and the corresponding y value, it is going to sit on the line. This point right over here represents a solution to this linear equation. Let me do this in a color you can see. So this point represents a solution to a linear equation."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "So you can literally take any, for any x value that you put here and the corresponding y value, it is going to sit on the line. This point right over here represents a solution to this linear equation. Let me do this in a color you can see. So this point represents a solution to a linear equation. This point represents a solution to a linear equation. This point is not a solution to a linear equation. So if x is equal to five, then y is not going to be equal to three."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "So this point represents a solution to a linear equation. This point represents a solution to a linear equation. This point is not a solution to a linear equation. So if x is equal to five, then y is not going to be equal to three. If x is going to be equal to five, you go to the line to see what the solution to the linear equation is. If x is five, this shows us that y is going to be seven. And it's indeed, that's indeed the case."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "So if x is equal to five, then y is not going to be equal to three. If x is going to be equal to five, you go to the line to see what the solution to the linear equation is. If x is five, this shows us that y is going to be seven. And it's indeed, that's indeed the case. Two times five is 10 minus three is seven. The point, the point five comma seven is on, or it satisfies this linear equation. So if you take all of the xy pairs that satisfy it, you get a line."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "And it's indeed, that's indeed the case. Two times five is 10 minus three is seven. The point, the point five comma seven is on, or it satisfies this linear equation. So if you take all of the xy pairs that satisfy it, you get a line. That is why it's called a linear equation. Now this isn't the only way that we could write a linear equation. You could write a linear equation like, let me do this in a new color."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "So if you take all of the xy pairs that satisfy it, you get a line. That is why it's called a linear equation. Now this isn't the only way that we could write a linear equation. You could write a linear equation like, let me do this in a new color. You could write a linear equation like this. Four x minus three y is equal to 12. This also is a linear equation."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "You could write a linear equation like, let me do this in a new color. You could write a linear equation like this. Four x minus three y is equal to 12. This also is a linear equation. And we can see that if we were to graph the xy pairs that satisfy this, we would once again get a line, x and y. If x is equal to zero, then this goes away. You have negative three y is equal to 12."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "This also is a linear equation. And we can see that if we were to graph the xy pairs that satisfy this, we would once again get a line, x and y. If x is equal to zero, then this goes away. You have negative three y is equal to 12. See if negative three y equals 12, then y would be equal to negative four. Negative zero comma negative four. You can verify that."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "You have negative three y is equal to 12. See if negative three y equals 12, then y would be equal to negative four. Negative zero comma negative four. You can verify that. Four times zero minus three times negative four, well that's going to be equal to positive 12. And let's see, if y were to equal zero, if y were to equal zero, then this is going to be four times x is equal to 12. Well then x is equal to three."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "You can verify that. Four times zero minus three times negative four, well that's going to be equal to positive 12. And let's see, if y were to equal zero, if y were to equal zero, then this is going to be four times x is equal to 12. Well then x is equal to three. And so you have the point zero comma negative four. Zero comma negative four on this line. And you have the point three comma zero on this line."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "Well then x is equal to three. And so you have the point zero comma negative four. Zero comma negative four on this line. And you have the point three comma zero on this line. Three comma zero. Did I do that right? Yep."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "And you have the point three comma zero on this line. Three comma zero. Did I do that right? Yep. So zero comma negative four, and then three comma zero. These are both going to be on this line. Three comma zero."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "Yep. So zero comma negative four, and then three comma zero. These are both going to be on this line. Three comma zero. Is also on this line. So this line is going to look something like, I'll just try to hand draw it, something like that. So once again, all of the xy pairs that satisfy this, if you were to plot them out, it forms a line."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "Three comma zero. Is also on this line. So this line is going to look something like, I'll just try to hand draw it, something like that. So once again, all of the xy pairs that satisfy this, if you were to plot them out, it forms a line. Now what are some examples of, maybe you're saying, well isn't any equation a linear equation? And the simple answer is no. Not any equation is a linear equation."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "So once again, all of the xy pairs that satisfy this, if you were to plot them out, it forms a line. Now what are some examples of, maybe you're saying, well isn't any equation a linear equation? And the simple answer is no. Not any equation is a linear equation. I'll give you some examples of nonlinear equations. So nonlinear, whoops, let me write a little bit neater than that. Nonlinear equations."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "Not any equation is a linear equation. I'll give you some examples of nonlinear equations. So nonlinear, whoops, let me write a little bit neater than that. Nonlinear equations. Well, those could include something like y is equal to x squared. If you graph this, you'll see that this is going to be a curve. It could be something like x times y is equal to 12."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "Nonlinear equations. Well, those could include something like y is equal to x squared. If you graph this, you'll see that this is going to be a curve. It could be something like x times y is equal to 12. This is also not going to be a line. Or it could be something like five over x plus y is equal to 10. This also is not going to be a line."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "It could be something like x times y is equal to 12. This is also not going to be a line. Or it could be something like five over x plus y is equal to 10. This also is not going to be a line. So now, and at some point, I encourage you to try to graph these things. These are actually quite interesting. But given that we've now seen examples of linear equations and nonlinear equations, let's see if we can come up with a definition for linear equations."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "This also is not going to be a line. So now, and at some point, I encourage you to try to graph these things. These are actually quite interesting. But given that we've now seen examples of linear equations and nonlinear equations, let's see if we can come up with a definition for linear equations. One way to think about it is it's an equation that if you were to graph all the x and y pairs that satisfy this equation, you'll get a line. And that's actually literally where the word linear equation comes from. But another way to think about it is it's going to be an equation where every term is either going to be a constant."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "But given that we've now seen examples of linear equations and nonlinear equations, let's see if we can come up with a definition for linear equations. One way to think about it is it's an equation that if you were to graph all the x and y pairs that satisfy this equation, you'll get a line. And that's actually literally where the word linear equation comes from. But another way to think about it is it's going to be an equation where every term is either going to be a constant. So for example, 12 is a constant. It's not going to change based on the value of some variable. 12 is 12."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "But another way to think about it is it's going to be an equation where every term is either going to be a constant. So for example, 12 is a constant. It's not going to change based on the value of some variable. 12 is 12. Or negative three is negative three. So every term is either going to be a constant or it's going to be a constant times a variable raised to the first power. So this is a constant two times x to the first power."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "12 is 12. Or negative three is negative three. So every term is either going to be a constant or it's going to be a constant times a variable raised to the first power. So this is a constant two times x to the first power. This is the variable y raised to the first power. You could say this is just one y. We're not dividing by x or y."}, {"video_title": "Two-variable linear equations and their graphs   Algebra I   Khan Academy.mp3", "Sentence": "So this is a constant two times x to the first power. This is the variable y raised to the first power. You could say this is just one y. We're not dividing by x or y. We're not multiplying, or we don't have a term that has x to the second power or x to the third power or y to the fifth power. We just have y to the first power. We have x to the first power."}, {"video_title": "Independent and dependent systems   Algebra II   Khan Academy.mp3", "Sentence": "Is the system of linear equations below dependent or independent? And they give us two equations right here. Before I tackle this specific problem, let's just do a little bit of a review of what dependent or independent means. And actually I'll compare that to consistent and inconsistent. So just to start off with, if we're dealing with systems of linear equations in two dimensions, there's only three possibilities that the lines or the equations can have relative to each other. So let me draw the three possibilities. So let me draw three coordinate axes."}, {"video_title": "Independent and dependent systems   Algebra II   Khan Academy.mp3", "Sentence": "And actually I'll compare that to consistent and inconsistent. So just to start off with, if we're dealing with systems of linear equations in two dimensions, there's only three possibilities that the lines or the equations can have relative to each other. So let me draw the three possibilities. So let me draw three coordinate axes. So that's my first x-axis and y-axis. So x and y. Let me draw another one."}, {"video_title": "Independent and dependent systems   Algebra II   Khan Academy.mp3", "Sentence": "So let me draw three coordinate axes. So that's my first x-axis and y-axis. So x and y. Let me draw another one. That is x and that is y. Let me draw one more because there's only three possibilities in two dimensions. x and y, if we're dealing with linear equations."}, {"video_title": "Independent and dependent systems   Algebra II   Khan Academy.mp3", "Sentence": "Let me draw another one. That is x and that is y. Let me draw one more because there's only three possibilities in two dimensions. x and y, if we're dealing with linear equations. x and y. So you can have the situation where the lines just intersect in one point. So you could have one line like that and maybe the other line does something like that."}, {"video_title": "Independent and dependent systems   Algebra II   Khan Academy.mp3", "Sentence": "x and y, if we're dealing with linear equations. x and y. So you can have the situation where the lines just intersect in one point. So you could have one line like that and maybe the other line does something like that. And they intersect at one point. You could have the situation where the two lines are parallel. So you could have a situation, actually let me draw it over here, where you have one line that goes like that and the other line has the same slope but it's shifted."}, {"video_title": "Independent and dependent systems   Algebra II   Khan Academy.mp3", "Sentence": "So you could have one line like that and maybe the other line does something like that. And they intersect at one point. You could have the situation where the two lines are parallel. So you could have a situation, actually let me draw it over here, where you have one line that goes like that and the other line has the same slope but it's shifted. It has a different y-intercept so maybe it looks like this. And you have no points of intersection. And then you could have the situation where they're actually the same line."}, {"video_title": "Independent and dependent systems   Algebra II   Khan Academy.mp3", "Sentence": "So you could have a situation, actually let me draw it over here, where you have one line that goes like that and the other line has the same slope but it's shifted. It has a different y-intercept so maybe it looks like this. And you have no points of intersection. And then you could have the situation where they're actually the same line. So that both lines have the same slope and the same y-intercept. So really they are the same line. They intersect on an infinite number of points."}, {"video_title": "Independent and dependent systems   Algebra II   Khan Academy.mp3", "Sentence": "And then you could have the situation where they're actually the same line. So that both lines have the same slope and the same y-intercept. So really they are the same line. They intersect on an infinite number of points. Every point on either of those lines is also a point on the other line. So just to give you a little bit of the terminology here, and we learned this in the last video, this type of system where they don't intersect, where you have no solutions, this is an inconsistent system. And by definition, or I guess just taking the opposite of inconsistent, both of these would be considered consistent."}, {"video_title": "Independent and dependent systems   Algebra II   Khan Academy.mp3", "Sentence": "They intersect on an infinite number of points. Every point on either of those lines is also a point on the other line. So just to give you a little bit of the terminology here, and we learned this in the last video, this type of system where they don't intersect, where you have no solutions, this is an inconsistent system. And by definition, or I guess just taking the opposite of inconsistent, both of these would be considered consistent. But then within consistent, there's obviously a difference. Here we only have one solution. These are two different lines that intersect in one place."}, {"video_title": "Independent and dependent systems   Algebra II   Khan Academy.mp3", "Sentence": "And by definition, or I guess just taking the opposite of inconsistent, both of these would be considered consistent. But then within consistent, there's obviously a difference. Here we only have one solution. These are two different lines that intersect in one place. They're essentially the same exact line. And so we differentiate between these two scenarios by calling this one over here independent. And this one over here dependent."}, {"video_title": "Independent and dependent systems   Algebra II   Khan Academy.mp3", "Sentence": "These are two different lines that intersect in one place. They're essentially the same exact line. And so we differentiate between these two scenarios by calling this one over here independent. And this one over here dependent. So independent, both lines are doing their own thing. They're not dependent on each other. They're not the same line."}, {"video_title": "Independent and dependent systems   Algebra II   Khan Academy.mp3", "Sentence": "And this one over here dependent. So independent, both lines are doing their own thing. They're not dependent on each other. They're not the same line. They will intersect at one place. Dependent, they're the exact same line. Any point that satisfies one line will satisfy the other."}, {"video_title": "Independent and dependent systems   Algebra II   Khan Academy.mp3", "Sentence": "They're not the same line. They will intersect at one place. Dependent, they're the exact same line. Any point that satisfies one line will satisfy the other. Any point that satisfies one equation will satisfy the other. So with that said, let's see if this system of linear equations right here is dependent or independent. So they're kind of having us assume that it's going to be consistent, that we're either going to intersect in one place or we're going to intersect in an infinite number of places."}, {"video_title": "Independent and dependent systems   Algebra II   Khan Academy.mp3", "Sentence": "Any point that satisfies one line will satisfy the other. Any point that satisfies one equation will satisfy the other. So with that said, let's see if this system of linear equations right here is dependent or independent. So they're kind of having us assume that it's going to be consistent, that we're either going to intersect in one place or we're going to intersect in an infinite number of places. And the easiest way to do this, we already have this second equation here. It's already in slope-intercept form. We know the slope is negative 2."}, {"video_title": "Independent and dependent systems   Algebra II   Khan Academy.mp3", "Sentence": "So they're kind of having us assume that it's going to be consistent, that we're either going to intersect in one place or we're going to intersect in an infinite number of places. And the easiest way to do this, we already have this second equation here. It's already in slope-intercept form. We know the slope is negative 2. The y-intercept is 8. So let's put this first equation up here in slope-intercept form and see if it has a different slope or a different intercept or maybe it's the same line. So we have 4x plus 2y is equal to 16."}, {"video_title": "Independent and dependent systems   Algebra II   Khan Academy.mp3", "Sentence": "We know the slope is negative 2. The y-intercept is 8. So let's put this first equation up here in slope-intercept form and see if it has a different slope or a different intercept or maybe it's the same line. So we have 4x plus 2y is equal to 16. We can subtract 4x from both sides. What we want to do is isolate the y on the left-hand side. So let's subtract 4x from both sides."}, {"video_title": "Independent and dependent systems   Algebra II   Khan Academy.mp3", "Sentence": "So we have 4x plus 2y is equal to 16. We can subtract 4x from both sides. What we want to do is isolate the y on the left-hand side. So let's subtract 4x from both sides. The left-hand side, we are just left with a 2y. The right-hand side, we have a negative 4x plus 16. I just wrote the negative 4 in front of the 16 just so that we have it in the traditional slope-intercept form."}, {"video_title": "Independent and dependent systems   Algebra II   Khan Academy.mp3", "Sentence": "So let's subtract 4x from both sides. The left-hand side, we are just left with a 2y. The right-hand side, we have a negative 4x plus 16. I just wrote the negative 4 in front of the 16 just so that we have it in the traditional slope-intercept form. And now we can divide both sides of this equation by 2 so that we can isolate the y on the left-hand side. Divide both sides by 2. We are left with y is equal to negative 4 divided by 2 is negative 2x plus 16 over 2, plus 8."}, {"video_title": "Independent and dependent systems   Algebra II   Khan Academy.mp3", "Sentence": "I just wrote the negative 4 in front of the 16 just so that we have it in the traditional slope-intercept form. And now we can divide both sides of this equation by 2 so that we can isolate the y on the left-hand side. Divide both sides by 2. We are left with y is equal to negative 4 divided by 2 is negative 2x plus 16 over 2, plus 8. So all I did is algebraically manipulated this top equation up here. And when I did that, when I solved essentially for y, I got this right over here, which is the exact same thing as the second equation. We have the exact same slope, negative 2, negative 2, and we have the exact same y-intercept, 8 and 8."}, {"video_title": "Independent and dependent systems   Algebra II   Khan Academy.mp3", "Sentence": "We are left with y is equal to negative 4 divided by 2 is negative 2x plus 16 over 2, plus 8. So all I did is algebraically manipulated this top equation up here. And when I did that, when I solved essentially for y, I got this right over here, which is the exact same thing as the second equation. We have the exact same slope, negative 2, negative 2, and we have the exact same y-intercept, 8 and 8. If I were to graph these equations, that's my x-axis and that is my y-axis, both of them have a y-intercept at 8 and then have a slope of negative 2. So they look something... I'm just drawing an approximation of it, but they would look something like that."}, {"video_title": "Independent and dependent systems   Algebra II   Khan Academy.mp3", "Sentence": "We have the exact same slope, negative 2, negative 2, and we have the exact same y-intercept, 8 and 8. If I were to graph these equations, that's my x-axis and that is my y-axis, both of them have a y-intercept at 8 and then have a slope of negative 2. So they look something... I'm just drawing an approximation of it, but they would look something like that. So maybe this is the graph of this equation right here, this first equation. And then the second equation will be the exact same graph. It has the exact same y-intercept and the exact same slope."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "All right, now we have a very interesting situation. On both sides of the scale, we have our mystery mass. And now I'm calling the mystery mass having a mass of y, just to show you that it doesn't always have to be x. It can be any symbol, as long as you can keep track of that symbol. But all of these have the same mass. That's why I wrote y on all of them. And we also have the little one kilogram boxes on both sides of this scale."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "It can be any symbol, as long as you can keep track of that symbol. But all of these have the same mass. That's why I wrote y on all of them. And we also have the little one kilogram boxes on both sides of this scale. So the first thing I wanna do, we're gonna go step by step and try to figure out what this mystery mass is. But the first thing I wanna do is have you think about whether you can represent this algebraically, whether with a little bit of mathematic symbolry, you can represent what's going on in this scale. Over here, I have three y's and three of these boxes, and their total mass is equal to this one y."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "And we also have the little one kilogram boxes on both sides of this scale. So the first thing I wanna do, we're gonna go step by step and try to figure out what this mystery mass is. But the first thing I wanna do is have you think about whether you can represent this algebraically, whether with a little bit of mathematic symbolry, you can represent what's going on in this scale. Over here, I have three y's and three of these boxes, and their total mass is equal to this one y. And I think I have about, let's see, I have seven boxes right over here. So I'll give you a few seconds to do that. So let's think about the total mass over here."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "Over here, I have three y's and three of these boxes, and their total mass is equal to this one y. And I think I have about, let's see, I have seven boxes right over here. So I'll give you a few seconds to do that. So let's think about the total mass over here. We have three boxes of mass y. So they're going to have a mass of three y. And then you have three boxes with a mass of one kilogram."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "So let's think about the total mass over here. We have three boxes of mass y. So they're going to have a mass of three y. And then you have three boxes with a mass of one kilogram. So they're going to have a mass of three kilograms. Now over here, I have one box with a mass of y kilograms. So that's going to be my one y right over there."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "And then you have three boxes with a mass of one kilogram. So they're going to have a mass of three kilograms. Now over here, I have one box with a mass of y kilograms. So that's going to be my one y right over there. I could have written one y, but I don't need to. A y is the same thing as one y. So I have the y kilograms right there."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "So that's going to be my one y right over there. I could have written one y, but I don't need to. A y is the same thing as one y. So I have the y kilograms right there. And then I have seven of these, right? One, two, three, four, five, six, seven. Yep, seven of these."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "So I have the y kilograms right there. And then I have seven of these, right? One, two, three, four, five, six, seven. Yep, seven of these. So I have y plus seven kilograms on the right-hand side. And once again, it's balanced. The scale is balanced."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "Yep, seven of these. So I have y plus seven kilograms on the right-hand side. And once again, it's balanced. The scale is balanced. This total mass is equal to this total mass. So we can write an equal sign right over there. So that's a good starting point."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "The scale is balanced. This total mass is equal to this total mass. So we can write an equal sign right over there. So that's a good starting point. We were able to represent this situation, this real-life situation, you know, back in the day when people actually had to figure out the mass of things, if you were to go to the jewelry store or whatever. They actually did have problems like this. But we were able to represent it mathematically."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "So that's a good starting point. We were able to represent this situation, this real-life situation, you know, back in the day when people actually had to figure out the mass of things, if you were to go to the jewelry store or whatever. They actually did have problems like this. But we were able to represent it mathematically. Now, the next thing to do is, what are some reasonable next steps? How can we start to simplify this a little bit? And once again, I'll give you a few seconds to think about that."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "But we were able to represent it mathematically. Now, the next thing to do is, what are some reasonable next steps? How can we start to simplify this a little bit? And once again, I'll give you a few seconds to think about that. Well, the neat thing about algebra is there's actually multiple paths that you could go down. You might say, well, why don't we remove three of these yellow blocks from both sides? That would be completely legitimate."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "And once again, I'll give you a few seconds to think about that. Well, the neat thing about algebra is there's actually multiple paths that you could go down. You might say, well, why don't we remove three of these yellow blocks from both sides? That would be completely legitimate. You might say, well, why don't we remove one of these y's from both sides? That also would be legitimate. And we can do it in either order."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "That would be completely legitimate. You might say, well, why don't we remove one of these y's from both sides? That also would be legitimate. And we can do it in either order. So let's just pick one of them. Let's say that we first want to remove, let's say that we first want to remove a y from either side, just so that we feel a little bit more comfortable with all of our y's sitting on only one side. And so the best way, if we want all of our y's to sit on one side, we can remove a y from each side."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "And we can do it in either order. So let's just pick one of them. Let's say that we first want to remove, let's say that we first want to remove a y from either side, just so that we feel a little bit more comfortable with all of our y's sitting on only one side. And so the best way, if we want all of our y's to sit on one side, we can remove a y from each side. Remember, if we removed a y from only one side, that would unbalance the scale. The scale was already balanced. Whatever I have to do to one side, I have to do to the other."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "And so the best way, if we want all of our y's to sit on one side, we can remove a y from each side. Remember, if we removed a y from only one side, that would unbalance the scale. The scale was already balanced. Whatever I have to do to one side, I have to do to the other. So I'm going to remove a y. I'm going to remove y mass from both sides. Now, what would that look like algebraically? Well, I removed a y from both sides."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "Whatever I have to do to one side, I have to do to the other. So I'm going to remove a y. I'm going to remove y mass from both sides. Now, what would that look like algebraically? Well, I removed a y from both sides. So I subtracted y from the left-hand side, and I subtracted y from the right-hand side. That's exactly what I did. It had a mass of y. I don't know what that is, but I did take it away."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "Well, I removed a y from both sides. So I subtracted y from the left-hand side, and I subtracted y from the right-hand side. That's exactly what I did. It had a mass of y. I don't know what that is, but I did take it away. I lifted that little block. And so on the left-hand side, what am I left with? And you could think of it mathematically, but you could even look up here and see what you're left with."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "It had a mass of y. I don't know what that is, but I did take it away. I lifted that little block. And so on the left-hand side, what am I left with? And you could think of it mathematically, but you could even look up here and see what you're left with. If I have three of something, and I take away one of them, I'm left with two of that something. So I'm left with two y right over here. And you see it."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "And you could think of it mathematically, but you could even look up here and see what you're left with. If I have three of something, and I take away one of them, I'm left with two of that something. So I'm left with two y right over here. And you see it. I had three. I got rid of one, so I'm left with two. And I still have those three yellow blocks."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "And you see it. I had three. I got rid of one, so I'm left with two. And I still have those three yellow blocks. On the right-hand side, I had a y. I took away the y. And so now I have no y's left. And we see it visually right over here."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "And I still have those three yellow blocks. On the right-hand side, I had a y. I took away the y. And so now I have no y's left. And we see it visually right over here. But I still have seven of the yellow blocks. So I still have seven of the yellow blocks. And since I took away the exact same mass from both sides of the scale, the scale is still going to be balanced."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "And we see it visually right over here. But I still have seven of the yellow blocks. So I still have seven of the yellow blocks. And since I took away the exact same mass from both sides of the scale, the scale is still going to be balanced. It was balanced before. I took away the same thing from both sides. And so the scale is still balanced."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "And since I took away the exact same mass from both sides of the scale, the scale is still going to be balanced. It was balanced before. I took away the same thing from both sides. And so the scale is still balanced. So this is going to be equal to that. Now this is starting to look a little bit similar to what we saw in the last video. But I will ask you, what can we do from this point to simplify it further?"}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "And so the scale is still balanced. So this is going to be equal to that. Now this is starting to look a little bit similar to what we saw in the last video. But I will ask you, what can we do from this point to simplify it further? Or so even better, think of it so we can isolate these y's on the left-hand side. And I'll give you a few seconds to think about that. Well, if we want to isolate these y's on the left-hand side, these two y's, the best way is to get rid of this three, to get rid of these three blocks."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "But I will ask you, what can we do from this point to simplify it further? Or so even better, think of it so we can isolate these y's on the left-hand side. And I'll give you a few seconds to think about that. Well, if we want to isolate these y's on the left-hand side, these two y's, the best way is to get rid of this three, to get rid of these three blocks. So why don't we do that? Let's take three blocks from this side. But we can't just take it from that side if we want to keep it balanced."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "Well, if we want to isolate these y's on the left-hand side, these two y's, the best way is to get rid of this three, to get rid of these three blocks. So why don't we do that? Let's take three blocks from this side. But we can't just take it from that side if we want to keep it balanced. We have to do it to this side, too. We've got to take away three blocks. So we're subtracting three from that side and subtracting three from the right side."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "But we can't just take it from that side if we want to keep it balanced. We have to do it to this side, too. We've got to take away three blocks. So we're subtracting three from that side and subtracting three from the right side. So on the left-hand side, we're going to be left with just these two blocks of mass y. So our total mass is now going to be 2y. These 3 minus 3 is 0."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "So we're subtracting three from that side and subtracting three from the right side. So on the left-hand side, we're going to be left with just these two blocks of mass y. So our total mass is now going to be 2y. These 3 minus 3 is 0. And you see that here. We're just left with two y's right over here. And on the right-hand side, we got rid of three of the blocks."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "These 3 minus 3 is 0. And you see that here. We're just left with two y's right over here. And on the right-hand side, we got rid of three of the blocks. So we only have four of them left. So you have two of these y masses is equal to 4 kilograms. Because we did the same thing to both sides, the scale is still balanced."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "And on the right-hand side, we got rid of three of the blocks. So we only have four of them left. So you have two of these y masses is equal to 4 kilograms. Because we did the same thing to both sides, the scale is still balanced. And now, how do we solve this? And you might be able to solve this in your head. I have 2 times something is equal to 4."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "Because we did the same thing to both sides, the scale is still balanced. And now, how do we solve this? And you might be able to solve this in your head. I have 2 times something is equal to 4. You could kind of think about what that is. But if we want to stay true to what we've been doing before, let's think about it. I have 2 of something is equal to something else."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "I have 2 times something is equal to 4. You could kind of think about what that is. But if we want to stay true to what we've been doing before, let's think about it. I have 2 of something is equal to something else. What if I multiplied both sides by 2? Sorry, what if I multiplied both sides by 1 half? Or in other ways, dividing both sides by 2."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "I have 2 of something is equal to something else. What if I multiplied both sides by 2? Sorry, what if I multiplied both sides by 1 half? Or in other ways, dividing both sides by 2. If I multiply this side by 1 half, if I essentially take away half of the mass, or I only leave half of the mass, then I'm only going to have one block here. And if I take away half of the mass over here, I'm going to have to take away two of these blocks right over there. And what I just did, you could say I multiplied both sides by 1 half."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "Or in other ways, dividing both sides by 2. If I multiply this side by 1 half, if I essentially take away half of the mass, or I only leave half of the mass, then I'm only going to have one block here. And if I take away half of the mass over here, I'm going to have to take away two of these blocks right over there. And what I just did, you could say I multiplied both sides by 1 half. Or just for the sake of a little change, you could say I divided both sides by 2. And on the left-hand side, I'm left with a mass of y. And on the right-hand side, I'm left with a mass of 4 divided by 2 is 2."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "And what I just did, you could say I multiplied both sides by 1 half. Or just for the sake of a little change, you could say I divided both sides by 2. And on the left-hand side, I'm left with a mass of y. And on the right-hand side, I'm left with a mass of 4 divided by 2 is 2. And once again, I can still write this equal sign because the scale is balanced. I did the exact same thing to both sides. I left half of what was on the left-hand side and half of what was on the right-hand side."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "And on the right-hand side, I'm left with a mass of 4 divided by 2 is 2. And once again, I can still write this equal sign because the scale is balanced. I did the exact same thing to both sides. I left half of what was on the left-hand side and half of what was on the right-hand side. It was balanced before, half of each side, so it's going to be balanced again. But there we've done it. We've solved something that's actually not so easy to solve, or might not look so easy at first."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "I left half of what was on the left-hand side and half of what was on the right-hand side. It was balanced before, half of each side, so it's going to be balanced again. But there we've done it. We've solved something that's actually not so easy to solve, or might not look so easy at first. We figured out that our mystery mass, y, is 2 kilograms. And you can verify this. This is the really fun thing about algebra, is that once you get to this point, you can go back and think about whether the original problem we saw made sense."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "We've solved something that's actually not so easy to solve, or might not look so easy at first. We figured out that our mystery mass, y, is 2 kilograms. And you can verify this. This is the really fun thing about algebra, is that once you get to this point, you can go back and think about whether the original problem we saw made sense. Let's do that. Let's think about whether the original problem made sense. And to do that, I want you to calculate, now that we know what the mass, y, is, is 2 kilograms, what was the total mass on each side?"}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "This is the really fun thing about algebra, is that once you get to this point, you can go back and think about whether the original problem we saw made sense. Let's do that. Let's think about whether the original problem made sense. And to do that, I want you to calculate, now that we know what the mass, y, is, is 2 kilograms, what was the total mass on each side? Well, let's calculate it. You have 2, I'll write it right over here. This is 2 kilograms."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "And to do that, I want you to calculate, now that we know what the mass, y, is, is 2 kilograms, what was the total mass on each side? Well, let's calculate it. You have 2, I'll write it right over here. This is 2 kilograms. I'll do it in purple color. So this is a 2. So we had 6 kilograms plus these 3."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "This is 2 kilograms. I'll do it in purple color. So this is a 2. So we had 6 kilograms plus these 3. You had 9 kilograms on the left-hand side. And on the right-hand side, I had these 7 plus 2 here. 7 plus 2 is 9 kilograms."}, {"video_title": "Worked example using recursive formula for arithmetic sequence   High School Math   Khan Academy.mp3", "Sentence": "We are told b of one is equal to negative seven, and b of n is equal to b of n minus one plus 12, and they're asking us to find the fourth term in the sequence. So what we have up here, which you could use a function definition, it's really defining the terms of a sequence, so especially if you were to input whole numbers in here, it's the index on your sequence. So what we really want to do is, we want to figure out what is b of four going to be equal to. Well, if we just blindly apply this, we would say, all right, b of four, so b of n is equal to b of n minus one plus 12, so it's gonna be b of four minus one plus 12, well, four minus one is just three, so it's going to be equal to b of three plus 12. All I did is said, okay, well, we're not trying to figure out, or we're not immediately trying to figure out what b of one is, we're trying to figure out what b of four is, so n is equal to four, so b of four is going to be equal to b of four minus one, or b of three plus 12. Well, to evaluate this, we have to figure out what b of three is, so let's write that down. That's what's fun about a recursive definition, we have to keep recursing backwards."}, {"video_title": "Worked example using recursive formula for arithmetic sequence   High School Math   Khan Academy.mp3", "Sentence": "Well, if we just blindly apply this, we would say, all right, b of four, so b of n is equal to b of n minus one plus 12, so it's gonna be b of four minus one plus 12, well, four minus one is just three, so it's going to be equal to b of three plus 12. All I did is said, okay, well, we're not trying to figure out, or we're not immediately trying to figure out what b of one is, we're trying to figure out what b of four is, so n is equal to four, so b of four is going to be equal to b of four minus one, or b of three plus 12. Well, to evaluate this, we have to figure out what b of three is, so let's write that down. That's what's fun about a recursive definition, we have to keep recursing backwards. So b of three, well, if n is three, that's going to be equal to b of, now n minus one is two, b of two plus 12. Well, we don't know what b of two is, let's keep going. So we need to figure out b of two, if we use the same definition, b of two is going to be equal to b of two minus one plus 12, so b of two minus one, that's b of one plus 12."}, {"video_title": "Worked example using recursive formula for arithmetic sequence   High School Math   Khan Academy.mp3", "Sentence": "That's what's fun about a recursive definition, we have to keep recursing backwards. So b of three, well, if n is three, that's going to be equal to b of, now n minus one is two, b of two plus 12. Well, we don't know what b of two is, let's keep going. So we need to figure out b of two, if we use the same definition, b of two is going to be equal to b of two minus one plus 12, so b of two minus one, that's b of one plus 12. But we don't know what b of one is, so let's figure that out. B of one is equal to, well, here, we can finally use this top clause, so b of one is equal to negative seven. So now, we can go and fill everything back in."}, {"video_title": "Worked example using recursive formula for arithmetic sequence   High School Math   Khan Academy.mp3", "Sentence": "So we need to figure out b of two, if we use the same definition, b of two is going to be equal to b of two minus one plus 12, so b of two minus one, that's b of one plus 12. But we don't know what b of one is, so let's figure that out. B of one is equal to, well, here, we can finally use this top clause, so b of one is equal to negative seven. So now, we can go and fill everything back in. If b of one is equal to negative seven, then we know that this right over here is negative seven, and now we can figure out that b of two is equal to negative seven plus 12, which is equal to five. Well, if b of two is equal to five, well, then this is equal to five right over here, and then now we know that b of three is equal to five plus 12, which is equal to 17. Well, if we know that b of three is equal to 17, then we're ready to calculate what b of four is going to be."}, {"video_title": "Order of operations PEMDAS   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "Now that we've got the basics of order of operations out of the way, let's try to tackle a really hairy and beastly problem. So here we have all sorts of parentheses and numbers flying around, but in any of these order of operations problems, you really just have to take a deep breath. And remember, we're going to do parentheses first, parentheses, p for parentheses, then exponents. Don't worry if you don't know what exponents are, because this has no exponents in them. Then you're going to do multiplication and division. They're at the same level. Then you do addition and subtraction."}, {"video_title": "Order of operations PEMDAS   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "Don't worry if you don't know what exponents are, because this has no exponents in them. Then you're going to do multiplication and division. They're at the same level. Then you do addition and subtraction. So some people remember PEMDAS. But if you remember PEMDAS, remember multiplication, division, same level, addition and subtraction, also at the same level. So let's figure what order of operations say that this should evaluate to."}, {"video_title": "Order of operations PEMDAS   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "Then you do addition and subtraction. So some people remember PEMDAS. But if you remember PEMDAS, remember multiplication, division, same level, addition and subtraction, also at the same level. So let's figure what order of operations say that this should evaluate to. So the first thing we're going to do is our parentheses, and we have a lot of parentheses here. We have this expression in parentheses right there, and then even within that, we have these parentheses. So our order of operations say, look, do your parentheses first, but in order to evaluate this outer parentheses, this orange thing, we're going to have to evaluate this thing in yellow right there."}, {"video_title": "Order of operations PEMDAS   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "So let's figure what order of operations say that this should evaluate to. So the first thing we're going to do is our parentheses, and we have a lot of parentheses here. We have this expression in parentheses right there, and then even within that, we have these parentheses. So our order of operations say, look, do your parentheses first, but in order to evaluate this outer parentheses, this orange thing, we're going to have to evaluate this thing in yellow right there. So let's evaluate this whole thing. So how can we simplify it? Well, if we look at just inside of it, the first thing we want to do is simplify the parentheses inside the parentheses."}, {"video_title": "Order of operations PEMDAS   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "So our order of operations say, look, do your parentheses first, but in order to evaluate this outer parentheses, this orange thing, we're going to have to evaluate this thing in yellow right there. So let's evaluate this whole thing. So how can we simplify it? Well, if we look at just inside of it, the first thing we want to do is simplify the parentheses inside the parentheses. So you see this 5 minus 2 right there. We're going to do that first no matter what, and that's easy to evaluate. 5 minus 2 is 3."}, {"video_title": "Order of operations PEMDAS   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "Well, if we look at just inside of it, the first thing we want to do is simplify the parentheses inside the parentheses. So you see this 5 minus 2 right there. We're going to do that first no matter what, and that's easy to evaluate. 5 minus 2 is 3. And so this simplifies to, I'll do it step by step. Once you get the hang of it, you can do multiple steps at once. So this is going to be 7 plus 3 times the 5 minus 2, which is 3."}, {"video_title": "Order of operations PEMDAS   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "5 minus 2 is 3. And so this simplifies to, I'll do it step by step. Once you get the hang of it, you can do multiple steps at once. So this is going to be 7 plus 3 times the 5 minus 2, which is 3. And then you have, and all of those have parentheses around it. And of course, you have all this stuff on either side, the divide for, no, whoops, that's not what I want. I wanted to copy and paste that right there."}, {"video_title": "Order of operations PEMDAS   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "So this is going to be 7 plus 3 times the 5 minus 2, which is 3. And then you have, and all of those have parentheses around it. And of course, you have all this stuff on either side, the divide for, no, whoops, that's not what I want. I wanted to copy and paste that right there. So copy, and then, no, that's giving me the wrong thing. It would have been easier, let me just rewrite it. That's the easiest thing."}, {"video_title": "Order of operations PEMDAS   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "I wanted to copy and paste that right there. So copy, and then, no, that's giving me the wrong thing. It would have been easier, let me just rewrite it. That's the easiest thing. I'm having technical difficulties. So divided by 4 times 2, and on this side, you had that 7 times 2 plus this thing in orange parentheses there. Now, at any step, you just look again."}, {"video_title": "Order of operations PEMDAS   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "That's the easiest thing. I'm having technical difficulties. So divided by 4 times 2, and on this side, you had that 7 times 2 plus this thing in orange parentheses there. Now, at any step, you just look again. We always want to do parentheses first. We keep wanting to do this until there's really no parentheses left. So we have to evaluate this parentheses in orange here."}, {"video_title": "Order of operations PEMDAS   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "Now, at any step, you just look again. We always want to do parentheses first. We keep wanting to do this until there's really no parentheses left. So we have to evaluate this parentheses in orange here. So we have to evaluate this thing first. But in order to evaluate this thing, we have to look inside of it. And when you look inside of it, you have 7 plus 3 times 3."}, {"video_title": "Order of operations PEMDAS   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "So we have to evaluate this parentheses in orange here. So we have to evaluate this thing first. But in order to evaluate this thing, we have to look inside of it. And when you look inside of it, you have 7 plus 3 times 3. So if you just had 7 plus 3 times 3, how would you evaluate it? Well, look back to your order of operations. We're inside the parentheses here, so inside of it, there are no longer any parentheses."}, {"video_title": "Order of operations PEMDAS   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "And when you look inside of it, you have 7 plus 3 times 3. So if you just had 7 plus 3 times 3, how would you evaluate it? Well, look back to your order of operations. We're inside the parentheses here, so inside of it, there are no longer any parentheses. So the next thing we should do is, there are no exponents, there is multiplication. So we do that before we do any addition or subtraction. So we want to do the 3 times 3 before we add the 7."}, {"video_title": "Order of operations PEMDAS   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "We're inside the parentheses here, so inside of it, there are no longer any parentheses. So the next thing we should do is, there are no exponents, there is multiplication. So we do that before we do any addition or subtraction. So we want to do the 3 times 3 before we add the 7. So this is going to be 7 plus, and the 3 times 3 we want to do first. We want to do the multiplication first. 7 plus 9."}, {"video_title": "Order of operations PEMDAS   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "So we want to do the 3 times 3 before we add the 7. So this is going to be 7 plus, and the 3 times 3 we want to do first. We want to do the multiplication first. 7 plus 9. That's going to be in the orange parentheses. And then you have the 7 times 2 plus that on the left-hand side. You have the divided by 4 times 2 on the right-hand side."}, {"video_title": "Order of operations PEMDAS   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "7 plus 9. That's going to be in the orange parentheses. And then you have the 7 times 2 plus that on the left-hand side. You have the divided by 4 times 2 on the right-hand side. And now this, the thing in parentheses, because we still want to do the parentheses first, pretty easy to evaluate. What's 7 plus 9? 7 plus 9 is 16."}, {"video_title": "Order of operations PEMDAS   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "You have the divided by 4 times 2 on the right-hand side. And now this, the thing in parentheses, because we still want to do the parentheses first, pretty easy to evaluate. What's 7 plus 9? 7 plus 9 is 16. And so everything we have simplifies to 7 times 2 plus 16 divided by 4 times 2. Now, we don't have any parentheses left, so we don't have to worry about the p in PEMDAS. We have no e, no exponents in this, so then we go straight to multiplication and division."}, {"video_title": "Order of operations PEMDAS   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "7 plus 9 is 16. And so everything we have simplifies to 7 times 2 plus 16 divided by 4 times 2. Now, we don't have any parentheses left, so we don't have to worry about the p in PEMDAS. We have no e, no exponents in this, so then we go straight to multiplication and division. We have a multiplication, we have some multiplication going on there, we have some division going on here, and a multiplication there. So we should do these next, before we do this addition right there. So we could do this multiplication, we could do that multiplication."}, {"video_title": "Order of operations PEMDAS   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "We have no e, no exponents in this, so then we go straight to multiplication and division. We have a multiplication, we have some multiplication going on there, we have some division going on here, and a multiplication there. So we should do these next, before we do this addition right there. So we could do this multiplication, we could do that multiplication. 7 times 2 is 14. We're going to wait to do that addition. And then here we have a 16 divided by 4 times 2."}, {"video_title": "Order of operations PEMDAS   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "So we could do this multiplication, we could do that multiplication. 7 times 2 is 14. We're going to wait to do that addition. And then here we have a 16 divided by 4 times 2. That gets priority of the addition, so we're going to do that before we do the addition. But how do we evaluate that? Do we do the division first, do we do the multiplication first?"}, {"video_title": "Order of operations PEMDAS   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "And then here we have a 16 divided by 4 times 2. That gets priority of the addition, so we're going to do that before we do the addition. But how do we evaluate that? Do we do the division first, do we do the multiplication first? And remember, I told you in the last video, when you have multiple operations of the same level, in this case division and multiplication, they're at the same level, you're safest going left to right, or you should go left to right. So you do 16 divided by 4 is 4, so this thing right here simplifies 16 divided by 4 times 2, it simplifies to 4 times 2, that's this thing in green right there. And then we're going to want to do the multiplication next."}, {"video_title": "Order of operations PEMDAS   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "Do we do the division first, do we do the multiplication first? And remember, I told you in the last video, when you have multiple operations of the same level, in this case division and multiplication, they're at the same level, you're safest going left to right, or you should go left to right. So you do 16 divided by 4 is 4, so this thing right here simplifies 16 divided by 4 times 2, it simplifies to 4 times 2, that's this thing in green right there. And then we're going to want to do the multiplication next. So this is going to simplify to, because multiplication takes priority of addition, this simplifies to 8, and so you get 14, this 14 right here, plus 8. And what's 14 plus 8? That is 22."}, {"video_title": "Scientific notation examples   Pre-Algebra   Khan Academy.mp3", "Sentence": "There are two whole Khan Academy videos on what scientific notation is, why we even worry about it, and it also goes through a few examples. What I want to do in this video is just use the ck12.org Algebra 1 book to do some more scientific notation examples. So let's take some things that are written in scientific notation. Just a reminder, scientific notation is useful because it allows us to write really large or really small numbers in ways that are easy for our brains to 1. write down and 2. understand. So let's write down some numbers. Let's have 3.102 times 10 to the second. I want to write it as just a numerical value."}, {"video_title": "Scientific notation examples   Pre-Algebra   Khan Academy.mp3", "Sentence": "Just a reminder, scientific notation is useful because it allows us to write really large or really small numbers in ways that are easy for our brains to 1. write down and 2. understand. So let's write down some numbers. Let's have 3.102 times 10 to the second. I want to write it as just a numerical value. It's written in scientific notation already. It's written as a product with a power of 10. So how do I write this as just a numeral?"}, {"video_title": "Scientific notation examples   Pre-Algebra   Khan Academy.mp3", "Sentence": "I want to write it as just a numerical value. It's written in scientific notation already. It's written as a product with a power of 10. So how do I write this as just a numeral? Well, there's a slow way and a fast way. The slow way is to say, well, this is the same thing as 3.102 times 100, which means if you multiply 3.102 times 100, it'll be 3102 with two zeros behind it. And then we have one, two, three numbers behind the decimal point."}, {"video_title": "Scientific notation examples   Pre-Algebra   Khan Academy.mp3", "Sentence": "So how do I write this as just a numeral? Well, there's a slow way and a fast way. The slow way is to say, well, this is the same thing as 3.102 times 100, which means if you multiply 3.102 times 100, it'll be 3102 with two zeros behind it. And then we have one, two, three numbers behind the decimal point. And that would be the right answer. This is equal to 310.2. Now, a faster way to do this is just to say, well, look, right now I have only the 3 in front of the decimal point."}, {"video_title": "Scientific notation examples   Pre-Algebra   Khan Academy.mp3", "Sentence": "And then we have one, two, three numbers behind the decimal point. And that would be the right answer. This is equal to 310.2. Now, a faster way to do this is just to say, well, look, right now I have only the 3 in front of the decimal point. When I take something to the times 10 to the second power, I'm essentially shifting the decimal point 2 to the right. So 3.102 times 10 to the second power is the same thing as, if I shift the decimal point 1 and then 2, because this is 10 to the second power, it's the same thing as 310.2. So this might be a faster way of viewing it."}, {"video_title": "Scientific notation examples   Pre-Algebra   Khan Academy.mp3", "Sentence": "Now, a faster way to do this is just to say, well, look, right now I have only the 3 in front of the decimal point. When I take something to the times 10 to the second power, I'm essentially shifting the decimal point 2 to the right. So 3.102 times 10 to the second power is the same thing as, if I shift the decimal point 1 and then 2, because this is 10 to the second power, it's the same thing as 310.2. So this might be a faster way of viewing it. Every time you multiply it by 10, you shift the decimal to the right by 1. Let's do another example. Let's say I had 7.4 times 10 to the fourth."}, {"video_title": "Scientific notation examples   Pre-Algebra   Khan Academy.mp3", "Sentence": "So this might be a faster way of viewing it. Every time you multiply it by 10, you shift the decimal to the right by 1. Let's do another example. Let's say I had 7.4 times 10 to the fourth. Well, let's just do this the fast way. Let's shift the decimal 4 to the right. So 7.4 times 10 to the fourth times 10 to the 1, you're going to get 74."}, {"video_title": "Scientific notation examples   Pre-Algebra   Khan Academy.mp3", "Sentence": "Let's say I had 7.4 times 10 to the fourth. Well, let's just do this the fast way. Let's shift the decimal 4 to the right. So 7.4 times 10 to the fourth times 10 to the 1, you're going to get 74. Then times 10 to the second, you're going to get 740. We're going to have to add a 0 there because we have to shift the decimal again. 10 to the third, you're going to have 7,400."}, {"video_title": "Scientific notation examples   Pre-Algebra   Khan Academy.mp3", "Sentence": "So 7.4 times 10 to the fourth times 10 to the 1, you're going to get 74. Then times 10 to the second, you're going to get 740. We're going to have to add a 0 there because we have to shift the decimal again. 10 to the third, you're going to have 7,400. And then 10 to the fourth, you're going to have 74,000. Notice I just took this decimal and went 1, 2, 3, 4 spaces. So this is equal to 74,000."}, {"video_title": "Scientific notation examples   Pre-Algebra   Khan Academy.mp3", "Sentence": "10 to the third, you're going to have 7,400. And then 10 to the fourth, you're going to have 74,000. Notice I just took this decimal and went 1, 2, 3, 4 spaces. So this is equal to 74,000. And when I had 74 and I had to shift the decimal 1 more to the right, I had to throw a 0 here. I'm multiplying it by 10. Another way to think about it is I need 10 spaces between the leading digit and the decimal."}, {"video_title": "Scientific notation examples   Pre-Algebra   Khan Academy.mp3", "Sentence": "So this is equal to 74,000. And when I had 74 and I had to shift the decimal 1 more to the right, I had to throw a 0 here. I'm multiplying it by 10. Another way to think about it is I need 10 spaces between the leading digit and the decimal. So right here I only have 1 space. I'll need 4 spaces. So 1, 2, 3, 4."}, {"video_title": "Scientific notation examples   Pre-Algebra   Khan Academy.mp3", "Sentence": "Another way to think about it is I need 10 spaces between the leading digit and the decimal. So right here I only have 1 space. I'll need 4 spaces. So 1, 2, 3, 4. Let's do a few more examples. I think the more examples, the more you'll get what's going on. So I have 1.75 times 10 to the negative 3."}, {"video_title": "Scientific notation examples   Pre-Algebra   Khan Academy.mp3", "Sentence": "So 1, 2, 3, 4. Let's do a few more examples. I think the more examples, the more you'll get what's going on. So I have 1.75 times 10 to the negative 3. This is in scientific notation, and I want to just write the numerical value of this. So when you take something to the negative times 10 to the negative power, you shift the decimal to the left. So this is 1.75."}, {"video_title": "Scientific notation examples   Pre-Algebra   Khan Academy.mp3", "Sentence": "So I have 1.75 times 10 to the negative 3. This is in scientific notation, and I want to just write the numerical value of this. So when you take something to the negative times 10 to the negative power, you shift the decimal to the left. So this is 1.75. So if you do it times 10 to the negative 1 power, you will go 1 to the left. But if you do times 10 to the negative 2 power, you'll go 2 to the left. And you'd have to put a 0 here."}, {"video_title": "Scientific notation examples   Pre-Algebra   Khan Academy.mp3", "Sentence": "So this is 1.75. So if you do it times 10 to the negative 1 power, you will go 1 to the left. But if you do times 10 to the negative 2 power, you'll go 2 to the left. And you'd have to put a 0 here. And if you do times 10 to the negative 3, you'd go 3 to the left. And you would have to add another 0. So you take this decimal and go 1, 2, 3 to the left."}, {"video_title": "Scientific notation examples   Pre-Algebra   Khan Academy.mp3", "Sentence": "And you'd have to put a 0 here. And if you do times 10 to the negative 3, you'd go 3 to the left. And you would have to add another 0. So you take this decimal and go 1, 2, 3 to the left. So our answer would be 0.00175 is the same thing as 1.75 times 10 to the negative 3. And another way to check that you got the right answer is if you have a 1 right here, if you count the 1, 1 including the 0's to the right of the decimal should be the same as the negative exponent here. So you have 1, 2, 3 numbers behind the decimal."}, {"video_title": "Scientific notation examples   Pre-Algebra   Khan Academy.mp3", "Sentence": "So you take this decimal and go 1, 2, 3 to the left. So our answer would be 0.00175 is the same thing as 1.75 times 10 to the negative 3. And another way to check that you got the right answer is if you have a 1 right here, if you count the 1, 1 including the 0's to the right of the decimal should be the same as the negative exponent here. So you have 1, 2, 3 numbers behind the decimal. So you should have, that's the same thing as to the negative 3 power. You're doing 1,000th. So this is 1,000th right there."}, {"video_title": "Scientific notation examples   Pre-Algebra   Khan Academy.mp3", "Sentence": "So you have 1, 2, 3 numbers behind the decimal. So you should have, that's the same thing as to the negative 3 power. You're doing 1,000th. So this is 1,000th right there. Let's do another example. Actually, let's mix it up. Let's start with something that's written as a numeral and then write it in scientific notation."}, {"video_title": "Scientific notation examples   Pre-Algebra   Khan Academy.mp3", "Sentence": "So this is 1,000th right there. Let's do another example. Actually, let's mix it up. Let's start with something that's written as a numeral and then write it in scientific notation. So let's say I have 120,000. So that's just this numerical value. And I want to write it in scientific notation."}, {"video_title": "Scientific notation examples   Pre-Algebra   Khan Academy.mp3", "Sentence": "Let's start with something that's written as a numeral and then write it in scientific notation. So let's say I have 120,000. So that's just this numerical value. And I want to write it in scientific notation. So this I can write as, I take the leading digit, 1.2 times 10 to the, and I just count how many digits there are behind the leading digit. 1, 2, 3, 4, 5. So 1.2 times 10 to the 5th."}, {"video_title": "Scientific notation examples   Pre-Algebra   Khan Academy.mp3", "Sentence": "And I want to write it in scientific notation. So this I can write as, I take the leading digit, 1.2 times 10 to the, and I just count how many digits there are behind the leading digit. 1, 2, 3, 4, 5. So 1.2 times 10 to the 5th. And if you want to kind of internalize why that makes sense, 10 to the 5th is 10,000. So 1.2, sorry, 1.2, 10 to the 5th is 100,000. So it's 1.2 times 1, 1, 2, 3, 4, 5."}, {"video_title": "Scientific notation examples   Pre-Algebra   Khan Academy.mp3", "Sentence": "So 1.2 times 10 to the 5th. And if you want to kind of internalize why that makes sense, 10 to the 5th is 10,000. So 1.2, sorry, 1.2, 10 to the 5th is 100,000. So it's 1.2 times 1, 1, 2, 3, 4, 5. You have five 0's. That's 10 to the 5th. So 1.2 times 100,000 is going to be 120,000."}, {"video_title": "Scientific notation examples   Pre-Algebra   Khan Academy.mp3", "Sentence": "So it's 1.2 times 1, 1, 2, 3, 4, 5. You have five 0's. That's 10 to the 5th. So 1.2 times 100,000 is going to be 120,000. It's going to be 1 and 1 5th times 100,000. So 120. Hopefully that's sinking in."}, {"video_title": "Scientific notation examples   Pre-Algebra   Khan Academy.mp3", "Sentence": "So 1.2 times 100,000 is going to be 120,000. It's going to be 1 and 1 5th times 100,000. So 120. Hopefully that's sinking in. So let's do another one. Let's say the numerical value is 1,765,244. I want to write this in scientific notation."}, {"video_title": "Scientific notation examples   Pre-Algebra   Khan Academy.mp3", "Sentence": "Hopefully that's sinking in. So let's do another one. Let's say the numerical value is 1,765,244. I want to write this in scientific notation. So I take the leading digit, 1, put a decimal sign. Everything else goes behind the decimal. 7, 6, 5, 2, 4, 4."}, {"video_title": "Scientific notation examples   Pre-Algebra   Khan Academy.mp3", "Sentence": "I want to write this in scientific notation. So I take the leading digit, 1, put a decimal sign. Everything else goes behind the decimal. 7, 6, 5, 2, 4, 4. And then you count how many digits there were between the leading digit and I guess you could imagine the first decimal sign and you could have numbers that keep go over here. So between the leading digit and the decimal sign. And you have 1, 2, 3, 4, 5, 6 digits."}, {"video_title": "Scientific notation examples   Pre-Algebra   Khan Academy.mp3", "Sentence": "7, 6, 5, 2, 4, 4. And then you count how many digits there were between the leading digit and I guess you could imagine the first decimal sign and you could have numbers that keep go over here. So between the leading digit and the decimal sign. And you have 1, 2, 3, 4, 5, 6 digits. So this is times 10 to the 6th. And 10 to the 6th is just a million. So it's 1.765,244 times a million, which makes sense."}, {"video_title": "Scientific notation examples   Pre-Algebra   Khan Academy.mp3", "Sentence": "And you have 1, 2, 3, 4, 5, 6 digits. So this is times 10 to the 6th. And 10 to the 6th is just a million. So it's 1.765,244 times a million, which makes sense. Roughly 1.7 times a million is roughly 1.7 million. This is a little bit more than 1.7 million. So it makes sense."}, {"video_title": "Scientific notation examples   Pre-Algebra   Khan Academy.mp3", "Sentence": "So it's 1.765,244 times a million, which makes sense. Roughly 1.7 times a million is roughly 1.7 million. This is a little bit more than 1.7 million. So it makes sense. Let's do another one. How do I write 12 in scientific notation? Same drill."}, {"video_title": "Scientific notation examples   Pre-Algebra   Khan Academy.mp3", "Sentence": "So it makes sense. Let's do another one. How do I write 12 in scientific notation? Same drill. It's equal to 1.2 times, well we only have one digit between the one and the decimal spot or the decimal point. So it's 1.2 times 10 to the first power. Or 1.2 times 10, which is definitely equal to 12."}, {"video_title": "Scientific notation examples   Pre-Algebra   Khan Academy.mp3", "Sentence": "Same drill. It's equal to 1.2 times, well we only have one digit between the one and the decimal spot or the decimal point. So it's 1.2 times 10 to the first power. Or 1.2 times 10, which is definitely equal to 12. Let's do a couple of examples where we're taking 10 to a negative power. So let's say we had 0.00281. And we want to write this in scientific notation."}, {"video_title": "Scientific notation examples   Pre-Algebra   Khan Academy.mp3", "Sentence": "Or 1.2 times 10, which is definitely equal to 12. Let's do a couple of examples where we're taking 10 to a negative power. So let's say we had 0.00281. And we want to write this in scientific notation. So what you do is you just have to think, well, how many digits are there to get to include the leading numeral in the value? So what I mean there is count 1, 2, 3. So what we want to do is we move the decimal 1, 2, 3 spaces."}, {"video_title": "Scientific notation examples   Pre-Algebra   Khan Academy.mp3", "Sentence": "And we want to write this in scientific notation. So what you do is you just have to think, well, how many digits are there to get to include the leading numeral in the value? So what I mean there is count 1, 2, 3. So what we want to do is we move the decimal 1, 2, 3 spaces. So one way you could think about it is you could multiply. To move the decimal to the right 3 spaces, you would multiply it by 10 to the third. But if you're multiplying something by 10 to the third, you're changing its value."}, {"video_title": "Scientific notation examples   Pre-Algebra   Khan Academy.mp3", "Sentence": "So what we want to do is we move the decimal 1, 2, 3 spaces. So one way you could think about it is you could multiply. To move the decimal to the right 3 spaces, you would multiply it by 10 to the third. But if you're multiplying something by 10 to the third, you're changing its value. So you also have to multiply by 10 to the negative 3. Only this way will you not change the value. If I multiply by 10 to the 3 times 10 to the negative 3, 3 minus 3 is 0."}, {"video_title": "Scientific notation examples   Pre-Algebra   Khan Academy.mp3", "Sentence": "But if you're multiplying something by 10 to the third, you're changing its value. So you also have to multiply by 10 to the negative 3. Only this way will you not change the value. If I multiply by 10 to the 3 times 10 to the negative 3, 3 minus 3 is 0. This is just like multiplying it by 1. So what is this going to equal? If I take the decimal and I move it 3 spaces to the right, this part right here is going to be equal to 2.81."}, {"video_title": "Scientific notation examples   Pre-Algebra   Khan Academy.mp3", "Sentence": "If I multiply by 10 to the 3 times 10 to the negative 3, 3 minus 3 is 0. This is just like multiplying it by 1. So what is this going to equal? If I take the decimal and I move it 3 spaces to the right, this part right here is going to be equal to 2.81. And then we're left with this one, times 10 to the negative 3. Now, a very quick way to do it is just to say, look, let me count, including the leading numeral, how many spaces I have behind the decimal. 1, 2, 3."}, {"video_title": "Scientific notation examples   Pre-Algebra   Khan Academy.mp3", "Sentence": "If I take the decimal and I move it 3 spaces to the right, this part right here is going to be equal to 2.81. And then we're left with this one, times 10 to the negative 3. Now, a very quick way to do it is just to say, look, let me count, including the leading numeral, how many spaces I have behind the decimal. 1, 2, 3. So it's going to be 2.81 times 10 to the negative 1, 2, 3 power. Let's do one more like that. Let me actually scroll up here."}, {"video_title": "Scientific notation examples   Pre-Algebra   Khan Academy.mp3", "Sentence": "1, 2, 3. So it's going to be 2.81 times 10 to the negative 1, 2, 3 power. Let's do one more like that. Let me actually scroll up here. Let's do one more like that. Let's say I have 0 point, let's say I have 1, 2, 3, 4, 5, 6. How many 0's do I have in this problem?"}, {"video_title": "Scientific notation examples   Pre-Algebra   Khan Academy.mp3", "Sentence": "Let me actually scroll up here. Let's do one more like that. Let's say I have 0 point, let's say I have 1, 2, 3, 4, 5, 6. How many 0's do I have in this problem? Well, I'll just make up something. 0, 2, 7. And you want to write that in scientific notation."}, {"video_title": "Scientific notation examples   Pre-Algebra   Khan Academy.mp3", "Sentence": "How many 0's do I have in this problem? Well, I'll just make up something. 0, 2, 7. And you want to write that in scientific notation. Well, you count all the digits up to the 2 behind the decimal. So 1, 2, 3, 4, 5, 6, 7, 8. So this is going to be 2.7 times 10 to the negative 8 power."}, {"video_title": "Scientific notation examples   Pre-Algebra   Khan Academy.mp3", "Sentence": "And you want to write that in scientific notation. Well, you count all the digits up to the 2 behind the decimal. So 1, 2, 3, 4, 5, 6, 7, 8. So this is going to be 2.7 times 10 to the negative 8 power. Now, let's do another one where we start with the scientific notation value and we want to go to the numeric value, just to mix things up. So let's say you have 2.9 times 10 to the negative 5. So one way to think about it is this leading numeral plus all the 0's to the left of the decimal spot is going to be 5 digits."}, {"video_title": "Scientific notation examples   Pre-Algebra   Khan Academy.mp3", "Sentence": "So this is going to be 2.7 times 10 to the negative 8 power. Now, let's do another one where we start with the scientific notation value and we want to go to the numeric value, just to mix things up. So let's say you have 2.9 times 10 to the negative 5. So one way to think about it is this leading numeral plus all the 0's to the left of the decimal spot is going to be 5 digits. So you have a 2 and a 9, and then you're going to have 5, and then you're going to have 4 more 0's. 1, 2, 3, 4, and then you're going to have your decimal. And how did I say no 4 0's?"}, {"video_title": "Scientific notation examples   Pre-Algebra   Khan Academy.mp3", "Sentence": "So one way to think about it is this leading numeral plus all the 0's to the left of the decimal spot is going to be 5 digits. So you have a 2 and a 9, and then you're going to have 5, and then you're going to have 4 more 0's. 1, 2, 3, 4, and then you're going to have your decimal. And how did I say no 4 0's? Because I'm counting this as 1, 2, 3, 4, 5 spaces behind the decimal, including the leading numeral. And so it's 0.000029. And just to verify, do the other technique."}, {"video_title": "Scientific notation examples   Pre-Algebra   Khan Academy.mp3", "Sentence": "And how did I say no 4 0's? Because I'm counting this as 1, 2, 3, 4, 5 spaces behind the decimal, including the leading numeral. And so it's 0.000029. And just to verify, do the other technique. How do I write this in scientific notation? I count all of the digits, all of the leading 0's behind the decimal, including the leading non-zero numeral. So I have 1, 2, 3, 4, 5 digits."}, {"video_title": "Scientific notation examples   Pre-Algebra   Khan Academy.mp3", "Sentence": "And just to verify, do the other technique. How do I write this in scientific notation? I count all of the digits, all of the leading 0's behind the decimal, including the leading non-zero numeral. So I have 1, 2, 3, 4, 5 digits. So it's 10 to the negative 5. And so it'll be 2.9 times 10 to the negative 5. And once again, this isn't just some type of black magic here."}, {"video_title": "Scientific notation examples   Pre-Algebra   Khan Academy.mp3", "Sentence": "So I have 1, 2, 3, 4, 5 digits. So it's 10 to the negative 5. And so it'll be 2.9 times 10 to the negative 5. And once again, this isn't just some type of black magic here. This actually makes a lot of sense. If I wanted to get this number to 2.9, what I would have to do is move the decimal over 1, 2, 3, 4, 5 spots, like that. And to get the decimal to move over to the right by 5 spots, I'll have to, let's say with 0, 0, 0, 0, 2, 9."}, {"video_title": "Scientific notation examples   Pre-Algebra   Khan Academy.mp3", "Sentence": "And once again, this isn't just some type of black magic here. This actually makes a lot of sense. If I wanted to get this number to 2.9, what I would have to do is move the decimal over 1, 2, 3, 4, 5 spots, like that. And to get the decimal to move over to the right by 5 spots, I'll have to, let's say with 0, 0, 0, 0, 2, 9. If I multiply it by 10 to the 5th, I'm also going to have to multiply it by 10 to the negative 5. Because I don't want to change the number. This right here is just multiplying something by 1."}, {"video_title": "Scientific notation examples   Pre-Algebra   Khan Academy.mp3", "Sentence": "And to get the decimal to move over to the right by 5 spots, I'll have to, let's say with 0, 0, 0, 0, 2, 9. If I multiply it by 10 to the 5th, I'm also going to have to multiply it by 10 to the negative 5. Because I don't want to change the number. This right here is just multiplying something by 1. 10 to the 5th times 10 to the negative 5 is 1. So this right here, this part right here, is essentially going to move the decimal 5 to the right. 1, 2, 3, 4, 5."}, {"video_title": "Scientific notation examples   Pre-Algebra   Khan Academy.mp3", "Sentence": "This right here is just multiplying something by 1. 10 to the 5th times 10 to the negative 5 is 1. So this right here, this part right here, is essentially going to move the decimal 5 to the right. 1, 2, 3, 4, 5. So this will be 2.5. And then we're going to be left with times 10 to the negative 5. Anyway, hopefully you found that scientific notation drill useful."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "The way you verify that is you substitute x is equal to 0. If you get x is equal to 0, remember x is equal to 0, that means that's where we're going to intercept the y axis. If x is equal to 0, this equation becomes y is equal to m times 0 plus b. m times 0 is just going to be 0. I don't care what m is, so then y is going to be equal to b. So the point 0, b is going to be on that line. So 0, b. The line will intercept the y axis at the point y is equal to b."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "I don't care what m is, so then y is going to be equal to b. So the point 0, b is going to be on that line. So 0, b. The line will intercept the y axis at the point y is equal to b. We'll see that with actual numbers in the next few videos. Just to verify for you that m is really the slope, let's just try some numbers out. We know the point 0, b is on the line."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "The line will intercept the y axis at the point y is equal to b. We'll see that with actual numbers in the next few videos. Just to verify for you that m is really the slope, let's just try some numbers out. We know the point 0, b is on the line. What happens when x is equal to 1? When x is equal to 1, you get y is equal to m times 1, or it's equal to m plus b. We also know that the point 1, m plus b is also on the line."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "We know the point 0, b is on the line. What happens when x is equal to 1? When x is equal to 1, you get y is equal to m times 1, or it's equal to m plus b. We also know that the point 1, m plus b is also on the line. This is just the y value. So what's the slope between that point and that point? Let's take this as the end point."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "We also know that the point 1, m plus b is also on the line. This is just the y value. So what's the slope between that point and that point? Let's take this as the end point. So you have m plus b, our change in y, m plus b minus b over our change in x. Over 1 minus 0. This is our change in y over change in x."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "Let's take this as the end point. So you have m plus b, our change in y, m plus b minus b over our change in x. Over 1 minus 0. This is our change in y over change in x. We're using two points. That's our end point, that's our starting point. So if you simplify this, b minus b is 0, 1 minus 0 is 1."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "This is our change in y over change in x. We're using two points. That's our end point, that's our starting point. So if you simplify this, b minus b is 0, 1 minus 0 is 1. So you get m over 1, or it's equal to m. So hopefully you're satisfied, and hopefully I didn't confuse you by staying in the abstract with all of these variables here. But this is definitely going to be the slope, and this is definitely going to be the y-intercept. Now given that, what I want to do in this exercise is look at these graphs, and then use the already drawn graphs to figure out the equation."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "So if you simplify this, b minus b is 0, 1 minus 0 is 1. So you get m over 1, or it's equal to m. So hopefully you're satisfied, and hopefully I didn't confuse you by staying in the abstract with all of these variables here. But this is definitely going to be the slope, and this is definitely going to be the y-intercept. Now given that, what I want to do in this exercise is look at these graphs, and then use the already drawn graphs to figure out the equation. So we're going to look at these, figure out the slopes, figure out the y-intercepts, and then know the equation. So let's do this line A first. So what is A's slope?"}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "Now given that, what I want to do in this exercise is look at these graphs, and then use the already drawn graphs to figure out the equation. So we're going to look at these, figure out the slopes, figure out the y-intercepts, and then know the equation. So let's do this line A first. So what is A's slope? So let's start at some arbitrary point. Let's start right over there. And then let us see, and we want to get to even numbers."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "So what is A's slope? So let's start at some arbitrary point. Let's start right over there. And then let us see, and we want to get to even numbers. So let's see, if we run 1, 2, 3. So if our delta x is equal to 3, our delta y is equal to, we go down by 2, it's equal to negative 2. So for A, change in y for a change in x."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "And then let us see, and we want to get to even numbers. So let's see, if we run 1, 2, 3. So if our delta x is equal to 3, our delta y is equal to, we go down by 2, it's equal to negative 2. So for A, change in y for a change in x. When our change in x is 3, our change in y is negative 2. So our slope is negative 2 thirds. When we go over by 3, we're going to go down by 2."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "So for A, change in y for a change in x. When our change in x is 3, our change in y is negative 2. So our slope is negative 2 thirds. When we go over by 3, we're going to go down by 2. Or if we go over by 1, we're going to go down by 2 thirds. You can't exactly see it there, but you definitely see it when you go over by 3. So that's our slope."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "When we go over by 3, we're going to go down by 2. Or if we go over by 1, we're going to go down by 2 thirds. You can't exactly see it there, but you definitely see it when you go over by 3. So that's our slope. We've essentially done half of that problem. Now we have to figure out the y-intercept. So that right there is our m. Now what is our b, our y-intercept?"}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "So that's our slope. We've essentially done half of that problem. Now we have to figure out the y-intercept. So that right there is our m. Now what is our b, our y-intercept? Well, where does this intersect the y-axis? Well, we already said the slope is 2 thirds, so this is the point y is equal to 2. When we go over by 1 to the right, we would have gone down by 2 thirds."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "So that right there is our m. Now what is our b, our y-intercept? Well, where does this intersect the y-axis? Well, we already said the slope is 2 thirds, so this is the point y is equal to 2. When we go over by 1 to the right, we would have gone down by 2 thirds. So this right here must be the point 1 and 1 third. Or another way to say it is 4 thirds. That's the point y is equal to 4 thirds right there."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "When we go over by 1 to the right, we would have gone down by 2 thirds. So this right here must be the point 1 and 1 third. Or another way to say it is 4 thirds. That's the point y is equal to 4 thirds right there. A little bit more than 1, but 1 and 1 third. So we can say b is equal to 4 thirds. And so we'll know that the equation is y is equal to m, negative 2 thirds, x, plus b, plus 4 thirds."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "That's the point y is equal to 4 thirds right there. A little bit more than 1, but 1 and 1 third. So we can say b is equal to 4 thirds. And so we'll know that the equation is y is equal to m, negative 2 thirds, x, plus b, plus 4 thirds. That's equation A. Let's do equation B. Hopefully we won't have to deal with as many fractions here. Equation B, let's figure out its slope first."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "And so we'll know that the equation is y is equal to m, negative 2 thirds, x, plus b, plus 4 thirds. That's equation A. Let's do equation B. Hopefully we won't have to deal with as many fractions here. Equation B, let's figure out its slope first. Let's start at some reasonable point. Let's see. We could start at that point."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "Equation B, let's figure out its slope first. Let's start at some reasonable point. Let's see. We could start at that point. And then if we, let me do it right here. B, equation B. When our delta x is equal to, let me write it this way."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "We could start at that point. And then if we, let me do it right here. B, equation B. When our delta x is equal to, let me write it this way. Delta x. So our delta x could be 1. When we move over 1 to the right, what happens to our delta y?"}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "When our delta x is equal to, let me write it this way. Delta x. So our delta x could be 1. When we move over 1 to the right, what happens to our delta y? We go up by 3. Delta x, delta y. Our change in y is 3."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "When we move over 1 to the right, what happens to our delta y? We go up by 3. Delta x, delta y. Our change in y is 3. So delta y over delta x, when we go to the right, our change in x is 1. Our change in x is 1, our change in y is positive 3. So our slope is equal to 3."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "Our change in y is 3. So delta y over delta x, when we go to the right, our change in x is 1. Our change in x is 1, our change in y is positive 3. So our slope is equal to 3. What is our y intercept? Well when x is equal to 0, y is equal to 1. So b is equal to 1."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "So our slope is equal to 3. What is our y intercept? Well when x is equal to 0, y is equal to 1. So b is equal to 1. So this was a lot easier. Here the equation is y is equal to 3x plus 1. Let's do that last line there."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "So b is equal to 1. So this was a lot easier. Here the equation is y is equal to 3x plus 1. Let's do that last line there. Line C. Line C. Alright. So let's do the y intercept first. You see immediately the y intercept when x is equal to 0, y is negative 2."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "Let's do that last line there. Line C. Line C. Alright. So let's do the y intercept first. You see immediately the y intercept when x is equal to 0, y is negative 2. So b is equal to negative 2. And then what is the slope? M is equal to change in y over change in x."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "You see immediately the y intercept when x is equal to 0, y is negative 2. So b is equal to negative 2. And then what is the slope? M is equal to change in y over change in x. So let's see. Let's start at that y intercept. And if we go over to the right by 1, 2, 3, 4."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "M is equal to change in y over change in x. So let's see. Let's start at that y intercept. And if we go over to the right by 1, 2, 3, 4. So our change in x is equal to 4. What is our change in y? Our change in y is positive 2."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "And if we go over to the right by 1, 2, 3, 4. So our change in x is equal to 4. What is our change in y? Our change in y is positive 2. Change in y is equal to 2. So change in y is 2 when change in x is 4. So the slope is equal to 1 half."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "Our change in y is positive 2. Change in y is equal to 2. So change in y is 2 when change in x is 4. So the slope is equal to 1 half. 2 over 4. So the equation here is y is equal to 1 half x. That's our slope."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "So the slope is equal to 1 half. 2 over 4. So the equation here is y is equal to 1 half x. That's our slope. Minus 2. And we're done. Now let's go the other way."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "That's our slope. Minus 2. And we're done. Now let's go the other way. Let's look at some equations of lines knowing that this is the slope and this is the y intercept. That's the m, that's the b. And actually graph them."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "Now let's go the other way. Let's look at some equations of lines knowing that this is the slope and this is the y intercept. That's the m, that's the b. And actually graph them. So let's do this first line. I already started circling it in orange. The y intercept is 5."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "And actually graph them. So let's do this first line. I already started circling it in orange. The y intercept is 5. When x is equal to 0, y is equal to 5. You can verify that on the equation. So when x is equal to 0, y is equal to 1, 2, 3, 4, 5."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "The y intercept is 5. When x is equal to 0, y is equal to 5. You can verify that on the equation. So when x is equal to 0, y is equal to 1, 2, 3, 4, 5. That's the y intercept. And then the slope is 2. That means when I move 1 in the x direction, I move up 2 in the y direction."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "So when x is equal to 0, y is equal to 1, 2, 3, 4, 5. That's the y intercept. And then the slope is 2. That means when I move 1 in the x direction, I move up 2 in the y direction. So I move 1 in the x direction, I move up 2 in the y direction. If I move back 1 in the x direction, I move up 2 in the y direction. If I move back 1 in the x direction, I move down 2 in the y direction."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "That means when I move 1 in the x direction, I move up 2 in the y direction. So I move 1 in the x direction, I move up 2 in the y direction. If I move back 1 in the x direction, I move up 2 in the y direction. If I move back 1 in the x direction, I move down 2 in the y direction. I keep doing that. So we'll keep doing that. So this line is going to look, I can't draw lines too neatly, but this is going to be my best shot."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "If I move back 1 in the x direction, I move down 2 in the y direction. I keep doing that. So we'll keep doing that. So this line is going to look, I can't draw lines too neatly, but this is going to be my best shot. It's going to look something like that. And we'll just keep going on, on and on and on. So that's our first line."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "So this line is going to look, I can't draw lines too neatly, but this is going to be my best shot. It's going to look something like that. And we'll just keep going on, on and on and on. So that's our first line. I can just keep going down like that. Now let's do this second line. Y is equal to negative.2x plus seven."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "So that's our first line. I can just keep going down like that. Now let's do this second line. Y is equal to negative.2x plus seven. So let me write that. Y is equal to negative 0.2x plus seven. So it's always easier to think in fractions."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "Y is equal to negative.2x plus seven. So let me write that. Y is equal to negative 0.2x plus seven. So it's always easier to think in fractions. So.2 is the same thing as 1 5th. So we could write Y is equal to negative 1 5th x plus seven. So we know it's Y intercept, it's seven."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "So it's always easier to think in fractions. So.2 is the same thing as 1 5th. So we could write Y is equal to negative 1 5th x plus seven. So we know it's Y intercept, it's seven. So this is one, two, three, four, five, six. That's our Y intercept when X is equal to zero. And this tells us that for every five we move to the right, we move down one."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "So we know it's Y intercept, it's seven. So this is one, two, three, four, five, six. That's our Y intercept when X is equal to zero. And this tells us that for every five we move to the right, we move down one. So we could view this as negative one over five, that delta Y over delta X is equal to negative one over five. For every five we move to the right, we move down one. So every five, one, two, three, four, five."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "And this tells us that for every five we move to the right, we move down one. So we could view this as negative one over five, that delta Y over delta X is equal to negative one over five. For every five we move to the right, we move down one. So every five, one, two, three, four, five. We moved five to the right, that means we must move down one. We move five to the right, one, two, three, four, five, we must move down one. If you go backwards, if you move five backwards, so if you view instead of this, you view this as one over negative five, these are obviously equivalent numbers."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "So every five, one, two, three, four, five. We moved five to the right, that means we must move down one. We move five to the right, one, two, three, four, five, we must move down one. If you go backwards, if you move five backwards, so if you view instead of this, you view this as one over negative five, these are obviously equivalent numbers. So if you go back five, that's negative five, one, two, three, four, five, then you move up one. You move up one. You go back five, one, two, three, four, five, you move up one."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "If you go backwards, if you move five backwards, so if you view instead of this, you view this as one over negative five, these are obviously equivalent numbers. So if you go back five, that's negative five, one, two, three, four, five, then you move up one. You move up one. You go back five, one, two, three, four, five, you move up one. So the line is going to look like this. I should just connect the dots. I think you get the idea."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "You go back five, one, two, three, four, five, you move up one. So the line is going to look like this. I should just connect the dots. I think you get the idea. I just have to connect those dots. I could have drawn it a little bit straighter. Now let's do this one, Y is equal to negative X. Y is equal to negative X."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "I think you get the idea. I just have to connect those dots. I could have drawn it a little bit straighter. Now let's do this one, Y is equal to negative X. Y is equal to negative X. Where's the B term? I don't see any B term. You remember we're saying Y is equal to MX plus B."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "Now let's do this one, Y is equal to negative X. Y is equal to negative X. Where's the B term? I don't see any B term. You remember we're saying Y is equal to MX plus B. Where's the B? Well, the B is zero. You could view this as plus zero."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "You remember we're saying Y is equal to MX plus B. Where's the B? Well, the B is zero. You could view this as plus zero. So here is B is zero. When X is zero, Y is zero. So that's our Y intercept right there at the origin."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "You could view this as plus zero. So here is B is zero. When X is zero, Y is zero. So that's our Y intercept right there at the origin. And then the slope, once again, you just see a negative sign that you could view that as negative one X plus zero. So slope is negative one. When you move to the right by one, when change in X is one, change in Y is negative one."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "So that's our Y intercept right there at the origin. And then the slope, once again, you just see a negative sign that you could view that as negative one X plus zero. So slope is negative one. When you move to the right by one, when change in X is one, change in Y is negative one. When you move up by one in X, you go down by one in Y. Or if you go down by one in X, you're gonna go up by one in Y. X and Y are gonna have opposite signs. They go in opposite directions."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "When you move to the right by one, when change in X is one, change in Y is negative one. When you move up by one in X, you go down by one in Y. Or if you go down by one in X, you're gonna go up by one in Y. X and Y are gonna have opposite signs. They go in opposite directions. So the line is going to look like that. It's going to look like that. You can almost imagine it splitting the second and fourth quadrants."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "They go in opposite directions. So the line is going to look like that. It's going to look like that. You can almost imagine it splitting the second and fourth quadrants. Now I'll do one more. Let's do this last one right here. Y is equal to 3.75."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "You can almost imagine it splitting the second and fourth quadrants. Now I'll do one more. Let's do this last one right here. Y is equal to 3.75. So now you're saying, gee, we're looking for Y is equal to MX plus B. Where is this X term? It's completely gone."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "Y is equal to 3.75. So now you're saying, gee, we're looking for Y is equal to MX plus B. Where is this X term? It's completely gone. Well, the reality here is this could be rewritten as Y is equal to zero X plus 3.75. Now it makes sense. The slope is zero."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "It's completely gone. Well, the reality here is this could be rewritten as Y is equal to zero X plus 3.75. Now it makes sense. The slope is zero. No matter how much we change our X, Y does not change. Delta Y over delta X is equal to zero. I don't care how much you change your X."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "The slope is zero. No matter how much we change our X, Y does not change. Delta Y over delta X is equal to zero. I don't care how much you change your X. So our Y intercept is 3.75. So one, two, 3.75 is right around there. Just wanna get close, 3 3 4."}, {"video_title": "Slope-intercept equation from a graph examples   Algebra I   Khan Academy.mp3", "Sentence": "I don't care how much you change your X. So our Y intercept is 3.75. So one, two, 3.75 is right around there. Just wanna get close, 3 3 4. And then as I change X, Y will not change. Y is always going to be 3.75. So it's just gonna be a horizontal line at Y is equal to 3.75."}, {"video_title": "How many bags of potato chips do people eat   Algebra II   Khan Academy.mp3", "Sentence": "Everyone except for this gentleman right over here. This is Arbegla, and he is the king's top advisor and also chief party planner. And he seems somewhat threatened by your ability to solve these otherwise unsolvable problems, or at least from his point of view, because he keeps over-ordering or under-ordering things like cupcakes. And so he says, king, that cupcake problem was easy. Ask them about the potato chip issue, because we can never get the potato chips right. And so the king says, Arbegla, that's a good idea. We need to get the potato chips right."}, {"video_title": "How many bags of potato chips do people eat   Algebra II   Khan Academy.mp3", "Sentence": "And so he says, king, that cupcake problem was easy. Ask them about the potato chip issue, because we can never get the potato chips right. And so the king says, Arbegla, that's a good idea. We need to get the potato chips right. So he comes to you and says, how do we figure out, on average, how many potato chips we need to order? And to do that, we have to figure out how much, on average, does each man eat and how much each woman eats. And you say, well, what about the children?"}, {"video_title": "How many bags of potato chips do people eat   Algebra II   Khan Academy.mp3", "Sentence": "We need to get the potato chips right. So he comes to you and says, how do we figure out, on average, how many potato chips we need to order? And to do that, we have to figure out how much, on average, does each man eat and how much each woman eats. And you say, well, what about the children? He say, well, in our kingdom, the king says, in our kingdom, we forbid potato chips for children. You say, oh, well, that's all in good. Well, tell me what happened at the previous parties."}, {"video_title": "How many bags of potato chips do people eat   Algebra II   Khan Academy.mp3", "Sentence": "And you say, well, what about the children? He say, well, in our kingdom, the king says, in our kingdom, we forbid potato chips for children. You say, oh, well, that's all in good. Well, tell me what happened at the previous parties. And so the king says, you might remember, at the last party, in fact, the last two parties, we had 500 adults. At the last party, 200 of them were men, 200 men, and 300 of them were women, 300 were women, and in total, they ate 1,200 bags of potato chips. 1,200 bags of potato chips."}, {"video_title": "How many bags of potato chips do people eat   Algebra II   Khan Academy.mp3", "Sentence": "Well, tell me what happened at the previous parties. And so the king says, you might remember, at the last party, in fact, the last two parties, we had 500 adults. At the last party, 200 of them were men, 200 men, and 300 of them were women, 300 were women, and in total, they ate 1,200 bags of potato chips. 1,200 bags of potato chips. And you say, what about the party before that? He says, that one, we had a bigger skew towards women. We only had 100 men, 100 men, and we had 400 women, 400 women, and that time, we actually had fewer bags consumed, 1,100, 1,100 bags of potato chips."}, {"video_title": "How many bags of potato chips do people eat   Algebra II   Khan Academy.mp3", "Sentence": "1,200 bags of potato chips. And you say, what about the party before that? He says, that one, we had a bigger skew towards women. We only had 100 men, 100 men, and we had 400 women, 400 women, and that time, we actually had fewer bags consumed, 1,100, 1,100 bags of potato chips. So you say, okay, king and our bigla, this seems like a fairly straightforward thing. Let me define some variables to represent our unknowns. So you go ahead and you say, well, let's let, let's let, let's let m equal the number of bags eaten, eaten by each man, by each man, and you could think of it on average, or maybe everyone, all the men in that kingdom are completely identical, or maybe this is the average, number of bags eaten by each man, and let's let w equal the number of bags eaten by each woman, each woman."}, {"video_title": "How many bags of potato chips do people eat   Algebra II   Khan Academy.mp3", "Sentence": "We only had 100 men, 100 men, and we had 400 women, 400 women, and that time, we actually had fewer bags consumed, 1,100, 1,100 bags of potato chips. So you say, okay, king and our bigla, this seems like a fairly straightforward thing. Let me define some variables to represent our unknowns. So you go ahead and you say, well, let's let, let's let, let's let m equal the number of bags eaten, eaten by each man, by each man, and you could think of it on average, or maybe everyone, all the men in that kingdom are completely identical, or maybe this is the average, number of bags eaten by each man, and let's let w equal the number of bags eaten by each woman, each woman. And so with that, with these definitions of our variables, let's think about how we can represent this first piece of information, this piece of information in green. Well, let's think about the total number of bags that the men ate. You had 200, you had 200 men, 200, let me scroll over a little bit."}, {"video_title": "How many bags of potato chips do people eat   Algebra II   Khan Academy.mp3", "Sentence": "So you go ahead and you say, well, let's let, let's let, let's let m equal the number of bags eaten, eaten by each man, by each man, and you could think of it on average, or maybe everyone, all the men in that kingdom are completely identical, or maybe this is the average, number of bags eaten by each man, and let's let w equal the number of bags eaten by each woman, each woman. And so with that, with these definitions of our variables, let's think about how we can represent this first piece of information, this piece of information in green. Well, let's think about the total number of bags that the men ate. You had 200, you had 200 men, 200, let me scroll over a little bit. You had 200 men, and they each ate m bags, m bags per man, so the men at this first party collectively ate 200 times m bags. If m is 10 bags per man, then this would be 2,000. If m was five bags per man, then this would be 5,000."}, {"video_title": "How many bags of potato chips do people eat   Algebra II   Khan Academy.mp3", "Sentence": "You had 200, you had 200 men, 200, let me scroll over a little bit. You had 200 men, and they each ate m bags, m bags per man, so the men at this first party collectively ate 200 times m bags. If m is 10 bags per man, then this would be 2,000. If m was five bags per man, then this would be 5,000. We don't know what m is, but 200 times m is the total eaten by the men. Same logic, total eaten by the women is 300, 300 women times the number of bags eaten by each woman, and so if you add the total eaten by the men and the women, you get the 1,200 bags. You get the 1,200 bags."}, {"video_title": "How many bags of potato chips do people eat   Algebra II   Khan Academy.mp3", "Sentence": "If m was five bags per man, then this would be 5,000. We don't know what m is, but 200 times m is the total eaten by the men. Same logic, total eaten by the women is 300, 300 women times the number of bags eaten by each woman, and so if you add the total eaten by the men and the women, you get the 1,200 bags. You get the 1,200 bags. So this is this information written algebraically given these variable definitions. Now let's do the same thing with the second party, the information that they gave us right over here. Let's think about how we can represent this algebraically."}, {"video_title": "How many bags of potato chips do people eat   Algebra II   Khan Academy.mp3", "Sentence": "You get the 1,200 bags. So this is this information written algebraically given these variable definitions. Now let's do the same thing with the second party, the information that they gave us right over here. Let's think about how we can represent this algebraically. Well, similar logic. What was the total that the men ate at that party? It was 100 men times m bags per man, and we're assuming that m is the same across parties, that men on average always eat the same number of bags."}, {"video_title": "How many bags of potato chips do people eat   Algebra II   Khan Academy.mp3", "Sentence": "Let's think about how we can represent this algebraically. Well, similar logic. What was the total that the men ate at that party? It was 100 men times m bags per man, and we're assuming that m is the same across parties, that men on average always eat the same number of bags. And how many did the women eat at that second party? Well, you had 400 women, and on average, they ate w bags per woman. So this is how 400 times w is the total number that the women ate."}, {"video_title": "How many bags of potato chips do people eat   Algebra II   Khan Academy.mp3", "Sentence": "It was 100 men times m bags per man, and we're assuming that m is the same across parties, that men on average always eat the same number of bags. And how many did the women eat at that second party? Well, you had 400 women, and on average, they ate w bags per woman. So this is how 400 times w is the total number that the women ate. You add those two together, you have the total number that all the adults ate. So this is going to be 1,100 bags. So it looks pretty similar now."}, {"video_title": "How many bags of potato chips do people eat   Algebra II   Khan Academy.mp3", "Sentence": "So this is how 400 times w is the total number that the women ate. You add those two together, you have the total number that all the adults ate. So this is going to be 1,100 bags. So it looks pretty similar now. You have a system of two equations with two unknowns, and so you try your best to solve it. But when you solve it, you see something interesting. Last time it was very convenient."}, {"video_title": "How many bags of potato chips do people eat   Algebra II   Khan Academy.mp3", "Sentence": "So it looks pretty similar now. You have a system of two equations with two unknowns, and so you try your best to solve it. But when you solve it, you see something interesting. Last time it was very convenient. You had a, I think it was a 500 here for 500 adults, and you had another 500, and so it seemed like it was pretty easy to cancel out one of the variables. Here it seems a little bit more difficult. What's multiplying by the m's, it's different here."}, {"video_title": "How many bags of potato chips do people eat   Algebra II   Khan Academy.mp3", "Sentence": "Last time it was very convenient. You had a, I think it was a 500 here for 500 adults, and you had another 500, and so it seemed like it was pretty easy to cancel out one of the variables. Here it seems a little bit more difficult. What's multiplying by the m's, it's different here. The coefficient on the w is different over here. But you say, well, maybe I can change one of these equations so it makes it a little bit easier to cancel out with the other equation. So what if, for example, I were to take this blue equation right over here and multiply it by negative two?"}, {"video_title": "How many bags of potato chips do people eat   Algebra II   Khan Academy.mp3", "Sentence": "What's multiplying by the m's, it's different here. The coefficient on the w is different over here. But you say, well, maybe I can change one of these equations so it makes it a little bit easier to cancel out with the other equation. So what if, for example, I were to take this blue equation right over here and multiply it by negative two? And you might say, well, Sal, why are we multiplying it by negative two? Well, if we were to multiply it by negative two, this 100m would become a negative 200m. And if it was a negative 200m, then that would cancel out with a positive 200m when we add the two."}, {"video_title": "How many bags of potato chips do people eat   Algebra II   Khan Academy.mp3", "Sentence": "So what if, for example, I were to take this blue equation right over here and multiply it by negative two? And you might say, well, Sal, why are we multiplying it by negative two? Well, if we were to multiply it by negative two, this 100m would become a negative 200m. And if it was a negative 200m, then that would cancel out with a positive 200m when we add the two. So let's see what happens. So let's just multiply, let's multiply this blue equation by negative two. We're gonna multiply by negative two."}, {"video_title": "How many bags of potato chips do people eat   Algebra II   Khan Academy.mp3", "Sentence": "And if it was a negative 200m, then that would cancel out with a positive 200m when we add the two. So let's see what happens. So let's just multiply, let's multiply this blue equation by negative two. We're gonna multiply by negative two. Let me scroll over to the left a little bit. So what happens? Remember, when we multiply an equation, we can't just do one term."}, {"video_title": "How many bags of potato chips do people eat   Algebra II   Khan Academy.mp3", "Sentence": "We're gonna multiply by negative two. Let me scroll over to the left a little bit. So what happens? Remember, when we multiply an equation, we can't just do one term. We have to do, and we can't just do one side of the equation. We have to do the entire equation in order for the equality to hold true. So negative two times 100m is negative 200m."}, {"video_title": "How many bags of potato chips do people eat   Algebra II   Khan Academy.mp3", "Sentence": "Remember, when we multiply an equation, we can't just do one term. We have to do, and we can't just do one side of the equation. We have to do the entire equation in order for the equality to hold true. So negative two times 100m is negative 200m. Negative two times 400w, and there's a positive right over here, so it becomes negative 800w. And then negative two, now we did the left-hand side, but we also have to do the right-hand side. Negative two times 1,100 is negative 2,200."}, {"video_title": "How many bags of potato chips do people eat   Algebra II   Khan Academy.mp3", "Sentence": "So negative two times 100m is negative 200m. Negative two times 400w, and there's a positive right over here, so it becomes negative 800w. And then negative two, now we did the left-hand side, but we also have to do the right-hand side. Negative two times 1,100 is negative 2,200. So just to be clear, this equation that I just wrote here essentially has the same information we just manipulated. We just changed this equation, multiplied both sides by negative two. But it's kind of the same constraint."}, {"video_title": "How many bags of potato chips do people eat   Algebra II   Khan Academy.mp3", "Sentence": "Negative two times 1,100 is negative 2,200. So just to be clear, this equation that I just wrote here essentially has the same information we just manipulated. We just changed this equation, multiplied both sides by negative two. But it's kind of the same constraint. What makes this interesting is now we can rewrite this green equation. Let me do it over here, this first one. 200m plus 300w is equal to 1,200."}, {"video_title": "How many bags of potato chips do people eat   Algebra II   Khan Academy.mp3", "Sentence": "But it's kind of the same constraint. What makes this interesting is now we can rewrite this green equation. Let me do it over here, this first one. 200m plus 300w is equal to 1,200. And the whole reason why I multiplied by negative two is so that if I were to add these two things, I might be able to get rid of that variable over there. And so let's do that. Let's add the left-hand sides, and let's add the right-hand sides."}, {"video_title": "How many bags of potato chips do people eat   Algebra II   Khan Academy.mp3", "Sentence": "200m plus 300w is equal to 1,200. And the whole reason why I multiplied by negative two is so that if I were to add these two things, I might be able to get rid of that variable over there. And so let's do that. Let's add the left-hand sides, and let's add the right-hand sides. And you could literally view it as, we're starting with this blue equation. We're adding this quantity, the left-hand side of the yellow equation to the left-hand side of the blue. And then 1,200 is the exact same thing that we're adding to the right-hand side."}, {"video_title": "How many bags of potato chips do people eat   Algebra II   Khan Academy.mp3", "Sentence": "Let's add the left-hand sides, and let's add the right-hand sides. And you could literally view it as, we're starting with this blue equation. We're adding this quantity, the left-hand side of the yellow equation to the left-hand side of the blue. And then 1,200 is the exact same thing that we're adding to the right-hand side. We know that this is equal to this, so we can add this to the left-hand side and this to the right-hand side. So let's see what happens. So the good thing is, the whole reason why we multiplied it by negative two is so that these two characters cancel out."}, {"video_title": "How many bags of potato chips do people eat   Algebra II   Khan Academy.mp3", "Sentence": "And then 1,200 is the exact same thing that we're adding to the right-hand side. We know that this is equal to this, so we can add this to the left-hand side and this to the right-hand side. So let's see what happens. So the good thing is, the whole reason why we multiplied it by negative two is so that these two characters cancel out. You add those two together, you just get zero m, or just zero. You have negative 800w plus 300w. Well, that's negative 500w."}, {"video_title": "How many bags of potato chips do people eat   Algebra II   Khan Academy.mp3", "Sentence": "So the good thing is, the whole reason why we multiplied it by negative two is so that these two characters cancel out. You add those two together, you just get zero m, or just zero. You have negative 800w plus 300w. Well, that's negative 500w. And then on the right-hand side, you have negative 2,200 plus 1,200. So that's negative 1,000. And now this is pretty straightforward."}, {"video_title": "How many bags of potato chips do people eat   Algebra II   Khan Academy.mp3", "Sentence": "Well, that's negative 500w. And then on the right-hand side, you have negative 2,200 plus 1,200. So that's negative 1,000. And now this is pretty straightforward. One equation and one unknown, a fairly straightforward equation. We divide both sides by the coefficient of w, multiplying w, so divide by negative 500 on the left, divide by negative 500 on the right. And we are left with w is equal to, w is equal to two."}, {"video_title": "How many bags of potato chips do people eat   Algebra II   Khan Academy.mp3", "Sentence": "And now this is pretty straightforward. One equation and one unknown, a fairly straightforward equation. We divide both sides by the coefficient of w, multiplying w, so divide by negative 500 on the left, divide by negative 500 on the right. And we are left with w is equal to, w is equal to two. On average, women ate two bags of potato chips at these parties. We're assuming that that's constant across the parties. So let's think about how you would then figure out how many bags, on average, each man ate."}, {"video_title": "How many bags of potato chips do people eat   Algebra II   Khan Academy.mp3", "Sentence": "And we are left with w is equal to, w is equal to two. On average, women ate two bags of potato chips at these parties. We're assuming that that's constant across the parties. So let's think about how you would then figure out how many bags, on average, each man ate. Well, to do that, we just go back to either one of these equations. In the last video, or set of videos, I went to the first equation. I'll show you that the second equation should also work."}, {"video_title": "How many bags of potato chips do people eat   Algebra II   Khan Academy.mp3", "Sentence": "So let's think about how you would then figure out how many bags, on average, each man ate. Well, to do that, we just go back to either one of these equations. In the last video, or set of videos, I went to the first equation. I'll show you that the second equation should also work. Either one should work. So let's substitute back into the second equation. And you could either pick this version of it or this one, but I'll pick the original one."}, {"video_title": "How many bags of potato chips do people eat   Algebra II   Khan Academy.mp3", "Sentence": "I'll show you that the second equation should also work. Either one should work. So let's substitute back into the second equation. And you could either pick this version of it or this one, but I'll pick the original one. So you have 100 times m, which we're trying to figure out, plus 400 times, well, we now know that w is equal to two. 400 times two is equal to 1,100. Is equal to 1,100."}, {"video_title": "How many bags of potato chips do people eat   Algebra II   Khan Academy.mp3", "Sentence": "And you could either pick this version of it or this one, but I'll pick the original one. So you have 100 times m, which we're trying to figure out, plus 400 times, well, we now know that w is equal to two. 400 times two is equal to 1,100. Is equal to 1,100. So you have 100m plus 800. 800 is equal to 1,100. And now to solve for m, we can subtract 800 from both sides."}, {"video_title": "How many bags of potato chips do people eat   Algebra II   Khan Academy.mp3", "Sentence": "Is equal to 1,100. So you have 100m plus 800. 800 is equal to 1,100. And now to solve for m, we can subtract 800 from both sides. Subtract 800 from both sides. And we are left with 100m. 100m is equal to 300."}, {"video_title": "How many bags of potato chips do people eat   Algebra II   Khan Academy.mp3", "Sentence": "And now to solve for m, we can subtract 800 from both sides. Subtract 800 from both sides. And we are left with 100m. 100m is equal to 300. And now divide both sides by 100. 100 and 100. And we are left with m, which is, on average, the number of bags of chips each man eats is equal to three."}, {"video_title": "Multi-step inequalities 2   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Solve for x. And we have 5x plus 7 is greater than 3 times x plus 1. So let's just try to isolate x on one side of this inequality. But before we do that, let's just simplify this right-hand side. So we get 5x plus 7 is greater than, let's distribute this 3. So 3 times x plus 1 is the same thing as 3 times x plus 3 times 1. So it's going to be 3x plus 3 times 1 is 3."}, {"video_title": "Multi-step inequalities 2   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "But before we do that, let's just simplify this right-hand side. So we get 5x plus 7 is greater than, let's distribute this 3. So 3 times x plus 1 is the same thing as 3 times x plus 3 times 1. So it's going to be 3x plus 3 times 1 is 3. Now if we want to put our x's on the left-hand side, we can subtract 3x from both sides. That'll get rid of this 3x on the right-hand side. So let's do that."}, {"video_title": "Multi-step inequalities 2   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So it's going to be 3x plus 3 times 1 is 3. Now if we want to put our x's on the left-hand side, we can subtract 3x from both sides. That'll get rid of this 3x on the right-hand side. So let's do that. Let's subtract 3x from both sides. And we get, on the left-hand side, 5x minus 3x is 2x plus 7 is greater than 3x minus 3x. Those cancel out."}, {"video_title": "Multi-step inequalities 2   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So let's do that. Let's subtract 3x from both sides. And we get, on the left-hand side, 5x minus 3x is 2x plus 7 is greater than 3x minus 3x. Those cancel out. That was the whole point behind subtracting 3x from both sides. Is greater than 3. Now we can subtract 7 from both sides to get rid of this positive 7 right over here."}, {"video_title": "Multi-step inequalities 2   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Those cancel out. That was the whole point behind subtracting 3x from both sides. Is greater than 3. Now we can subtract 7 from both sides to get rid of this positive 7 right over here. So let's subtract 7 from both sides. And we get, on the left-hand side, 2x plus 7 minus 7 is just 2x. Is greater than 3 minus 7, which is negative 4."}, {"video_title": "Multi-step inequalities 2   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Now we can subtract 7 from both sides to get rid of this positive 7 right over here. So let's subtract 7 from both sides. And we get, on the left-hand side, 2x plus 7 minus 7 is just 2x. Is greater than 3 minus 7, which is negative 4. And then let's see. We have 2x is greater than negative 4. If we just wanted x over here, we can divide both sides by 2."}, {"video_title": "Multi-step inequalities 2   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Is greater than 3 minus 7, which is negative 4. And then let's see. We have 2x is greater than negative 4. If we just wanted x over here, we can divide both sides by 2. Since 2 is a positive number, we don't have to swap the inequality. So let's just divide both sides by 2. And we get x is greater than negative 4 divided by 2 is negative 2."}, {"video_title": "Multi-step inequalities 2   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "If we just wanted x over here, we can divide both sides by 2. Since 2 is a positive number, we don't have to swap the inequality. So let's just divide both sides by 2. And we get x is greater than negative 4 divided by 2 is negative 2. So the solution will look like this. Draw the number line. I can draw a straighter number line than that."}, {"video_title": "Multi-step inequalities 2   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "And we get x is greater than negative 4 divided by 2 is negative 2. So the solution will look like this. Draw the number line. I can draw a straighter number line than that. There we go. Still not that great, but it'll serve our purposes. Let's say that's negative 3, negative 2, negative 1, 0, 1, 2, 3. x is greater than negative 2."}, {"video_title": "Multi-step inequalities 2   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "I can draw a straighter number line than that. There we go. Still not that great, but it'll serve our purposes. Let's say that's negative 3, negative 2, negative 1, 0, 1, 2, 3. x is greater than negative 2. It does not include negative 2. It is not greater than or equal to negative 2. So we have to exclude negative 2."}, {"video_title": "Multi-step inequalities 2   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Let's say that's negative 3, negative 2, negative 1, 0, 1, 2, 3. x is greater than negative 2. It does not include negative 2. It is not greater than or equal to negative 2. So we have to exclude negative 2. And we exclude negative 2 by drawing an open circle at negative 2. But then all of the values greater than that are valid x's that would satisfy this inequality. So anything above it will work."}, {"video_title": "Multi-step inequalities 2   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So we have to exclude negative 2. And we exclude negative 2 by drawing an open circle at negative 2. But then all of the values greater than that are valid x's that would satisfy this inequality. So anything above it will work. And let's just try something that should work. And let's try something else that shouldn't work. So 0 should work."}, {"video_title": "Multi-step inequalities 2   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So anything above it will work. And let's just try something that should work. And let's try something else that shouldn't work. So 0 should work. It is greater than negative 2. It's right over here. So let's verify that."}, {"video_title": "Multi-step inequalities 2   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So 0 should work. It is greater than negative 2. It's right over here. So let's verify that. 5 times 0 plus 7 should be greater than 3 times 0 plus 1. So this is 7, because this is just a 0. 7 should be greater than 3."}, {"video_title": "Multi-step inequalities 2   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So let's verify that. 5 times 0 plus 7 should be greater than 3 times 0 plus 1. So this is 7, because this is just a 0. 7 should be greater than 3. 3 times 1. So 7 should be greater than 3. And it definitely is."}, {"video_title": "Multi-step inequalities 2   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "7 should be greater than 3. 3 times 1. So 7 should be greater than 3. And it definitely is. Now let's try something that should not work. Let's try negative 3. So 5 times negative 3 plus 7."}, {"video_title": "Multi-step inequalities 2   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "And it definitely is. Now let's try something that should not work. Let's try negative 3. So 5 times negative 3 plus 7. Let's see if it's greater than 3 times negative 3 plus 1. So this is negative 15 plus 7 is negative 8. That is negative 8."}, {"video_title": "Multi-step inequalities 2   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So 5 times negative 3 plus 7. Let's see if it's greater than 3 times negative 3 plus 1. So this is negative 15 plus 7 is negative 8. That is negative 8. Let's see if that is greater than negative 3 plus 1 is negative 2 times 3 is negative 6. Negative 8 is not greater than negative 6. Negative 8 is more negative than negative 6."}, {"video_title": "Multi-step inequalities 2   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "That is negative 8. Let's see if that is greater than negative 3 plus 1 is negative 2 times 3 is negative 6. Negative 8 is not greater than negative 6. Negative 8 is more negative than negative 6. It's less than. So it's good that negative 3 didn't work, because we didn't include that in our solution set. So we tried something that is in our solution set, and it did work, and something that's not."}, {"video_title": "Exponent properties involving quotients (examples)   8th grade   Khan Academy.mp3", "Sentence": "Let's say I were to ask you what 5 to the 6th power divided by 5 to the 2nd power is. Well, we could just go to the basic definition of what an exponent represents and say, well, 5 to the 6th power, that's going to be 5 times 5 times 5 times 5 times 5. One more 5. Times 5. 5 times itself 6 times. When 5 is squared, that's just 5 times itself 2 times. So it's going to be 5 times 5."}, {"video_title": "Exponent properties involving quotients (examples)   8th grade   Khan Academy.mp3", "Sentence": "Times 5. 5 times itself 6 times. When 5 is squared, that's just 5 times itself 2 times. So it's going to be 5 times 5. Well, we know how to simplify a fraction or a rational expression like this. We can divide the numerator and the denominator by 1 5, and then these will cancel out. And we could do it by another 5, or this 5 and this 5 will cancel out."}, {"video_title": "Exponent properties involving quotients (examples)   8th grade   Khan Academy.mp3", "Sentence": "So it's going to be 5 times 5. Well, we know how to simplify a fraction or a rational expression like this. We can divide the numerator and the denominator by 1 5, and then these will cancel out. And we could do it by another 5, or this 5 and this 5 will cancel out. And what are we going to be left with? 5 times 5 times 5 times 5 over, well, you could say over 1. Or you could say that this is just 5 to the 4th power."}, {"video_title": "Exponent properties involving quotients (examples)   8th grade   Khan Academy.mp3", "Sentence": "And we could do it by another 5, or this 5 and this 5 will cancel out. And what are we going to be left with? 5 times 5 times 5 times 5 over, well, you could say over 1. Or you could say that this is just 5 to the 4th power. Now notice what happens. Essentially, we started with 6 in the numerator, 6 5's multiplied by themselves in the numerator. And then we subtracted out."}, {"video_title": "Exponent properties involving quotients (examples)   8th grade   Khan Academy.mp3", "Sentence": "Or you could say that this is just 5 to the 4th power. Now notice what happens. Essentially, we started with 6 in the numerator, 6 5's multiplied by themselves in the numerator. And then we subtracted out. We were able to cancel out the 2 in the denominator. So this really was equal to 5 to the 6th power minus 2. So we were able to subtract the exponent in the denominator from the exponent in the numerator."}, {"video_title": "Exponent properties involving quotients (examples)   8th grade   Khan Academy.mp3", "Sentence": "And then we subtracted out. We were able to cancel out the 2 in the denominator. So this really was equal to 5 to the 6th power minus 2. So we were able to subtract the exponent in the denominator from the exponent in the numerator. And let's remember how this relates to multiplication. If I had 5 to the, let me do this in different colors, 5 to the 6th times 5 to the 2nd power, we saw in the last video that this is equal to 5 to the 6 plus, I'm trying to make it color coded for you, 6 plus 2 power. Now we see a new property."}, {"video_title": "Exponent properties involving quotients (examples)   8th grade   Khan Academy.mp3", "Sentence": "So we were able to subtract the exponent in the denominator from the exponent in the numerator. And let's remember how this relates to multiplication. If I had 5 to the, let me do this in different colors, 5 to the 6th times 5 to the 2nd power, we saw in the last video that this is equal to 5 to the 6 plus, I'm trying to make it color coded for you, 6 plus 2 power. Now we see a new property. And in the next video, we're going to see that these aren't really different properties, that they're kind of same sides of the same coin when we learn about negative exponents. But now in this video, we just saw that 5 to the 6th power divided by 5 to the 2nd power is going to be equal to 5 to the, it's time consuming to make it color coded for you, 6 minus 2 power. Or 5 to the 4th power here, it's going to be 5 to the 8th."}, {"video_title": "Exponent properties involving quotients (examples)   8th grade   Khan Academy.mp3", "Sentence": "Now we see a new property. And in the next video, we're going to see that these aren't really different properties, that they're kind of same sides of the same coin when we learn about negative exponents. But now in this video, we just saw that 5 to the 6th power divided by 5 to the 2nd power is going to be equal to 5 to the, it's time consuming to make it color coded for you, 6 minus 2 power. Or 5 to the 4th power here, it's going to be 5 to the 8th. So when you multiply exponents with the same base, you add the exponents, when you divide with the same base, you subtract the denominator exponent from the numerator exponent. Let's do a bunch more of these examples right here. What is 6 to the 7th power divided by 6 to the 3rd power?"}, {"video_title": "Exponent properties involving quotients (examples)   8th grade   Khan Academy.mp3", "Sentence": "Or 5 to the 4th power here, it's going to be 5 to the 8th. So when you multiply exponents with the same base, you add the exponents, when you divide with the same base, you subtract the denominator exponent from the numerator exponent. Let's do a bunch more of these examples right here. What is 6 to the 7th power divided by 6 to the 3rd power? Well once again, we can just use this property. This is going to be 6 to the 7 minus 3 power, which is equal to 6 to the 4th power. And you could multiply it out this way, like we did in the first problem, and verify that it indeed will be 6 to the 4th power."}, {"video_title": "Exponent properties involving quotients (examples)   8th grade   Khan Academy.mp3", "Sentence": "What is 6 to the 7th power divided by 6 to the 3rd power? Well once again, we can just use this property. This is going to be 6 to the 7 minus 3 power, which is equal to 6 to the 4th power. And you could multiply it out this way, like we did in the first problem, and verify that it indeed will be 6 to the 4th power. Now let's try something interesting. And this will be a good segue into the next video. Let's say we have 3 to the 4th power divided by 3 to the 10th power."}, {"video_title": "Exponent properties involving quotients (examples)   8th grade   Khan Academy.mp3", "Sentence": "And you could multiply it out this way, like we did in the first problem, and verify that it indeed will be 6 to the 4th power. Now let's try something interesting. And this will be a good segue into the next video. Let's say we have 3 to the 4th power divided by 3 to the 10th power. Well if we just go from basic principles, this would be 3 times 3 times 3 times 3, all of that over 3 times 3. We're going to have 10 of these. 3 times 3 times 3 times 3 times 3 times 3."}, {"video_title": "Exponent properties involving quotients (examples)   8th grade   Khan Academy.mp3", "Sentence": "Let's say we have 3 to the 4th power divided by 3 to the 10th power. Well if we just go from basic principles, this would be 3 times 3 times 3 times 3, all of that over 3 times 3. We're going to have 10 of these. 3 times 3 times 3 times 3 times 3 times 3. How many is that? 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. Well if we do what we did in the last video, this 3 cancels with that 3."}, {"video_title": "Exponent properties involving quotients (examples)   8th grade   Khan Academy.mp3", "Sentence": "3 times 3 times 3 times 3 times 3 times 3. How many is that? 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. Well if we do what we did in the last video, this 3 cancels with that 3. Those 3's cancel, those 3's cancel, those 3's cancel. And we're left with 1 over 1, 2, 3, 4, 5, 6 3's. So 1 over 3 to the 6th power."}, {"video_title": "Exponent properties involving quotients (examples)   8th grade   Khan Academy.mp3", "Sentence": "Well if we do what we did in the last video, this 3 cancels with that 3. Those 3's cancel, those 3's cancel, those 3's cancel. And we're left with 1 over 1, 2, 3, 4, 5, 6 3's. So 1 over 3 to the 6th power. We have 1 over all of these 3's down here. But that property that I just told you would have told you that, look, this should also be equal to 3 to the 4 minus 10 power. Well what's 4 minus 10?"}, {"video_title": "Exponent properties involving quotients (examples)   8th grade   Khan Academy.mp3", "Sentence": "So 1 over 3 to the 6th power. We have 1 over all of these 3's down here. But that property that I just told you would have told you that, look, this should also be equal to 3 to the 4 minus 10 power. Well what's 4 minus 10? We're going to get a negative number. This is 3 to the negative 6th power. So using the property we just saw, you'd get 3 to the negative 6th power."}, {"video_title": "Exponent properties involving quotients (examples)   8th grade   Khan Academy.mp3", "Sentence": "Well what's 4 minus 10? We're going to get a negative number. This is 3 to the negative 6th power. So using the property we just saw, you'd get 3 to the negative 6th power. Just multiplying them out, you get 1 over 3 to the 6th power. And the fun part about all of this is these are the same quantity. So now you're learning a little bit about what it means to take a negative exponent."}, {"video_title": "Exponent properties involving quotients (examples)   8th grade   Khan Academy.mp3", "Sentence": "So using the property we just saw, you'd get 3 to the negative 6th power. Just multiplying them out, you get 1 over 3 to the 6th power. And the fun part about all of this is these are the same quantity. So now you're learning a little bit about what it means to take a negative exponent. 3 to the negative 6th power is equal to 1 over 3 to the 6th power, and I'm going to do many, many more examples of this in the next video. But if you take anything to the negative power, so a to the negative b power is equal to 1 over a to the b. That's one thing that we just established just now."}, {"video_title": "Exponent properties involving quotients (examples)   8th grade   Khan Academy.mp3", "Sentence": "So now you're learning a little bit about what it means to take a negative exponent. 3 to the negative 6th power is equal to 1 over 3 to the 6th power, and I'm going to do many, many more examples of this in the next video. But if you take anything to the negative power, so a to the negative b power is equal to 1 over a to the b. That's one thing that we just established just now. And earlier in this video, we saw that if I have a to the b over a to the c, that this is equal to a to the b minus c. That's the other property we've been using. Now using what we've just learned and what we learned in the last video, let's do some more complicated problems. Let's say I have a to the third, b to the fourth power over a squared b and all of that to the third power."}, {"video_title": "Exponent properties involving quotients (examples)   8th grade   Khan Academy.mp3", "Sentence": "That's one thing that we just established just now. And earlier in this video, we saw that if I have a to the b over a to the c, that this is equal to a to the b minus c. That's the other property we've been using. Now using what we've just learned and what we learned in the last video, let's do some more complicated problems. Let's say I have a to the third, b to the fourth power over a squared b and all of that to the third power. Well, we can use the property we just learned to simplify the inside. This is going to be equal to a to the third divided by a squared. That's a to the 3 minus 2 power, right?"}, {"video_title": "Exponent properties involving quotients (examples)   8th grade   Khan Academy.mp3", "Sentence": "Let's say I have a to the third, b to the fourth power over a squared b and all of that to the third power. Well, we can use the property we just learned to simplify the inside. This is going to be equal to a to the third divided by a squared. That's a to the 3 minus 2 power, right? So this would simplify to just an a. And you can imagine, this is a times a times a divided by a times a. You'll just have an a on top."}, {"video_title": "Exponent properties involving quotients (examples)   8th grade   Khan Academy.mp3", "Sentence": "That's a to the 3 minus 2 power, right? So this would simplify to just an a. And you can imagine, this is a times a times a divided by a times a. You'll just have an a on top. And then the b, b to the fourth divided by b, well, that's just going to be b to the third. This is b to the first power. 4 minus 1 is 3."}, {"video_title": "Exponent properties involving quotients (examples)   8th grade   Khan Academy.mp3", "Sentence": "You'll just have an a on top. And then the b, b to the fourth divided by b, well, that's just going to be b to the third. This is b to the first power. 4 minus 1 is 3. And then all of that in parentheses to the third power. I don't want to forget about this third power out here. This third power is this one."}, {"video_title": "Exponent properties involving quotients (examples)   8th grade   Khan Academy.mp3", "Sentence": "4 minus 1 is 3. And then all of that in parentheses to the third power. I don't want to forget about this third power out here. This third power is this one. Let me color code it. That third power is that one right there. And then this a in orange is that a right there."}, {"video_title": "Exponent properties involving quotients (examples)   8th grade   Khan Academy.mp3", "Sentence": "This third power is this one. Let me color code it. That third power is that one right there. And then this a in orange is that a right there. And I think we understand what maps to what. And now we can use the property that when you multiply something and take the third power, that's equivalent of taking each of these. This is equal to a to the third power times b to the third to the third power."}, {"video_title": "Exponent properties involving quotients (examples)   8th grade   Khan Academy.mp3", "Sentence": "And then this a in orange is that a right there. And I think we understand what maps to what. And now we can use the property that when you multiply something and take the third power, that's equivalent of taking each of these. This is equal to a to the third power times b to the third to the third power. And then this is going to be equal to a to the third power. I'll do it like this. a to the third power."}, {"video_title": "Exponent properties involving quotients (examples)   8th grade   Khan Academy.mp3", "Sentence": "This is equal to a to the third power times b to the third to the third power. And then this is going to be equal to a to the third power. I'll do it like this. a to the third power. That's just a to the third right there. And then what is this? Times b to the 3 times 3 power."}, {"video_title": "Exponent properties involving quotients (examples)   8th grade   Khan Academy.mp3", "Sentence": "a to the third power. That's just a to the third right there. And then what is this? Times b to the 3 times 3 power. Times b to the ninth. And we would have simplified this about as far as you can go. Let's do one more of these."}, {"video_title": "Exponent properties involving quotients (examples)   8th grade   Khan Academy.mp3", "Sentence": "Times b to the 3 times 3 power. Times b to the ninth. And we would have simplified this about as far as you can go. Let's do one more of these. Because I think they're good practice and super valuable experience, I think, later on. Let's say I have 25xy to the sixth over 20y to the fifth x squared. So once again, we can rearrange the numerators and denominators."}, {"video_title": "Exponent properties involving quotients (examples)   8th grade   Khan Academy.mp3", "Sentence": "Let's do one more of these. Because I think they're good practice and super valuable experience, I think, later on. Let's say I have 25xy to the sixth over 20y to the fifth x squared. So once again, we can rearrange the numerators and denominators. So this you could rewrite as 25 over 20 times x over x squared. We could have made this bottom 20x squared y to the fifth. It doesn't matter the order we do it in."}, {"video_title": "Exponent properties involving quotients (examples)   8th grade   Khan Academy.mp3", "Sentence": "So once again, we can rearrange the numerators and denominators. So this you could rewrite as 25 over 20 times x over x squared. We could have made this bottom 20x squared y to the fifth. It doesn't matter the order we do it in. Times y to the sixth over y to the fifth. And let's use our newly learned exponent properties and actually just simplifying fractions. 25 over 20, if you divide them both by 5, you're going to get this is equal to 5 over 4. x divided by x squared."}, {"video_title": "Exponent properties involving quotients (examples)   8th grade   Khan Academy.mp3", "Sentence": "It doesn't matter the order we do it in. Times y to the sixth over y to the fifth. And let's use our newly learned exponent properties and actually just simplifying fractions. 25 over 20, if you divide them both by 5, you're going to get this is equal to 5 over 4. x divided by x squared. Well, there's two ways you could think about it. That you could view as x to the negative 1. You have a first power here."}, {"video_title": "Exponent properties involving quotients (examples)   8th grade   Khan Academy.mp3", "Sentence": "25 over 20, if you divide them both by 5, you're going to get this is equal to 5 over 4. x divided by x squared. Well, there's two ways you could think about it. That you could view as x to the negative 1. You have a first power here. 1 minus 2 is negative 1. So this right here is equal to x to the negative 1 power. Or it could also be equal to 1 over x."}, {"video_title": "Exponent properties involving quotients (examples)   8th grade   Khan Academy.mp3", "Sentence": "You have a first power here. 1 minus 2 is negative 1. So this right here is equal to x to the negative 1 power. Or it could also be equal to 1 over x. These are equivalent. So let's say that this is equivalent to 1 over x. Just like that."}, {"video_title": "Exponent properties involving quotients (examples)   8th grade   Khan Academy.mp3", "Sentence": "Or it could also be equal to 1 over x. These are equivalent. So let's say that this is equivalent to 1 over x. Just like that. And it would be x over x times x. One of those sets of x's would cancel out. And you're just left with 1 over x."}, {"video_title": "Exponent properties involving quotients (examples)   8th grade   Khan Academy.mp3", "Sentence": "Just like that. And it would be x over x times x. One of those sets of x's would cancel out. And you're just left with 1 over x. And then finally, y to the sixth over y to the fifth. That's y. That right there is y to the sixth minus 5 power."}, {"video_title": "Exponent properties involving quotients (examples)   8th grade   Khan Academy.mp3", "Sentence": "And you're just left with 1 over x. And then finally, y to the sixth over y to the fifth. That's y. That right there is y to the sixth minus 5 power. Which is just y to the first power. Or just y. So times y."}, {"video_title": "Exponent properties involving quotients (examples)   8th grade   Khan Academy.mp3", "Sentence": "That right there is y to the sixth minus 5 power. Which is just y to the first power. Or just y. So times y. So if you want to write it all out as just one combined rational expression, you have 5 times 1 times y. Which would be 5y. All of that over 4 times x."}, {"video_title": "Exponent properties involving quotients (examples)   8th grade   Khan Academy.mp3", "Sentence": "So times y. So if you want to write it all out as just one combined rational expression, you have 5 times 1 times y. Which would be 5y. All of that over 4 times x. This is y over 1. So 4 times x times 1. All of that over 4x."}, {"video_title": "The Pythagorean theorem intro   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "Pythagorean Theorem. Which is fun on its own, but you'll see as you learn more and more mathematics, it's one of those cornerstone theorems of really all of math. It's useful in geometry, it's kind of the backbone of trigonometry. You're also going to use it to calculate distances between points. So it's a good thing to really make sure we know well. So enough talk on my end, let me tell you what the Pythagorean Theorem is. So if we have a triangle, and the triangle has to be a right triangle."}, {"video_title": "The Pythagorean theorem intro   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "You're also going to use it to calculate distances between points. So it's a good thing to really make sure we know well. So enough talk on my end, let me tell you what the Pythagorean Theorem is. So if we have a triangle, and the triangle has to be a right triangle. So it has to be a right triangle, which means that one of the three angles in the triangle have to be 90 degrees. And you specify that it's 90 degrees by drawing that little box right there. So that right there is, let me do this in a different color, that right there is a 90 degree angle."}, {"video_title": "The Pythagorean theorem intro   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "So if we have a triangle, and the triangle has to be a right triangle. So it has to be a right triangle, which means that one of the three angles in the triangle have to be 90 degrees. And you specify that it's 90 degrees by drawing that little box right there. So that right there is, let me do this in a different color, that right there is a 90 degree angle. Or we could call it a right angle. And a triangle that has a right angle in it is called a right triangle. So this is called a right triangle."}, {"video_title": "The Pythagorean theorem intro   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "So that right there is, let me do this in a different color, that right there is a 90 degree angle. Or we could call it a right angle. And a triangle that has a right angle in it is called a right triangle. So this is called a right triangle. Now, with the Pythagorean Theorem, if we know two sides of a right triangle, we can always figure out the third side. And before I show you how to do that, let me give you one more piece of terminology. The longest side of a right triangle is the side opposite the 90 degree angle, or opposite the right angle."}, {"video_title": "The Pythagorean theorem intro   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "So this is called a right triangle. Now, with the Pythagorean Theorem, if we know two sides of a right triangle, we can always figure out the third side. And before I show you how to do that, let me give you one more piece of terminology. The longest side of a right triangle is the side opposite the 90 degree angle, or opposite the right angle. So in this case, it is this side right here. This is the longest side. The way to figure out where that right triangle is, and it opens into that longest side, that longest side is called the hypotenuse."}, {"video_title": "The Pythagorean theorem intro   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "The longest side of a right triangle is the side opposite the 90 degree angle, or opposite the right angle. So in this case, it is this side right here. This is the longest side. The way to figure out where that right triangle is, and it opens into that longest side, that longest side is called the hypotenuse. And it's good to know because we'll keep referring to it. And just so we always are good at identifying the hypotenuse, let me draw a couple of more right triangles. So let's say I have a triangle that looks like that."}, {"video_title": "The Pythagorean theorem intro   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "The way to figure out where that right triangle is, and it opens into that longest side, that longest side is called the hypotenuse. And it's good to know because we'll keep referring to it. And just so we always are good at identifying the hypotenuse, let me draw a couple of more right triangles. So let's say I have a triangle that looks like that. A triangle that looks, let me draw it a little bit nicer. So let's say I have a triangle that looks like that. And I will tell you that this angle right here is 90 degrees."}, {"video_title": "The Pythagorean theorem intro   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "So let's say I have a triangle that looks like that. A triangle that looks, let me draw it a little bit nicer. So let's say I have a triangle that looks like that. And I will tell you that this angle right here is 90 degrees. In this situation, this is the hypotenuse because it is opposite the 90 degree angle. It is the longest side. Let me do one more, just so that we're good at recognizing the hypotenuse."}, {"video_title": "The Pythagorean theorem intro   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "And I will tell you that this angle right here is 90 degrees. In this situation, this is the hypotenuse because it is opposite the 90 degree angle. It is the longest side. Let me do one more, just so that we're good at recognizing the hypotenuse. So let's say that that is my triangle. And this is the 90 degree angle right there. And I think you know how to do this already."}, {"video_title": "The Pythagorean theorem intro   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "Let me do one more, just so that we're good at recognizing the hypotenuse. So let's say that that is my triangle. And this is the 90 degree angle right there. And I think you know how to do this already. You go right, what it opens into, that is the hypotenuse. That is the longest side. So that is the hypotenuse."}, {"video_title": "The Pythagorean theorem intro   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "And I think you know how to do this already. You go right, what it opens into, that is the hypotenuse. That is the longest side. So that is the hypotenuse. So once you have identified the hypotenuse, and let's say that that has length c, and now we're going to learn what the Pythagorean theorem tells us. So let's say that c is equal to the length of the hypotenuse. So let's call this c, that side is c. Let's call this side right over here a."}, {"video_title": "The Pythagorean theorem intro   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "So that is the hypotenuse. So once you have identified the hypotenuse, and let's say that that has length c, and now we're going to learn what the Pythagorean theorem tells us. So let's say that c is equal to the length of the hypotenuse. So let's call this c, that side is c. Let's call this side right over here a. And let's call this side over here b. So the Pythagorean theorem tells us that a squared, so one of the shorter sides, the length of one of the shorter sides squared, plus the length of the other shorter side squared, is going to be equal to the length of the hypotenuse squared. Now, let's do that with an actual problem, and you'll see that it's actually not so bad."}, {"video_title": "The Pythagorean theorem intro   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "So let's call this c, that side is c. Let's call this side right over here a. And let's call this side over here b. So the Pythagorean theorem tells us that a squared, so one of the shorter sides, the length of one of the shorter sides squared, plus the length of the other shorter side squared, is going to be equal to the length of the hypotenuse squared. Now, let's do that with an actual problem, and you'll see that it's actually not so bad. So let's say that I have a triangle that looks like this. Let me draw it. Let's say that this is my triangle."}, {"video_title": "The Pythagorean theorem intro   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "Now, let's do that with an actual problem, and you'll see that it's actually not so bad. So let's say that I have a triangle that looks like this. Let me draw it. Let's say that this is my triangle. Looks something like this. And let's say that they tell us that this is the right angle, that this length right here, let me do this in different colors, this length right here is 3, and that this length right here is 4. And they want us to figure out that length right there."}, {"video_title": "The Pythagorean theorem intro   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "Let's say that this is my triangle. Looks something like this. And let's say that they tell us that this is the right angle, that this length right here, let me do this in different colors, this length right here is 3, and that this length right here is 4. And they want us to figure out that length right there. Now, the first thing you want to do before you even apply the Pythagorean theorem is to make sure you have your hypotenuse straight. You make sure you know what you're solving for. And in this circumstance, we're solving for the hypotenuse."}, {"video_title": "The Pythagorean theorem intro   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "And they want us to figure out that length right there. Now, the first thing you want to do before you even apply the Pythagorean theorem is to make sure you have your hypotenuse straight. You make sure you know what you're solving for. And in this circumstance, we're solving for the hypotenuse. And we know that because this side over here, it is the side opposite the right angle. If we look at the Pythagorean theorem, this is c. This is the longest side. So now we're ready to apply."}, {"video_title": "The Pythagorean theorem intro   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "And in this circumstance, we're solving for the hypotenuse. And we know that because this side over here, it is the side opposite the right angle. If we look at the Pythagorean theorem, this is c. This is the longest side. So now we're ready to apply. We're ready to apply the Pythagorean theorem. It tells us that 4 squared, one of the shorter sides, plus 3 squared, the square of another of the shorter sides, is going to be equal to this longer side squared. The hypotenuse squared is going to be equal to c squared."}, {"video_title": "The Pythagorean theorem intro   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "So now we're ready to apply. We're ready to apply the Pythagorean theorem. It tells us that 4 squared, one of the shorter sides, plus 3 squared, the square of another of the shorter sides, is going to be equal to this longer side squared. The hypotenuse squared is going to be equal to c squared. And then you just solve for c. So 4 squared is the same thing as 4 times 4. That is 16. And 3 squared is the same thing as 3 times 3."}, {"video_title": "The Pythagorean theorem intro   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "The hypotenuse squared is going to be equal to c squared. And then you just solve for c. So 4 squared is the same thing as 4 times 4. That is 16. And 3 squared is the same thing as 3 times 3. So that is 9. And that is going to be equal to c squared. Now, what is 16 plus 9?"}, {"video_title": "The Pythagorean theorem intro   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "And 3 squared is the same thing as 3 times 3. So that is 9. And that is going to be equal to c squared. Now, what is 16 plus 9? It's 25. So 25 is equal to c squared. And we can take the positive square root of both sides."}, {"video_title": "The Pythagorean theorem intro   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "Now, what is 16 plus 9? It's 25. So 25 is equal to c squared. And we can take the positive square root of both sides. c, I guess, just if you look at it mathematically, could be negative 5 as well. But we're dealing with distances, so we only care about the positive roots. So you take the principal root of both sides, and you get 5 is equal to c, or the length of the longest side is equal to 5."}, {"video_title": "The Pythagorean theorem intro   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "And we can take the positive square root of both sides. c, I guess, just if you look at it mathematically, could be negative 5 as well. But we're dealing with distances, so we only care about the positive roots. So you take the principal root of both sides, and you get 5 is equal to c, or the length of the longest side is equal to 5. Now, you could use the Pythagorean theorem, if we give you two of the sides to figure out the third side, no matter what the third side is. So let's do another one right over here. Let's say that our triangle looks like this."}, {"video_title": "The Pythagorean theorem intro   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "So you take the principal root of both sides, and you get 5 is equal to c, or the length of the longest side is equal to 5. Now, you could use the Pythagorean theorem, if we give you two of the sides to figure out the third side, no matter what the third side is. So let's do another one right over here. Let's say that our triangle looks like this. And that is our right angle. Let's say that this side over here has length 12. And let's say that this side over here has length 6."}, {"video_title": "The Pythagorean theorem intro   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "Let's say that our triangle looks like this. And that is our right angle. Let's say that this side over here has length 12. And let's say that this side over here has length 6. And we want to figure out this length right over there. Now, like I said, the first thing you want to do is identify the hypotenuse. And that's going to be the side opposite the right angle."}, {"video_title": "The Pythagorean theorem intro   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "And let's say that this side over here has length 6. And we want to figure out this length right over there. Now, like I said, the first thing you want to do is identify the hypotenuse. And that's going to be the side opposite the right angle. We have the right angle here. You go opposite the right angle. The longest side, the hypotenuse, is right there."}, {"video_title": "The Pythagorean theorem intro   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "And that's going to be the side opposite the right angle. We have the right angle here. You go opposite the right angle. The longest side, the hypotenuse, is right there. So if we think about the Pythagorean theorem, that a squared plus b squared is equal to c squared, 12 you could view as c. This is the hypotenuse. The c squared is the hypotenuse squared. So you could say 12 is equal to c. And then we could set these sides."}, {"video_title": "The Pythagorean theorem intro   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "The longest side, the hypotenuse, is right there. So if we think about the Pythagorean theorem, that a squared plus b squared is equal to c squared, 12 you could view as c. This is the hypotenuse. The c squared is the hypotenuse squared. So you could say 12 is equal to c. And then we could set these sides. It doesn't matter whether you call one of them a or one of them b. So let's just call this side right here. Let's say a is equal to 6."}, {"video_title": "The Pythagorean theorem intro   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "So you could say 12 is equal to c. And then we could set these sides. It doesn't matter whether you call one of them a or one of them b. So let's just call this side right here. Let's say a is equal to 6. And then we say b, this colored b, is equal to question mark. And now we can apply the Pythagorean theorem. a squared, which is 6 squared, plus the unknown b squared, is equal to the hypotenuse squared, is equal to c squared, is equal to 12 squared."}, {"video_title": "The Pythagorean theorem intro   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "Let's say a is equal to 6. And then we say b, this colored b, is equal to question mark. And now we can apply the Pythagorean theorem. a squared, which is 6 squared, plus the unknown b squared, is equal to the hypotenuse squared, is equal to c squared, is equal to 12 squared. And now we can solve for b. And notice the difference here. Now we're not solving for the hypotenuse."}, {"video_title": "The Pythagorean theorem intro   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "a squared, which is 6 squared, plus the unknown b squared, is equal to the hypotenuse squared, is equal to c squared, is equal to 12 squared. And now we can solve for b. And notice the difference here. Now we're not solving for the hypotenuse. We're solving for one of the shorter sides. In the last example, we solved for the hypotenuse. We solved for c. That's why it's always important to recognize that a squared plus b squared plus c squared, c is the length of the hypotenuse."}, {"video_title": "The Pythagorean theorem intro   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "Now we're not solving for the hypotenuse. We're solving for one of the shorter sides. In the last example, we solved for the hypotenuse. We solved for c. That's why it's always important to recognize that a squared plus b squared plus c squared, c is the length of the hypotenuse. Let's just solve for b here. So we get 6 squared is 36, plus b squared is equal to 12 squared. This 12 times 12 is 144."}, {"video_title": "The Pythagorean theorem intro   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "We solved for c. That's why it's always important to recognize that a squared plus b squared plus c squared, c is the length of the hypotenuse. Let's just solve for b here. So we get 6 squared is 36, plus b squared is equal to 12 squared. This 12 times 12 is 144. Now we can subtract 36 from both sides of this equation. Subtract 36. Those cancel out."}, {"video_title": "The Pythagorean theorem intro   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "This 12 times 12 is 144. Now we can subtract 36 from both sides of this equation. Subtract 36. Those cancel out. On the left-hand side, we're left with just a b squared is equal to 144 minus 36 is what? That is 144 minus 30 is 114. And then you subtract 6 is 108."}, {"video_title": "The Pythagorean theorem intro   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "Those cancel out. On the left-hand side, we're left with just a b squared is equal to 144 minus 36 is what? That is 144 minus 30 is 114. And then you subtract 6 is 108. So this is going to be 108. So that's what b squared is. Now we want to take the principal root, or the positive root, of both sides."}, {"video_title": "The Pythagorean theorem intro   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "And then you subtract 6 is 108. So this is going to be 108. So that's what b squared is. Now we want to take the principal root, or the positive root, of both sides. And you get b is equal to the square root, the principal root, of 108. Now let's see if we can simplify this a little bit. The square root of 108."}, {"video_title": "The Pythagorean theorem intro   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "Now we want to take the principal root, or the positive root, of both sides. And you get b is equal to the square root, the principal root, of 108. Now let's see if we can simplify this a little bit. The square root of 108. And what we could do is we could take the prime factorization of 108 and see how we can simplify this radical. So 108 is the same thing as 2 times 54, which is the same thing as 2 times 27, which is the same thing as 3 times 9. So we have the square root of 108 is the same thing as the square root of 2 times 2 times, actually I'm not done, 9 can be factorized into 3 times 3."}, {"video_title": "The Pythagorean theorem intro   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "The square root of 108. And what we could do is we could take the prime factorization of 108 and see how we can simplify this radical. So 108 is the same thing as 2 times 54, which is the same thing as 2 times 27, which is the same thing as 3 times 9. So we have the square root of 108 is the same thing as the square root of 2 times 2 times, actually I'm not done, 9 can be factorized into 3 times 3. So it's 2 times 2 times 3 times 3 times 3. And so we have a couple of perfect squares in here. Let me rewrite it a little bit neater."}, {"video_title": "The Pythagorean theorem intro   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "So we have the square root of 108 is the same thing as the square root of 2 times 2 times, actually I'm not done, 9 can be factorized into 3 times 3. So it's 2 times 2 times 3 times 3 times 3. And so we have a couple of perfect squares in here. Let me rewrite it a little bit neater. And this is all an exercise in simplifying radicals that you will bump into a lot while doing the Pythagorean theorem, so it doesn't hurt to do it right here. So this is the same thing as the square root of 2 times 2 times 3 times 3 times the square root of that last 3 right over there. And this is the same thing, and you know you wouldn't have to do all of this in your head, well all of this on paper, you could do it in your head."}, {"video_title": "The Pythagorean theorem intro   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "Let me rewrite it a little bit neater. And this is all an exercise in simplifying radicals that you will bump into a lot while doing the Pythagorean theorem, so it doesn't hurt to do it right here. So this is the same thing as the square root of 2 times 2 times 3 times 3 times the square root of that last 3 right over there. And this is the same thing, and you know you wouldn't have to do all of this in your head, well all of this on paper, you could do it in your head. What is this? 2 times 2 is 4, 4 times 9, this is 36. This is the square root of 36 times the square root of 3."}, {"video_title": "The Pythagorean theorem intro   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "And this is the same thing, and you know you wouldn't have to do all of this in your head, well all of this on paper, you could do it in your head. What is this? 2 times 2 is 4, 4 times 9, this is 36. This is the square root of 36 times the square root of 3. The principal root of 36 is 6. So this simplifies to 6 square roots of 3. So the length of B, you could write the square root of 108, or you could say it's equal to 6 times the square root of 3."}, {"video_title": "Special products of the form (x+a)(x-a)   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "Well, one way to tackle it is the way that we've always tackled when we multiply binomials is just apply the distributive property twice. So first, we could take this entire yellow x plus three and multiply it times each of these two terms. So first, we can multiply it times this x, so that's going to be x times x plus three, and then we are going to multiply it times, we could say this negative three. So we could write minus three times, now that's going to be multiplied by x plus three again, x plus three, and then we apply the distributive property one more time, where we take this magenta x and we distribute it across this x plus three, so x times x is x squared, x times three is three x, and then we do it on this side. Negative three times x is negative three x, and negative three times three is negative nine. And what does this simplify to? Well, we're gonna get x squared, and we have three x minus three x, so these two characters cancel out, and we are just left with x squared minus nine."}, {"video_title": "Special products of the form (x+a)(x-a)   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "So we could write minus three times, now that's going to be multiplied by x plus three again, x plus three, and then we apply the distributive property one more time, where we take this magenta x and we distribute it across this x plus three, so x times x is x squared, x times three is three x, and then we do it on this side. Negative three times x is negative three x, and negative three times three is negative nine. And what does this simplify to? Well, we're gonna get x squared, and we have three x minus three x, so these two characters cancel out, and we are just left with x squared minus nine. And you might see a little pattern here. Notice, I added three and then I subtracted three, and I got this, I got the x squared, and then if you take three and multiply it by negative three, you are going to get a negative nine. And notice, the middle terms canceled out, and one thing you might ask is, well, will that always be the case if we add a number and then we subtract that same number like that?"}, {"video_title": "Special products of the form (x+a)(x-a)   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "Well, we're gonna get x squared, and we have three x minus three x, so these two characters cancel out, and we are just left with x squared minus nine. And you might see a little pattern here. Notice, I added three and then I subtracted three, and I got this, I got the x squared, and then if you take three and multiply it by negative three, you are going to get a negative nine. And notice, the middle terms canceled out, and one thing you might ask is, well, will that always be the case if we add a number and then we subtract that same number like that? And we could try it out. Let's talk in general terms. So if we, instead of doing x plus three times x minus three, we could write the same thing as, instead of three, let's just say you have x plus, x plus a times x minus a, times x minus a."}, {"video_title": "Special products of the form (x+a)(x-a)   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "And notice, the middle terms canceled out, and one thing you might ask is, well, will that always be the case if we add a number and then we subtract that same number like that? And we could try it out. Let's talk in general terms. So if we, instead of doing x plus three times x minus three, we could write the same thing as, instead of three, let's just say you have x plus, x plus a times x minus a, times x minus a. And I encourage you to pause this video and work it all out. Just assume a is some number, like three or some other number, and apply the distributive property twice and see what you get. Well, let's work through it."}, {"video_title": "Special products of the form (x+a)(x-a)   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "So if we, instead of doing x plus three times x minus three, we could write the same thing as, instead of three, let's just say you have x plus, x plus a times x minus a, times x minus a. And I encourage you to pause this video and work it all out. Just assume a is some number, like three or some other number, and apply the distributive property twice and see what you get. Well, let's work through it. So first we can distribute this yellow x plus a onto the x and the negative a. So x plus a times x, or we could say x times x plus a. So that's going to be, that's going to be x times x plus a, and then we're going to have minus a, or this negative a times x plus a."}, {"video_title": "Special products of the form (x+a)(x-a)   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "Well, let's work through it. So first we can distribute this yellow x plus a onto the x and the negative a. So x plus a times x, or we could say x times x plus a. So that's going to be, that's going to be x times x plus a, and then we're going to have minus a, or this negative a times x plus a. So minus, and then we're going to have this minus a times x plus a, times x plus a, times x plus a. Notice, all I did is I distributed this yellow, I just distributed this big chunk of this expression, I just distributed it onto the x and onto this negative a. I'm multiplying it times the x and I'm multiplying it by the negative a. And now we can apply the distributive property again."}, {"video_title": "Special products of the form (x+a)(x-a)   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "So that's going to be, that's going to be x times x plus a, and then we're going to have minus a, or this negative a times x plus a. So minus, and then we're going to have this minus a times x plus a, times x plus a, times x plus a. Notice, all I did is I distributed this yellow, I just distributed this big chunk of this expression, I just distributed it onto the x and onto this negative a. I'm multiplying it times the x and I'm multiplying it by the negative a. And now we can apply the distributive property again. X times x is x squared. X times a is ax. And then we get negative a times x is negative ax."}, {"video_title": "Special products of the form (x+a)(x-a)   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "And now we can apply the distributive property again. X times x is x squared. X times a is ax. And then we get negative a times x is negative ax. And then negative a times a is negative a squared. And notice, regardless of my choice of a, I'm going to have ax and then minus ax. So this is always going to cancel out."}, {"video_title": "Special products of the form (x+a)(x-a)   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "And then we get negative a times x is negative ax. And then negative a times a is negative a squared. And notice, regardless of my choice of a, I'm going to have ax and then minus ax. So this is always going to cancel out. It didn't just work for the case when a was three. For any a, if I have a times x and then I subtract a times x, that's just going to cancel out. So this is just going to cancel out."}, {"video_title": "Special products of the form (x+a)(x-a)   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "So this is always going to cancel out. It didn't just work for the case when a was three. For any a, if I have a times x and then I subtract a times x, that's just going to cancel out. So this is just going to cancel out. And what are we going to be left with? We are going to be left with x squared minus a squared. X squared minus a squared."}, {"video_title": "Special products of the form (x+a)(x-a)   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "So this is just going to cancel out. And what are we going to be left with? We are going to be left with x squared minus a squared. X squared minus a squared. And you could view this as a special case. When you have something x plus something times x minus that same something, it's going to be x squared minus that something squared. And this is a good one to know in general."}, {"video_title": "Special products of the form (x+a)(x-a)   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "X squared minus a squared. And you could view this as a special case. When you have something x plus something times x minus that same something, it's going to be x squared minus that something squared. And this is a good one to know in general. This is a good one to know in general. And we could use it to quickly figure out the products of other binomials that fit this pattern here. So if I were to say, quick, what is x plus 10 times x minus 10?"}, {"video_title": "Special products of the form (x+a)(x-a)   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "And this is a good one to know in general. This is a good one to know in general. And we could use it to quickly figure out the products of other binomials that fit this pattern here. So if I were to say, quick, what is x plus 10 times x minus 10? Well, you could say, all right, this fits the pattern. It's x plus a times x minus a. So it's going to be x squared minus a squared."}, {"video_title": "Percent word problem example 2   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "In the United States, 13 out of every 20 cans are recycled. What percent of cans are recycled? So 13 out of every 20 are recycled. So 13 20, it's 13 over 20, could also be viewed as 13, 13 divided by 20, or 13 divided by 20. And if we do this, we'll get a decimal, it's fairly straightforward to convert that decimal into a percentage. So 13 divided by 20, we have the smaller number, in this case, being divided by the larger number. So we're going to get a value less than one."}, {"video_title": "Percent word problem example 2   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "So 13 20, it's 13 over 20, could also be viewed as 13, 13 divided by 20, or 13 divided by 20. And if we do this, we'll get a decimal, it's fairly straightforward to convert that decimal into a percentage. So 13 divided by 20, we have the smaller number, in this case, being divided by the larger number. So we're going to get a value less than one. Since we're gonna get a value less than one, let's put a decimal right over here, and let's add a couple of zeros, as many zeros as we would need. And we could say, hey look, 20 goes into 13, zero times, zero times 20 is zero, and then 13 minus zero is 13, now you bring down a zero. 20 goes into 130, let's see, it goes in, five times 20 is 100, so six times 20 is 120, so it's gonna go six times."}, {"video_title": "Percent word problem example 2   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "So we're going to get a value less than one. Since we're gonna get a value less than one, let's put a decimal right over here, and let's add a couple of zeros, as many zeros as we would need. And we could say, hey look, 20 goes into 13, zero times, zero times 20 is zero, and then 13 minus zero is 13, now you bring down a zero. 20 goes into 130, let's see, it goes in, five times 20 is 100, so six times 20 is 120, so it's gonna go six times. Six times 20 is 120, you subtract, you get a 10, let's bring down another zero. 20 goes into 100 five times, five times 20 is 100, and we are done. So this written as a decimal, this written as a decimal is 0.65."}, {"video_title": "Percent word problem example 2   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "20 goes into 130, let's see, it goes in, five times 20 is 100, so six times 20 is 120, so it's gonna go six times. Six times 20 is 120, you subtract, you get a 10, let's bring down another zero. 20 goes into 100 five times, five times 20 is 100, and we are done. So this written as a decimal, this written as a decimal is 0.65. So as a decimal, it's 0.65, and if you wanna write it as a percentage, you essentially multiply this by 100, or another way you could say is you shift the decimal over two spots to the right. So this is going to be equal to 65%, 65%. Now, there's another way you could have done it, is you could have said, look, percent literally means per 100, so 13 out of 20, 13 out of 20 is going to be equal to what, over, is going to be equal to what over 100?"}, {"video_title": "Percent word problem example 2   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "So this written as a decimal, this written as a decimal is 0.65. So as a decimal, it's 0.65, and if you wanna write it as a percentage, you essentially multiply this by 100, or another way you could say is you shift the decimal over two spots to the right. So this is going to be equal to 65%, 65%. Now, there's another way you could have done it, is you could have said, look, percent literally means per 100, so 13 out of 20, 13 out of 20 is going to be equal to what, over, is going to be equal to what over 100? Well, to go from 20 to 100, you forget the denominator, to go from 20 to 100, you would multiply by five, so let's multiply the numerator by five as well. And 13 times five, let's see, that's 15 plus 50, which is 65. So this would have been a little bit of a faster way to do it, especially if you recognize it's pretty easy to go from 20 to 100, you multiply by five, so we would do the same thing with the 13, and so you'd get 65 over 100, which is the same thing as 65 per, let me write this percent symbol, 65%."}, {"video_title": "Similarity postulates   Similarity   Geometry   Khan Academy.mp3", "Sentence": "It looks something like this. ABC. I want to think about the minimum amount of information. I want to come up with a couple of postulates that we can use to determine whether another triangle is similar to triangle ABC. So we already know that if all three angles, all three of the corresponding angles are congruent to the corresponding angles on ABC, then we know that we're dealing with congruent triangles. For example, if this is 30 degrees, this angle is 90 degrees, and this angle right over here is 60 degrees, and we have another triangle that looks like this, that looks like this. It's clearly a smaller triangle, but its corresponding angles, so this is 30 degrees, this is 90 degrees, and this is 60 degrees, we know that XYZ, in this case, is going to be similar to ABC."}, {"video_title": "Similarity postulates   Similarity   Geometry   Khan Academy.mp3", "Sentence": "I want to come up with a couple of postulates that we can use to determine whether another triangle is similar to triangle ABC. So we already know that if all three angles, all three of the corresponding angles are congruent to the corresponding angles on ABC, then we know that we're dealing with congruent triangles. For example, if this is 30 degrees, this angle is 90 degrees, and this angle right over here is 60 degrees, and we have another triangle that looks like this, that looks like this. It's clearly a smaller triangle, but its corresponding angles, so this is 30 degrees, this is 90 degrees, and this is 60 degrees, we know that XYZ, in this case, is going to be similar to ABC. We would know from this, because corresponding angles are congruent, we would know that triangle ABC is similar to triangle XYZ. You've got to get the order right to make sure that you have the right corresponding angles. Y corresponds to the 90 degree angle, X corresponds to the 30 degree angle, A corresponds to the 30 degree angle, so A and X are the first two things, B and Y, which are the 90 degrees, are the second two, and then Z is the last one."}, {"video_title": "Similarity postulates   Similarity   Geometry   Khan Academy.mp3", "Sentence": "It's clearly a smaller triangle, but its corresponding angles, so this is 30 degrees, this is 90 degrees, and this is 60 degrees, we know that XYZ, in this case, is going to be similar to ABC. We would know from this, because corresponding angles are congruent, we would know that triangle ABC is similar to triangle XYZ. You've got to get the order right to make sure that you have the right corresponding angles. Y corresponds to the 90 degree angle, X corresponds to the 30 degree angle, A corresponds to the 30 degree angle, so A and X are the first two things, B and Y, which are the 90 degrees, are the second two, and then Z is the last one. That's what we know already, if you have three angles. But do you need three angles? If we only knew two of the angles, would that be enough?"}, {"video_title": "Similarity postulates   Similarity   Geometry   Khan Academy.mp3", "Sentence": "Y corresponds to the 90 degree angle, X corresponds to the 30 degree angle, A corresponds to the 30 degree angle, so A and X are the first two things, B and Y, which are the 90 degrees, are the second two, and then Z is the last one. That's what we know already, if you have three angles. But do you need three angles? If we only knew two of the angles, would that be enough? Sure, because if you know two angles for a triangle, you know the third. For example, if I have another triangle, if I have a triangle that looks like this, and if I told you that only two of the corresponding angles are congruent, maybe this angle right here is congruent to this angle, and that angle right there is congruent to that angle. Is that enough to say that these two triangles are similar?"}, {"video_title": "Similarity postulates   Similarity   Geometry   Khan Academy.mp3", "Sentence": "If we only knew two of the angles, would that be enough? Sure, because if you know two angles for a triangle, you know the third. For example, if I have another triangle, if I have a triangle that looks like this, and if I told you that only two of the corresponding angles are congruent, maybe this angle right here is congruent to this angle, and that angle right there is congruent to that angle. Is that enough to say that these two triangles are similar? Sure, because in a triangle, if you know two of the angles, then you know what the last angle has to be. If you know that this is 30 and you know that that is 90, then you know that this angle has to be 60 degrees. Whatever these two angles are, subtract them from 180, and that's going to be this angle."}, {"video_title": "Similarity postulates   Similarity   Geometry   Khan Academy.mp3", "Sentence": "Is that enough to say that these two triangles are similar? Sure, because in a triangle, if you know two of the angles, then you know what the last angle has to be. If you know that this is 30 and you know that that is 90, then you know that this angle has to be 60 degrees. Whatever these two angles are, subtract them from 180, and that's going to be this angle. In general, in order to show similarity, you don't have to show three corresponding angles are congruent. You really just have to show two. This will be the first of our similarity postulates."}, {"video_title": "Similarity postulates   Similarity   Geometry   Khan Academy.mp3", "Sentence": "Whatever these two angles are, subtract them from 180, and that's going to be this angle. In general, in order to show similarity, you don't have to show three corresponding angles are congruent. You really just have to show two. This will be the first of our similarity postulates. We've called it angle-angle. If you could show that two corresponding angles are congruent, then we're dealing with similar triangles. For example, just to put some numbers here, if this was 30 degrees, and we know that on this triangle this is 90 degrees right over here, we know that this triangle right over here is similar to that one there."}, {"video_title": "Similarity postulates   Similarity   Geometry   Khan Academy.mp3", "Sentence": "This will be the first of our similarity postulates. We've called it angle-angle. If you could show that two corresponding angles are congruent, then we're dealing with similar triangles. For example, just to put some numbers here, if this was 30 degrees, and we know that on this triangle this is 90 degrees right over here, we know that this triangle right over here is similar to that one there. You can really just go to the third angle in a pretty straightforward way. You say, hey, this third angle is 60 degrees, so all three angles are the same. That's one of our constraints for similarity."}, {"video_title": "Similarity postulates   Similarity   Geometry   Khan Academy.mp3", "Sentence": "For example, just to put some numbers here, if this was 30 degrees, and we know that on this triangle this is 90 degrees right over here, we know that this triangle right over here is similar to that one there. You can really just go to the third angle in a pretty straightforward way. You say, hey, this third angle is 60 degrees, so all three angles are the same. That's one of our constraints for similarity. The other thing we know about similarity is that the ratio between all of the sides are going to be the same. For example, if we have another triangle right over here, let me draw another triangle. I'll call this triangle x, y, and z."}, {"video_title": "Similarity postulates   Similarity   Geometry   Khan Academy.mp3", "Sentence": "That's one of our constraints for similarity. The other thing we know about similarity is that the ratio between all of the sides are going to be the same. For example, if we have another triangle right over here, let me draw another triangle. I'll call this triangle x, y, and z. Let's say that we know that the ratio between AB and xy, we know that AB over xy, so the ratio between this side and this side, notice we're not saying that they're congruent, we're just saying that they're ratio. We're looking at the ratio now. We're saying AB over xy, let's say that that is equal to BC over yz, and that is equal to AC over xz."}, {"video_title": "Similarity postulates   Similarity   Geometry   Khan Academy.mp3", "Sentence": "I'll call this triangle x, y, and z. Let's say that we know that the ratio between AB and xy, we know that AB over xy, so the ratio between this side and this side, notice we're not saying that they're congruent, we're just saying that they're ratio. We're looking at the ratio now. We're saying AB over xy, let's say that that is equal to BC over yz, and that is equal to AC over xz. Once again, this is one of the ways that we say, hey, this means similarity. If we have all three corresponding sides, the ratio between all three corresponding sides are the same, then we know we are dealing with similar triangles. This is what we call side-side-side similarity."}, {"video_title": "Similarity postulates   Similarity   Geometry   Khan Academy.mp3", "Sentence": "We're saying AB over xy, let's say that that is equal to BC over yz, and that is equal to AC over xz. Once again, this is one of the ways that we say, hey, this means similarity. If we have all three corresponding sides, the ratio between all three corresponding sides are the same, then we know we are dealing with similar triangles. This is what we call side-side-side similarity. You don't want to get these confused with side-side-side congruence. These are all of our similarity postulates, or axioms, or things that we're going to assume and then we're going to build off of them to solve problems and prove other things. Side-side-side, when we're talking about congruence, means that the corresponding sides are congruent."}, {"video_title": "Similarity postulates   Similarity   Geometry   Khan Academy.mp3", "Sentence": "This is what we call side-side-side similarity. You don't want to get these confused with side-side-side congruence. These are all of our similarity postulates, or axioms, or things that we're going to assume and then we're going to build off of them to solve problems and prove other things. Side-side-side, when we're talking about congruence, means that the corresponding sides are congruent. Side-side-side for similarity, we're saying that the ratio between corresponding sides are going to be the same. For example, if this right over here is 10, let's say this is 60, this right over here is 30, and this right over here is 30 square roots of 3. I just made those numbers right because we will soon learn what typical ratios are of the sides of 30, 60, 90 triangles."}, {"video_title": "Similarity postulates   Similarity   Geometry   Khan Academy.mp3", "Sentence": "Side-side-side, when we're talking about congruence, means that the corresponding sides are congruent. Side-side-side for similarity, we're saying that the ratio between corresponding sides are going to be the same. For example, if this right over here is 10, let's say this is 60, this right over here is 30, and this right over here is 30 square roots of 3. I just made those numbers right because we will soon learn what typical ratios are of the sides of 30, 60, 90 triangles. Let's say this one over here is 6, 3, and 3 square roots of 3. Notice, AB over xy, 30 square roots of 3 over 3 square roots of 3, this will be 10. What is BC over xy?"}, {"video_title": "Similarity postulates   Similarity   Geometry   Khan Academy.mp3", "Sentence": "I just made those numbers right because we will soon learn what typical ratios are of the sides of 30, 60, 90 triangles. Let's say this one over here is 6, 3, and 3 square roots of 3. Notice, AB over xy, 30 square roots of 3 over 3 square roots of 3, this will be 10. What is BC over xy? 30 divided by 3 is 10. What is 60 divided by 6? AC over xz, that's going to be 10."}, {"video_title": "Similarity postulates   Similarity   Geometry   Khan Academy.mp3", "Sentence": "What is BC over xy? 30 divided by 3 is 10. What is 60 divided by 6? AC over xz, that's going to be 10. In general, to go from the corresponding side here to the corresponding side there, we always multiply by 10 on every side. We're not saying the sides are the same for this side-side-side for similarity. We're saying that we're really just scaling them up by the same amount."}, {"video_title": "Similarity postulates   Similarity   Geometry   Khan Academy.mp3", "Sentence": "AC over xz, that's going to be 10. In general, to go from the corresponding side here to the corresponding side there, we always multiply by 10 on every side. We're not saying the sides are the same for this side-side-side for similarity. We're saying that we're really just scaling them up by the same amount. Or, another way to think about it, the ratio between corresponding sides are the same. Now what about if we had, let's start another triangle right over here. Let me draw it like this."}, {"video_title": "Similarity postulates   Similarity   Geometry   Khan Academy.mp3", "Sentence": "We're saying that we're really just scaling them up by the same amount. Or, another way to think about it, the ratio between corresponding sides are the same. Now what about if we had, let's start another triangle right over here. Let me draw it like this. Actually, I want to leave this here so we can have our list. Let me draw another triangle ABC. Let's draw another triangle ABC."}, {"video_title": "Similarity postulates   Similarity   Geometry   Khan Academy.mp3", "Sentence": "Let me draw it like this. Actually, I want to leave this here so we can have our list. Let me draw another triangle ABC. Let's draw another triangle ABC. This is A, B, and C. Let's say that we know that this side, when we go to another triangle, we know that xy is AB multiplied by some constant. I can write it over here. xy is equal to some constant times AB."}, {"video_title": "Similarity postulates   Similarity   Geometry   Khan Academy.mp3", "Sentence": "Let's draw another triangle ABC. This is A, B, and C. Let's say that we know that this side, when we go to another triangle, we know that xy is AB multiplied by some constant. I can write it over here. xy is equal to some constant times AB. Actually, let me make xy bigger. It doesn't have to be. That constant could be less than 1, in which case it would be a smaller value."}, {"video_title": "Similarity postulates   Similarity   Geometry   Khan Academy.mp3", "Sentence": "xy is equal to some constant times AB. Actually, let me make xy bigger. It doesn't have to be. That constant could be less than 1, in which case it would be a smaller value. Let me just do it that way. Let me just make xy look a little bit bigger. Let's say that this is x and that is y."}, {"video_title": "Similarity postulates   Similarity   Geometry   Khan Academy.mp3", "Sentence": "That constant could be less than 1, in which case it would be a smaller value. Let me just do it that way. Let me just make xy look a little bit bigger. Let's say that this is x and that is y. Let's say that we know that xy over AB is equal to some constant. Or, if you multiply both sides by AB, you would get xy is some scaled-up version of AB. Maybe AB is 5, xy is 10, then our constant would be 2."}, {"video_title": "Similarity postulates   Similarity   Geometry   Khan Academy.mp3", "Sentence": "Let's say that this is x and that is y. Let's say that we know that xy over AB is equal to some constant. Or, if you multiply both sides by AB, you would get xy is some scaled-up version of AB. Maybe AB is 5, xy is 10, then our constant would be 2. We scaled it up by a factor of 2. Let's say we also know that angle ABC is congruent to angle XYZ. I'll add another point over here."}, {"video_title": "Similarity postulates   Similarity   Geometry   Khan Academy.mp3", "Sentence": "Maybe AB is 5, xy is 10, then our constant would be 2. We scaled it up by a factor of 2. Let's say we also know that angle ABC is congruent to angle XYZ. I'll add another point over here. Let me draw another side right over here. This is Z. Let's say we also know that angle ABC is congruent to XYZ."}, {"video_title": "Similarity postulates   Similarity   Geometry   Khan Academy.mp3", "Sentence": "I'll add another point over here. Let me draw another side right over here. This is Z. Let's say we also know that angle ABC is congruent to XYZ. Let's say we know that the ratio between BC and YZ is also this constant. The ratio between BC and YZ is also equal to the same constant. In the example where this is 5 and 10, maybe this is 3 and 6."}, {"video_title": "Similarity postulates   Similarity   Geometry   Khan Academy.mp3", "Sentence": "Let's say we also know that angle ABC is congruent to XYZ. Let's say we know that the ratio between BC and YZ is also this constant. The ratio between BC and YZ is also equal to the same constant. In the example where this is 5 and 10, maybe this is 3 and 6. The constant, we're doubling the length of the side. Is this triangle, is triangle XYZ going to be similar? If you think about it, if you say that this is some multiple, if XY is the same multiple of AB as YZ is the multiple of BC, and the angle in between is congruent, there's only one triangle we can set up over here."}, {"video_title": "Similarity postulates   Similarity   Geometry   Khan Academy.mp3", "Sentence": "In the example where this is 5 and 10, maybe this is 3 and 6. The constant, we're doubling the length of the side. Is this triangle, is triangle XYZ going to be similar? If you think about it, if you say that this is some multiple, if XY is the same multiple of AB as YZ is the multiple of BC, and the angle in between is congruent, there's only one triangle we can set up over here. We're only constrained to one triangle right over here. We're completely constraining the length of this side. The length of this side is going to have to be that same scale as that over there."}, {"video_title": "Similarity postulates   Similarity   Geometry   Khan Academy.mp3", "Sentence": "If you think about it, if you say that this is some multiple, if XY is the same multiple of AB as YZ is the multiple of BC, and the angle in between is congruent, there's only one triangle we can set up over here. We're only constrained to one triangle right over here. We're completely constraining the length of this side. The length of this side is going to have to be that same scale as that over there. We call that side-angle-side similarity. Once again, we saw SSS and SAS in our congruence postulates, but we're saying something very different here. We're saying that in SAS, if the ratio between corresponding sides of the true triangle are the same, so AB and XY of one corresponding side, and then another corresponding side, that's that second side, so that's between BC and YZ, and the angle between them are congruent, then we're saying it's similar."}, {"video_title": "Similarity postulates   Similarity   Geometry   Khan Academy.mp3", "Sentence": "The length of this side is going to have to be that same scale as that over there. We call that side-angle-side similarity. Once again, we saw SSS and SAS in our congruence postulates, but we're saying something very different here. We're saying that in SAS, if the ratio between corresponding sides of the true triangle are the same, so AB and XY of one corresponding side, and then another corresponding side, that's that second side, so that's between BC and YZ, and the angle between them are congruent, then we're saying it's similar. For SAS for congruency, we said that the sides actually had to be congruent. Here we're saying that the ratio between the corresponding sides just has to be the same. For example, SAS, just to apply it, if I have, let me just show some examples here."}, {"video_title": "Similarity postulates   Similarity   Geometry   Khan Academy.mp3", "Sentence": "We're saying that in SAS, if the ratio between corresponding sides of the true triangle are the same, so AB and XY of one corresponding side, and then another corresponding side, that's that second side, so that's between BC and YZ, and the angle between them are congruent, then we're saying it's similar. For SAS for congruency, we said that the sides actually had to be congruent. Here we're saying that the ratio between the corresponding sides just has to be the same. For example, SAS, just to apply it, if I have, let me just show some examples here. Let's say I have a triangle here that is 3, 2, 4. Let's say we have another triangle here that has length 9, 6, and we also know that the angle in between are congruent, so that that angle is equal to that angle. What SAS in the similarity world tells you is that these triangles are definitely going to be similar triangles, that we're actually constraining, because there's actually only one triangle we can draw right over here."}, {"video_title": "Similarity postulates   Similarity   Geometry   Khan Academy.mp3", "Sentence": "For example, SAS, just to apply it, if I have, let me just show some examples here. Let's say I have a triangle here that is 3, 2, 4. Let's say we have another triangle here that has length 9, 6, and we also know that the angle in between are congruent, so that that angle is equal to that angle. What SAS in the similarity world tells you is that these triangles are definitely going to be similar triangles, that we're actually constraining, because there's actually only one triangle we can draw right over here. It's a triangle where all of the sides are going to have to be scaled up by the same amount. There's only one long side right here that we can actually draw, and that's going to have to be scaled up by 3 as well. This is the only possible triangle."}, {"video_title": "Similarity postulates   Similarity   Geometry   Khan Academy.mp3", "Sentence": "What SAS in the similarity world tells you is that these triangles are definitely going to be similar triangles, that we're actually constraining, because there's actually only one triangle we can draw right over here. It's a triangle where all of the sides are going to have to be scaled up by the same amount. There's only one long side right here that we can actually draw, and that's going to have to be scaled up by 3 as well. This is the only possible triangle. If you constrain this side, you're saying, look, this is 3 times that side, this is 3 times that side, and the angle between them is congruent. There's only one triangle we can make, and we know that there is a similar triangle there, where everything is scaled up by a factor of 3, so that one triangle we can draw has to be that one similar triangle. This is what we're talking about SAS."}, {"video_title": "Similarity postulates   Similarity   Geometry   Khan Academy.mp3", "Sentence": "This is the only possible triangle. If you constrain this side, you're saying, look, this is 3 times that side, this is 3 times that side, and the angle between them is congruent. There's only one triangle we can make, and we know that there is a similar triangle there, where everything is scaled up by a factor of 3, so that one triangle we can draw has to be that one similar triangle. This is what we're talking about SAS. We're not saying that this side is congruent to that side, or that side is congruent to that side. We're saying that they're scaled up by the same factor. If we had another triangle that looked like this, so maybe this is 9, this is 4, and the angle between them were congruent, you couldn't say that they're similar, because this side is scaled up by a factor of 3, this side is only scaled up by a factor of 2, so this one right over there, you could not say that it is necessarily similar."}, {"video_title": "Similarity postulates   Similarity   Geometry   Khan Academy.mp3", "Sentence": "This is what we're talking about SAS. We're not saying that this side is congruent to that side, or that side is congruent to that side. We're saying that they're scaled up by the same factor. If we had another triangle that looked like this, so maybe this is 9, this is 4, and the angle between them were congruent, you couldn't say that they're similar, because this side is scaled up by a factor of 3, this side is only scaled up by a factor of 2, so this one right over there, you could not say that it is necessarily similar. Likewise, if you had a triangle that had length 9 here and length 6 there, but you did not know that these two angles are the same, once again, you're not constraining this enough, and you would not know that those two triangles are necessarily similar, because you don't know that middle angle is the same. You might be saying, well, there were a few other postulates that we had. We had AAS when we dealt with congruency, but if you think about it, we've already shown that two angles by themselves are enough to show similarity, so why worry about an angle and angle and a side, or the ratio between the sides?"}, {"video_title": "Similarity postulates   Similarity   Geometry   Khan Academy.mp3", "Sentence": "If we had another triangle that looked like this, so maybe this is 9, this is 4, and the angle between them were congruent, you couldn't say that they're similar, because this side is scaled up by a factor of 3, this side is only scaled up by a factor of 2, so this one right over there, you could not say that it is necessarily similar. Likewise, if you had a triangle that had length 9 here and length 6 there, but you did not know that these two angles are the same, once again, you're not constraining this enough, and you would not know that those two triangles are necessarily similar, because you don't know that middle angle is the same. You might be saying, well, there were a few other postulates that we had. We had AAS when we dealt with congruency, but if you think about it, we've already shown that two angles by themselves are enough to show similarity, so why worry about an angle and angle and a side, or the ratio between the sides? Why even worry about that? We also had angle, side, angle, and congruence, but once again, we already know that two angles are enough, so we don't need to throw in this extra side, so we don't even need this right over here. So these are going to be our similarity postulates, and I want to remind you, side, side, side, this is different than the side, side, side for congruence."}, {"video_title": "Similarity postulates   Similarity   Geometry   Khan Academy.mp3", "Sentence": "We had AAS when we dealt with congruency, but if you think about it, we've already shown that two angles by themselves are enough to show similarity, so why worry about an angle and angle and a side, or the ratio between the sides? Why even worry about that? We also had angle, side, angle, and congruence, but once again, we already know that two angles are enough, so we don't need to throw in this extra side, so we don't even need this right over here. So these are going to be our similarity postulates, and I want to remind you, side, side, side, this is different than the side, side, side for congruence. We're talking about the ratio between corresponding sides. We're not saying that they're actually congruent, and here, side, angle, side, it's different than the side, angle, side for congruence. It's kind of related, but here we're talking about the ratio between the sides, not the actual measures."}, {"video_title": "Interpreting a parabola in context   Quadratic functions & equations   Algebra I   Khan Academy.mp3", "Sentence": "So what they want us to do is plot the point on the graph of f that corresponds to each of the following things. So pause the video and see if you can do that. And obviously, you can't draw on your screen. This is from an exercise on Khan Academy, but you can visually look at it, and even with your finger, point to the part of the graph of f that represents each of these things. All right, so the first thing here is the height of the platform. So the drone is at the height of the platform right when it takes off, because it says Adam flew his remote-controlled drone off of a platform. So what is the time that he's taking off, the drone, or the drone is taking off?"}, {"video_title": "Interpreting a parabola in context   Quadratic functions & equations   Algebra I   Khan Academy.mp3", "Sentence": "This is from an exercise on Khan Academy, but you can visually look at it, and even with your finger, point to the part of the graph of f that represents each of these things. All right, so the first thing here is the height of the platform. So the drone is at the height of the platform right when it takes off, because it says Adam flew his remote-controlled drone off of a platform. So what is the time that he's taking off, the drone, or the drone is taking off? Well, that's gonna be at time t equals zero right over here. And what is the height of the drone at that moment? It is 60 meters."}, {"video_title": "Interpreting a parabola in context   Quadratic functions & equations   Algebra I   Khan Academy.mp3", "Sentence": "So what is the time that he's taking off, the drone, or the drone is taking off? Well, that's gonna be at time t equals zero right over here. And what is the height of the drone at that moment? It is 60 meters. So that must be the height of the platform. So that point right over there tells us the height of the platform. And if they asked us what the height of the platform is, it would be 60 meters."}, {"video_title": "Interpreting a parabola in context   Quadratic functions & equations   Algebra I   Khan Academy.mp3", "Sentence": "It is 60 meters. So that must be the height of the platform. So that point right over there tells us the height of the platform. And if they asked us what the height of the platform is, it would be 60 meters. The next one is the drone's maximum height. So then as time goes on, we can see the drone starts going to a higher and higher and higher height, gets as high as 80 meters, and then it starts going down. So it looks like 80 meters at time 10 seconds, the drone hits a maximum height of 80 meters."}, {"video_title": "Interpreting a parabola in context   Quadratic functions & equations   Algebra I   Khan Academy.mp3", "Sentence": "And if they asked us what the height of the platform is, it would be 60 meters. The next one is the drone's maximum height. So then as time goes on, we can see the drone starts going to a higher and higher and higher height, gets as high as 80 meters, and then it starts going down. So it looks like 80 meters at time 10 seconds, the drone hits a maximum height of 80 meters. And then last but not least, they say the time when the drone landed on the ground. Now we can assume that the ground is when the height of the drone is at zero meters. And we can see that that happens right over here."}, {"video_title": "Interpreting a parabola in context   Quadratic functions & equations   Algebra I   Khan Academy.mp3", "Sentence": "So it looks like 80 meters at time 10 seconds, the drone hits a maximum height of 80 meters. And then last but not least, they say the time when the drone landed on the ground. Now we can assume that the ground is when the height of the drone is at zero meters. And we can see that that happens right over here. And that happens at time t equals 30 seconds. And so we've just marked it off. And I know some of y'all are thinking, wait, there's another time where the drone is, where the drone's height is at zero."}, {"video_title": "Interpreting a parabola in context   Quadratic functions & equations   Algebra I   Khan Academy.mp3", "Sentence": "And we can see that that happens right over here. And that happens at time t equals 30 seconds. And so we've just marked it off. And I know some of y'all are thinking, wait, there's another time where the drone is, where the drone's height is at zero. And that's right over here. That's at negative 10 seconds. Couldn't we say that that's also a time when the drone landed on the ground?"}, {"video_title": "Interpreting a parabola in context   Quadratic functions & equations   Algebra I   Khan Academy.mp3", "Sentence": "And I know some of y'all are thinking, wait, there's another time where the drone is, where the drone's height is at zero. And that's right over here. That's at negative 10 seconds. Couldn't we say that that's also a time when the drone landed on the ground? And this is an important point to realize, because if we're really trying to model the drone's behavior from time t equals zero, if t equals zero is right when you take off all the way to it lands, then this parabola that we're showing right over here, it actually, we would probably wanna restrict its domain to positive times. And so this negative time region right over here really doesn't make a lot of sense. We should probably consider the non-negative values of time when we're trying to think about these different things."}, {"video_title": "Determining whether real world model is linear or exponential.mp3", "Sentence": "And I'm using, this is an example from the Khan Academy exercises, and we're really trying to pick between whether a linear model or a linear function models this relationship, or an exponential model or exponential function will model this relationship. So like always, pause this video and see if you can figure it out on your own. Alright, so now let's think about this together. So as the time goes by, or on the time variable right over here, we see that we keep incrementing it by two. Go from zero to two, two to four, four to six, so on and so forth, it keeps going up by two. So if this is a linear relationship, given that our change in time is constant, our change in cost should increase by a constant amount. It doesn't have to be this constant, but it has to be a constant amount."}, {"video_title": "Determining whether real world model is linear or exponential.mp3", "Sentence": "So as the time goes by, or on the time variable right over here, we see that we keep incrementing it by two. Go from zero to two, two to four, four to six, so on and so forth, it keeps going up by two. So if this is a linear relationship, given that our change in time is constant, our change in cost should increase by a constant amount. It doesn't have to be this constant, but it has to be a constant amount. If we were dealing with an exponential relationship, we would multiply by the same amount for a constant change in time. Let's see what's going on here. Let's just first look at the difference between these numbers."}, {"video_title": "Determining whether real world model is linear or exponential.mp3", "Sentence": "It doesn't have to be this constant, but it has to be a constant amount. If we were dealing with an exponential relationship, we would multiply by the same amount for a constant change in time. Let's see what's going on here. Let's just first look at the difference between these numbers. To go from 30 to 36.9, you would have to add 6.9. Now to go from 36.9 to 44.1, what do you have to add? You have to add 7.2."}, {"video_title": "Determining whether real world model is linear or exponential.mp3", "Sentence": "Let's just first look at the difference between these numbers. To go from 30 to 36.9, you would have to add 6.9. Now to go from 36.9 to 44.1, what do you have to add? You have to add 7.2. Now to go from 44.1 to 51.1, you would have to add seven. Now to go to 51.1 to 57.9, you are adding 6.8. 6.8, and then finally going from 57.9 to 65.1, let's see, this is almost eight, 7.1, this is what?"}, {"video_title": "Determining whether real world model is linear or exponential.mp3", "Sentence": "You have to add 7.2. Now to go from 44.1 to 51.1, you would have to add seven. Now to go to 51.1 to 57.9, you are adding 6.8. 6.8, and then finally going from 57.9 to 65.1, let's see, this is almost eight, 7.1, this is what? 7.2, we're adding, plus 7.2. So you might say, hey, we're not adding the exact same amount every time. But remember, this is intended to be data that you might actually get from a real-world situation."}, {"video_title": "Determining whether real world model is linear or exponential.mp3", "Sentence": "6.8, and then finally going from 57.9 to 65.1, let's see, this is almost eight, 7.1, this is what? 7.2, we're adding, plus 7.2. So you might say, hey, we're not adding the exact same amount every time. But remember, this is intended to be data that you might actually get from a real-world situation. And the data that you get from a real-world situation will never be exactly a linear model or exactly an exponential model. But every time we add two years, it does look like we're getting pretty close to adding $7,000 in cost. $6,000 in cost, 6.9 is pretty close to seven."}, {"video_title": "Determining whether real world model is linear or exponential.mp3", "Sentence": "But remember, this is intended to be data that you might actually get from a real-world situation. And the data that you get from a real-world situation will never be exactly a linear model or exactly an exponential model. But every time we add two years, it does look like we're getting pretty close to adding $7,000 in cost. $6,000 in cost, 6.9 is pretty close to seven. That's pretty close to seven, that is seven. That's pretty close to seven, that's pretty close to seven. So this is looking like a linear model to me."}, {"video_title": "Determining whether real world model is linear or exponential.mp3", "Sentence": "$6,000 in cost, 6.9 is pretty close to seven. That's pretty close to seven, that is seven. That's pretty close to seven, that's pretty close to seven. So this is looking like a linear model to me. You could test whether it's an exponential model. You see, well, what factor am I multiplying each time? But that doesn't seem to be as, this doesn't seem to be growing exponentially."}, {"video_title": "Determining whether real world model is linear or exponential.mp3", "Sentence": "So this is looking like a linear model to me. You could test whether it's an exponential model. You see, well, what factor am I multiplying each time? But that doesn't seem to be as, this doesn't seem to be growing exponentially. It doesn't seem like we're multiplying by the same factor every time. It seems like we're multiplying by a slightly lower factor as we get to higher costs. So the linear model seems to be a pretty good thing."}, {"video_title": "Determining whether real world model is linear or exponential.mp3", "Sentence": "But that doesn't seem to be as, this doesn't seem to be growing exponentially. It doesn't seem like we're multiplying by the same factor every time. It seems like we're multiplying by a slightly lower factor as we get to higher costs. So the linear model seems to be a pretty good thing. If I see every time I increase by two years, I'm increasing cost by 6.9 or 7.2 or seven, it's pretty close to seven. So it's not exactly the cost, but the model predicts it pretty well. If you were to plot these on a coordinate plane and try to connect the dots, you could, it would look pretty close to a line or you could draw a line that gets pretty close to those dots."}, {"video_title": "Setting up a system of equations from context example.mp3", "Sentence": "So we're told Sanjay's dog weighs five times as much as his cat. His dog is also 20 kilograms heavier than his cat. Let C be the cat's weight and let D be the dog's weight. So pause this video and see if you can set up a system of equation, two linear equations with two unknowns that we could use to solve for C and D, but we don't have to in this video. All right, so let's do it together. So what I like to do is usually there's a sentence or two that describes each of the equations we wanna set up. So this first one tells us Sanjay's dog weighs five times as much as his cat."}, {"video_title": "Setting up a system of equations from context example.mp3", "Sentence": "So pause this video and see if you can set up a system of equation, two linear equations with two unknowns that we could use to solve for C and D, but we don't have to in this video. All right, so let's do it together. So what I like to do is usually there's a sentence or two that describes each of the equations we wanna set up. So this first one tells us Sanjay's dog weighs five times as much as his cat. So how much does his dog weigh? So his dog weighs D. So we know D is going to be equal to five times as much as his cat weighs. So his cat weighs C. So D is going to be equal to five times as much as his cat weighs."}, {"video_title": "Setting up a system of equations from context example.mp3", "Sentence": "So this first one tells us Sanjay's dog weighs five times as much as his cat. So how much does his dog weigh? So his dog weighs D. So we know D is going to be equal to five times as much as his cat weighs. So his cat weighs C. So D is going to be equal to five times as much as his cat weighs. So that's one linear equation using D and C. And so what's another one? Well, then we are told his dog is also 20 kilograms heavier than his cat. So we could say that the dog is going to be, the dog's weight is going to be equal to the cat's weight plus what?"}, {"video_title": "Setting up a system of equations from context example.mp3", "Sentence": "So his cat weighs C. So D is going to be equal to five times as much as his cat weighs. So that's one linear equation using D and C. And so what's another one? Well, then we are told his dog is also 20 kilograms heavier than his cat. So we could say that the dog is going to be, the dog's weight is going to be equal to the cat's weight plus what? Plus 20 kilograms. We're assuming everything's in kilograms so I don't have to write the units. But there you have it."}, {"video_title": "Setting up a system of equations from context example.mp3", "Sentence": "So we could say that the dog is going to be, the dog's weight is going to be equal to the cat's weight plus what? Plus 20 kilograms. We're assuming everything's in kilograms so I don't have to write the units. But there you have it. I have just set up two equations in two unknowns, two linear equations, based on the information given in this word problem, which we could then solve, and I encourage you to do so if you're curious. But sometimes the difficult part is just to find, is to re-express the information that you're given in a mathematical form. But as you see, as you get practice, it becomes somewhat intuitive."}, {"video_title": "Exponential growth functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "And really just show you how fast these things can grow. So let's just write an example exponential function here. So let's say we have y is equal to 3 to the x power. Notice, this isn't x to the third power, this is 3 to the x power. Our independent variable x is the actual exponent. So let's make a table here to see how quickly this thing grows. And maybe we'll graph it as well."}, {"video_title": "Exponential growth functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "Notice, this isn't x to the third power, this is 3 to the x power. Our independent variable x is the actual exponent. So let's make a table here to see how quickly this thing grows. And maybe we'll graph it as well. So let's take some x values here. So let's say, let's start with x is equal to negative 4, then we'll go to negative 3, negative 2, 0, 1, 2, 3, and 4. And let's figure out what our y values are going to be for each of these x values."}, {"video_title": "Exponential growth functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "And maybe we'll graph it as well. So let's take some x values here. So let's say, let's start with x is equal to negative 4, then we'll go to negative 3, negative 2, 0, 1, 2, 3, and 4. And let's figure out what our y values are going to be for each of these x values. Now, here y is going to be 3 to the negative 4 power, which is equal to 1 over 3 to the fourth power. 3 to the third is 27, times 3 again is 81. So this is equal to 1 over 81."}, {"video_title": "Exponential growth functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "And let's figure out what our y values are going to be for each of these x values. Now, here y is going to be 3 to the negative 4 power, which is equal to 1 over 3 to the fourth power. 3 to the third is 27, times 3 again is 81. So this is equal to 1 over 81. When x is equal to negative 3, y is 3. Let me do this in a different color, that color is hard to read. y is 3 to the negative 3 power."}, {"video_title": "Exponential growth functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "So this is equal to 1 over 81. When x is equal to negative 3, y is 3. Let me do this in a different color, that color is hard to read. y is 3 to the negative 3 power. Well, that's 1 over 3 to the third power, which is equal to 1 over 27. So we're going from a super small number to a less super small number. And then 3 to the negative 2 power is going to be 1 over 9, right?"}, {"video_title": "Exponential growth functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "y is 3 to the negative 3 power. Well, that's 1 over 3 to the third power, which is equal to 1 over 27. So we're going from a super small number to a less super small number. And then 3 to the negative 2 power is going to be 1 over 9, right? 1 over 3 squared. And then we have 3 to the 0 power, which is just equal to 1. So we're getting a little bit larger, a little bit larger, but we're going to see we're about to explode."}, {"video_title": "Exponential growth functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "And then 3 to the negative 2 power is going to be 1 over 9, right? 1 over 3 squared. And then we have 3 to the 0 power, which is just equal to 1. So we're getting a little bit larger, a little bit larger, but we're going to see we're about to explode. Now we have 3 to the first power, that's equal to 3. Then we have 3 to the second power, right? y is equal to 3 to the second power, that's 9."}, {"video_title": "Exponential growth functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "So we're getting a little bit larger, a little bit larger, but we're going to see we're about to explode. Now we have 3 to the first power, that's equal to 3. Then we have 3 to the second power, right? y is equal to 3 to the second power, that's 9. 3 to the third power, 27. 3 to the fourth power, 81. If we were to put the fifth power, 243."}, {"video_title": "Exponential growth functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "y is equal to 3 to the second power, that's 9. 3 to the third power, 27. 3 to the fourth power, 81. If we were to put the fifth power, 243. Let's graph this just to get an idea of how quickly we're exploding. So let me draw my axes here. So that's my x-axis and that is my y-axis."}, {"video_title": "Exponential growth functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "If we were to put the fifth power, 243. Let's graph this just to get an idea of how quickly we're exploding. So let me draw my axes here. So that's my x-axis and that is my y-axis. And let me just do it in increments of 5, because I really want to get the general shape of the graph here. So let me just draw as straight a line as I can. Let's say this is 5, 10, 15, actually I want to get to 81 that way."}, {"video_title": "Exponential growth functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "So that's my x-axis and that is my y-axis. And let me just do it in increments of 5, because I really want to get the general shape of the graph here. So let me just draw as straight a line as I can. Let's say this is 5, 10, 15, actually I want to get to 81 that way. I want to get to 81. Actually, well, that's good enough. Let me draw it a little bit differently than I've drawn it."}, {"video_title": "Exponential growth functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "Let's say this is 5, 10, 15, actually I want to get to 81 that way. I want to get to 81. Actually, well, that's good enough. Let me draw it a little bit differently than I've drawn it. So let me draw it down here, because all of these values you might notice are positive values, because I have a positive base. So let me draw it like this. Like that."}, {"video_title": "Exponential growth functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "Let me draw it a little bit differently than I've drawn it. So let me draw it down here, because all of these values you might notice are positive values, because I have a positive base. So let me draw it like this. Like that. Good enough. And then let's say I have 10, 20, 30, 40, 50, 60, 70, 80. That is 80 right there."}, {"video_title": "Exponential growth functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "Like that. Good enough. And then let's say I have 10, 20, 30, 40, 50, 60, 70, 80. That is 80 right there. That's 10, that's 30. That'll be good for approximation. And then let's say that this is negative 5."}, {"video_title": "Exponential growth functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "That is 80 right there. That's 10, that's 30. That'll be good for approximation. And then let's say that this is negative 5. This is positive 5 right here. And actually, let me stretch it out a little bit more. Say this is negative 1, negative 2, negative 3, negative 4."}, {"video_title": "Exponential growth functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "And then let's say that this is negative 5. This is positive 5 right here. And actually, let me stretch it out a little bit more. Say this is negative 1, negative 2, negative 3, negative 4. And then we have 1, 2, 3, and 4. So when x is equal to 0, we're equal to 1, right? When x is equal to 0, y is equal to 1, which is maybe right around there."}, {"video_title": "Exponential growth functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "Say this is negative 1, negative 2, negative 3, negative 4. And then we have 1, 2, 3, and 4. So when x is equal to 0, we're equal to 1, right? When x is equal to 0, y is equal to 1, which is maybe right around there. When x is equal to 1, y is equal to 3, which is maybe right around there. When x is equal to 2, y is equal to 9, which is right around there. When x is equal to 3, y is equal to 27, which is right around there."}, {"video_title": "Exponential growth functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "When x is equal to 0, y is equal to 1, which is maybe right around there. When x is equal to 1, y is equal to 3, which is maybe right around there. When x is equal to 2, y is equal to 9, which is right around there. When x is equal to 3, y is equal to 27, which is right around there. When x is equal to 4, y is equal to 81. y is equal to 81. So you see very quickly this is just exploding. If I did 5, it would go to 243, which wouldn't even fit on my screen."}, {"video_title": "Exponential growth functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "When x is equal to 3, y is equal to 27, which is right around there. When x is equal to 4, y is equal to 81. y is equal to 81. So you see very quickly this is just exploding. If I did 5, it would go to 243, which wouldn't even fit on my screen. When you go to negative 1, we get smaller and smaller. So at negative 1, we're at 1 ninth. Eventually you're not even going to see this."}, {"video_title": "Exponential growth functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "If I did 5, it would go to 243, which wouldn't even fit on my screen. When you go to negative 1, we get smaller and smaller. So at negative 1, we're at 1 ninth. Eventually you're not even going to see this. It's going to get closer and closer to 0. As this approaches larger and larger negative numbers, or I guess I should say smaller negative numbers, so 3 to the negative 1,000, 3 to the negative million, we're getting numbers closer and closer to 0 without actually ever approaching 0. So as we go from negative infinity, x is equal to negative infinity, we're getting very close to 0."}, {"video_title": "Exponential growth functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "Eventually you're not even going to see this. It's going to get closer and closer to 0. As this approaches larger and larger negative numbers, or I guess I should say smaller negative numbers, so 3 to the negative 1,000, 3 to the negative million, we're getting numbers closer and closer to 0 without actually ever approaching 0. So as we go from negative infinity, x is equal to negative infinity, we're getting very close to 0. We're slowly getting ourselves away from 0. And bam, once we start getting into the positive numbers, we just explode. We just explode, and we keep exploding at an ever-increasing rate."}, {"video_title": "Exponential growth functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "So as we go from negative infinity, x is equal to negative infinity, we're getting very close to 0. We're slowly getting ourselves away from 0. And bam, once we start getting into the positive numbers, we just explode. We just explode, and we keep exploding at an ever-increasing rate. So the idea here is just to show you that exponential functions are really, really dramatic. You can always construct a faster expanding function. For example, you could say y is equal to x to the x, even faster expanding."}, {"video_title": "Exponential growth functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "We just explode, and we keep exploding at an ever-increasing rate. So the idea here is just to show you that exponential functions are really, really dramatic. You can always construct a faster expanding function. For example, you could say y is equal to x to the x, even faster expanding. But out of the ones that we deal with in everyday lives, these are the fastest. So given that, let's do some word problems that just give us an appreciation for exponential functions. So let's say that someone sends out a chain letter."}, {"video_title": "Exponential growth functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "For example, you could say y is equal to x to the x, even faster expanding. But out of the ones that we deal with in everyday lives, these are the fastest. So given that, let's do some word problems that just give us an appreciation for exponential functions. So let's say that someone sends out a chain letter. Let's say in week 1, someone sends a chain letter to 10 people. And the chain letter says, you have to now send this chain letter to 10 more new people. And if you don't, you're going to have bad luck, and your hair is going to fall out, and you'll marry a frog or whatever else."}, {"video_title": "Exponential growth functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "So let's say that someone sends out a chain letter. Let's say in week 1, someone sends a chain letter to 10 people. And the chain letter says, you have to now send this chain letter to 10 more new people. And if you don't, you're going to have bad luck, and your hair is going to fall out, and you'll marry a frog or whatever else. So all of these people agree, and they go and each send it to 10 people the next week. So in week 2, they go out and each send it to 10 more people. So each of those original 10 people are each sending out 10 more of the letters."}, {"video_title": "Exponential growth functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "And if you don't, you're going to have bad luck, and your hair is going to fall out, and you'll marry a frog or whatever else. So all of these people agree, and they go and each send it to 10 people the next week. So in week 2, they go out and each send it to 10 more people. So each of those original 10 people are each sending out 10 more of the letters. So now 100 people have the letters, right? Each of those 10 sent out 10. So 100 letters were sent out."}, {"video_title": "Exponential growth functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "So each of those original 10 people are each sending out 10 more of the letters. So now 100 people have the letters, right? Each of those 10 sent out 10. So 100 letters were sent out. So let me see, sent. 10 were sent. Here, 100 were sent."}, {"video_title": "Exponential growth functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "So 100 letters were sent out. So let me see, sent. 10 were sent. Here, 100 were sent. In week 3, what's going to happen? Each of those 100 people who got this one, they are each going to send out 10, assuming that everyone is really into chain letters. So 1,000 people are going to get it."}, {"video_title": "Exponential growth functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "Here, 100 were sent. In week 3, what's going to happen? Each of those 100 people who got this one, they are each going to send out 10, assuming that everyone is really into chain letters. So 1,000 people are going to get it. And so the general pattern here is the people who receive it, so in week N, where N is the week we're talking about, how many people receive the letter? In week N, we have 10 to the Nth people receive I before E except after C. Receive the letter. So if I were to ask you, how many people are getting the letter on the 6th week?"}, {"video_title": "Exponential growth functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "So 1,000 people are going to get it. And so the general pattern here is the people who receive it, so in week N, where N is the week we're talking about, how many people receive the letter? In week N, we have 10 to the Nth people receive I before E except after C. Receive the letter. So if I were to ask you, how many people are getting the letter on the 6th week? On the 6th week. How many people are actually going to receive that letter? Well, what's 10 to the 6th power?"}, {"video_title": "Exponential growth functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "So if I were to ask you, how many people are getting the letter on the 6th week? On the 6th week. How many people are actually going to receive that letter? Well, what's 10 to the 6th power? 10 to the 6th is equal to 1 with 6 zeros, which is 1 million people are going to receive that letter in just 6 weeks, just sending out 10 letters each. And obviously in the real world, most people will chuck these in the basket so you don't have this good of a hit rate. But if you did, if every 10 people you sent it to also sent it to 10 people and so on and so forth, by the 6th week, you would have a million people."}, {"video_title": "Exponential growth functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "Well, what's 10 to the 6th power? 10 to the 6th is equal to 1 with 6 zeros, which is 1 million people are going to receive that letter in just 6 weeks, just sending out 10 letters each. And obviously in the real world, most people will chuck these in the basket so you don't have this good of a hit rate. But if you did, if every 10 people you sent it to also sent it to 10 people and so on and so forth, by the 6th week, you would have a million people. And by the 9th week, you would have a billion people. And frankly, the week after that, you would run out of people. I'll see you in the next video."}, {"video_title": "Solving equations with zero product property.mp3", "Sentence": "All right, now let's work through this together. So at first you might be tempted to multiply these things out, or there's multiple ways that you might have tried to approach it, but the key realization here is that you have two things being multiplied, and it's being equal to zero. So you have the first thing being multiplied is two x minus one. This expression is being multiplied by x plus four, and to get it to be equal to zero, one or both of these expressions needs to be equal to zero. Let me really reinforce that idea. If I had two variables, let's say a and b, and I told you a times b is equal to zero, well, can you get the product of two numbers to equal zero without at least one of them being equal to zero? And the simple answer is no."}, {"video_title": "Solving equations with zero product property.mp3", "Sentence": "This expression is being multiplied by x plus four, and to get it to be equal to zero, one or both of these expressions needs to be equal to zero. Let me really reinforce that idea. If I had two variables, let's say a and b, and I told you a times b is equal to zero, well, can you get the product of two numbers to equal zero without at least one of them being equal to zero? And the simple answer is no. If a is seven, the only way that you would get zero is if b is zero, or if b was five, the only way to get zero is if a is zero. So you see from this example, either, let me write this down, either a or b, or both, because zero times zero is zero, or both must be zero. The only way that you get the product of two quantities and you get zero is if one or both of them is equal to zero."}, {"video_title": "Solving equations with zero product property.mp3", "Sentence": "And the simple answer is no. If a is seven, the only way that you would get zero is if b is zero, or if b was five, the only way to get zero is if a is zero. So you see from this example, either, let me write this down, either a or b, or both, because zero times zero is zero, or both must be zero. The only way that you get the product of two quantities and you get zero is if one or both of them is equal to zero. I really want to reinforce this idea. I'm gonna put a red box around it so that it really gets stuck in your brain. I want you to think about why that is."}, {"video_title": "Solving equations with zero product property.mp3", "Sentence": "The only way that you get the product of two quantities and you get zero is if one or both of them is equal to zero. I really want to reinforce this idea. I'm gonna put a red box around it so that it really gets stuck in your brain. I want you to think about why that is. Try to come up with two numbers. Try to multiply them so that you get zero and you're gonna see that one of those numbers is gonna need to be, is going to need to be zero. So we're gonna use this idea right over here."}, {"video_title": "Solving equations with zero product property.mp3", "Sentence": "I want you to think about why that is. Try to come up with two numbers. Try to multiply them so that you get zero and you're gonna see that one of those numbers is gonna need to be, is going to need to be zero. So we're gonna use this idea right over here. Now, this might look a little bit different, but you could view two x minus one as our a, and you could view x plus four as our b. So either two x minus one needs to be equal to zero, or x plus four needs to be equal to zero, or both of them needs to be equal to zero. So I could write that as two x minus one needs to be equal to zero, or, or x plus four, or x, let me do that orange, actually let me do the two x minus one in that yellow color."}, {"video_title": "Solving equations with zero product property.mp3", "Sentence": "So we're gonna use this idea right over here. Now, this might look a little bit different, but you could view two x minus one as our a, and you could view x plus four as our b. So either two x minus one needs to be equal to zero, or x plus four needs to be equal to zero, or both of them needs to be equal to zero. So I could write that as two x minus one needs to be equal to zero, or, or x plus four, or x, let me do that orange, actually let me do the two x minus one in that yellow color. So either two x minus one is equal to zero, or x plus four is equal to zero. X plus four is equal to zero. And so let's solve each of these."}, {"video_title": "Solving equations with zero product property.mp3", "Sentence": "So I could write that as two x minus one needs to be equal to zero, or, or x plus four, or x, let me do that orange, actually let me do the two x minus one in that yellow color. So either two x minus one is equal to zero, or x plus four is equal to zero. X plus four is equal to zero. And so let's solve each of these. If two x minus one could be equal to zero, well, let's see, we could add one to both sides, and we get two x is equal to one, divide both sides by two, and this is just straightforward solving a linear equation. If this looks unfamiliar, I encourage you to watch videos on solving linear equations on Khan Academy. But you'll get x is equal to 1 1\u20442 as one solution, and this is interesting, because we're gonna have two solutions here, or over here, if we want to solve for x, we can subtract four from both sides, and we would get x is equal to negative four."}, {"video_title": "Solving equations with zero product property.mp3", "Sentence": "And so let's solve each of these. If two x minus one could be equal to zero, well, let's see, we could add one to both sides, and we get two x is equal to one, divide both sides by two, and this is just straightforward solving a linear equation. If this looks unfamiliar, I encourage you to watch videos on solving linear equations on Khan Academy. But you'll get x is equal to 1 1\u20442 as one solution, and this is interesting, because we're gonna have two solutions here, or over here, if we want to solve for x, we can subtract four from both sides, and we would get x is equal to negative four. So it's neat, in an equation like this, you can actually have two solutions. X could be equal to 1 1\u20442, or x could be equal to negative four. I think it's pretty interesting to substitute either one of these in."}, {"video_title": "Solving equations with zero product property.mp3", "Sentence": "But you'll get x is equal to 1 1\u20442 as one solution, and this is interesting, because we're gonna have two solutions here, or over here, if we want to solve for x, we can subtract four from both sides, and we would get x is equal to negative four. So it's neat, in an equation like this, you can actually have two solutions. X could be equal to 1 1\u20442, or x could be equal to negative four. I think it's pretty interesting to substitute either one of these in. If x is equal to 1 1\u20442, what is going to happen? Well, this is going to be two times 1 1\u20442 minus one, two times 1 1\u20442 minus one, that's going to be our first expression, and then our second expression is going to be 1 1\u20442 plus four and so what's this going to be equal to? Well, two times 1 1\u20442 is one, one minus one is zero, so I don't care what you have over here, zero times anything is going to be equal to zero."}, {"video_title": "Solving equations with zero product property.mp3", "Sentence": "I think it's pretty interesting to substitute either one of these in. If x is equal to 1 1\u20442, what is going to happen? Well, this is going to be two times 1 1\u20442 minus one, two times 1 1\u20442 minus one, that's going to be our first expression, and then our second expression is going to be 1 1\u20442 plus four and so what's this going to be equal to? Well, two times 1 1\u20442 is one, one minus one is zero, so I don't care what you have over here, zero times anything is going to be equal to zero. So when x equals 1 1\u20442, the first thing becomes zero, making everything, making the product equal zero, and likewise, if x equals negative four, it's pretty clear that this second expression is going to be zero, and even though this first expression isn't going to be zero in that case, anything times zero is going to be zero. Let's do one more example here. So let me delete out everything that I just wrote here."}, {"video_title": "Solving equations with zero product property.mp3", "Sentence": "Well, two times 1 1\u20442 is one, one minus one is zero, so I don't care what you have over here, zero times anything is going to be equal to zero. So when x equals 1 1\u20442, the first thing becomes zero, making everything, making the product equal zero, and likewise, if x equals negative four, it's pretty clear that this second expression is going to be zero, and even though this first expression isn't going to be zero in that case, anything times zero is going to be zero. Let's do one more example here. So let me delete out everything that I just wrote here. And so let's, I'm gonna involve a function. So let's say someone told you that f of x is equal to x minus five times five x plus two, times five x plus two, and someone said find the zeros of f of x. Well, the zeros are what are the x values that make f of x equal to zero?"}, {"video_title": "Solving equations with zero product property.mp3", "Sentence": "So let me delete out everything that I just wrote here. And so let's, I'm gonna involve a function. So let's say someone told you that f of x is equal to x minus five times five x plus two, times five x plus two, and someone said find the zeros of f of x. Well, the zeros are what are the x values that make f of x equal to zero? When does f of x equal zero? For what, for what x values does f of x equal zero? That's what people are really asking when they say find the zeros of f of x."}, {"video_title": "Solving equations with zero product property.mp3", "Sentence": "Well, the zeros are what are the x values that make f of x equal to zero? When does f of x equal zero? For what, for what x values does f of x equal zero? That's what people are really asking when they say find the zeros of f of x. So to do that, well, when does f of x equal zero? Well, f of x is equal to zero when this expression right over here is equal to zero, and so it sets up just like the equation we just saw. X minus five times five x plus two, when does that equal zero?"}, {"video_title": "Solving equations with zero product property.mp3", "Sentence": "That's what people are really asking when they say find the zeros of f of x. So to do that, well, when does f of x equal zero? Well, f of x is equal to zero when this expression right over here is equal to zero, and so it sets up just like the equation we just saw. X minus five times five x plus two, when does that equal zero? And like we saw before, this has, well, this is just like what we saw before, and I encourage you to pause the video and try to work it out on your own. So there's two situations where this could happen, where either the first expression equals zero, or the second expression, or maybe in some cases you'll have a situation where both expressions equal zero. So we could say either x minus five is equal to zero, or five x plus two is equal to zero."}, {"video_title": "Solving equations with zero product property.mp3", "Sentence": "X minus five times five x plus two, when does that equal zero? And like we saw before, this has, well, this is just like what we saw before, and I encourage you to pause the video and try to work it out on your own. So there's two situations where this could happen, where either the first expression equals zero, or the second expression, or maybe in some cases you'll have a situation where both expressions equal zero. So we could say either x minus five is equal to zero, or five x plus two is equal to zero. I'll write an or right over here. Now if we solve for x, you add five to both sides of this equation, you get x is equal to five. Here, let's see, to solve for x, you can subtract two from both sides, you get five x is equal to negative two, and you can divide both sides by five to solve for x, and you get x is equal to negative 2 5ths."}, {"video_title": "Solving equations with zero product property.mp3", "Sentence": "So we could say either x minus five is equal to zero, or five x plus two is equal to zero. I'll write an or right over here. Now if we solve for x, you add five to both sides of this equation, you get x is equal to five. Here, let's see, to solve for x, you can subtract two from both sides, you get five x is equal to negative two, and you can divide both sides by five to solve for x, and you get x is equal to negative 2 5ths. So here are two zeros. You input either one of these into f of x. If you input x equals five, if you take f of five, if you try to evaluate f of five, then this first expression's gonna be zero, and so a product of zero and something else, it doesn't matter that this is gonna be 27, zero times 27 is zero."}, {"video_title": "Products and exponents raised to an exponent properties   Algebra I   Khan Academy.mp3", "Sentence": "Let's say I have two numbers, a and b, and I'm going to raise it to, I could do it in the abstract, I could raise it to the c power, but I'll do it a little bit more concrete. Let's raise it to the fourth power. What is that going to be equal to? Well, that's going to be equal to, that's going to be equal to, I could write it like this, let me copy and paste this, copy and paste, that's going to be equal to a b times a b, times a b, times a b, times a b. What is that equal to? Well, when you just multiply a bunch of numbers like this, it doesn't matter what order you're going to multiply it in. This right over here is going to be equivalent to a times a times a times a, times, we have four b's as well that we're multiplying together, times b, times b, times b, times b."}, {"video_title": "Products and exponents raised to an exponent properties   Algebra I   Khan Academy.mp3", "Sentence": "Well, that's going to be equal to, that's going to be equal to, I could write it like this, let me copy and paste this, copy and paste, that's going to be equal to a b times a b, times a b, times a b, times a b. What is that equal to? Well, when you just multiply a bunch of numbers like this, it doesn't matter what order you're going to multiply it in. This right over here is going to be equivalent to a times a times a times a, times, we have four b's as well that we're multiplying together, times b, times b, times b, times b. What is that equal to? This right over here is a to the fourth power, and this right over here is b, b to the fourth power. You see, if you take the product of two numbers and you raise them to some exponent, that's equivalent to taking each of the numbers to that exponent and then taking their product."}, {"video_title": "Products and exponents raised to an exponent properties   Algebra I   Khan Academy.mp3", "Sentence": "This right over here is going to be equivalent to a times a times a times a, times, we have four b's as well that we're multiplying together, times b, times b, times b, times b. What is that equal to? This right over here is a to the fourth power, and this right over here is b, b to the fourth power. You see, if you take the product of two numbers and you raise them to some exponent, that's equivalent to taking each of the numbers to that exponent and then taking their product. Here I just used the example of four, but you could do this really with any arbitrary, you could do this with actually any integer, or actually any exponent this property holds. You could satisfy yourself by trying different values and using the same logic right over here, but this is a general property that, let me write it this way, that if I have a to the b, a to the b to the c power, that this is going to be equal to a to the c, a to the c times b to the c, times b to the c, times b to the c power. We'll use this throughout actually mathematics when we try to simplify things or rewrite an expression in a different way."}, {"video_title": "Products and exponents raised to an exponent properties   Algebra I   Khan Academy.mp3", "Sentence": "You see, if you take the product of two numbers and you raise them to some exponent, that's equivalent to taking each of the numbers to that exponent and then taking their product. Here I just used the example of four, but you could do this really with any arbitrary, you could do this with actually any integer, or actually any exponent this property holds. You could satisfy yourself by trying different values and using the same logic right over here, but this is a general property that, let me write it this way, that if I have a to the b, a to the b to the c power, that this is going to be equal to a to the c, a to the c times b to the c, times b to the c, times b to the c power. We'll use this throughout actually mathematics when we try to simplify things or rewrite an expression in a different way. Now let me introduce you another core idea here, and this is the idea of raising something to some power, and I'll just use the example of three, and then raising that to some power. What could this be simplified as? Well, let's think about it."}, {"video_title": "Products and exponents raised to an exponent properties   Algebra I   Khan Academy.mp3", "Sentence": "We'll use this throughout actually mathematics when we try to simplify things or rewrite an expression in a different way. Now let me introduce you another core idea here, and this is the idea of raising something to some power, and I'll just use the example of three, and then raising that to some power. What could this be simplified as? Well, let's think about it. This is the same thing as a to the third, this is the same thing, let me copy and paste that, is a to the third times a to the third. What is a to the third times? This is equal to a to the third times a to the third, and that's going to be equal to a to the three plus three power."}, {"video_title": "Products and exponents raised to an exponent properties   Algebra I   Khan Academy.mp3", "Sentence": "Well, let's think about it. This is the same thing as a to the third, this is the same thing, let me copy and paste that, is a to the third times a to the third. What is a to the third times? This is equal to a to the third times a to the third, and that's going to be equal to a to the three plus three power. We have the same base, and so we would add, and they're being multiplied, they're being raised to these two exponents, so it's going to be the sum of the exponents, which of course is going to be equal to a to the, that's a different color, a, it's going to be a to the sixth power. What just happened over here? Well, I took two a to the thirds, and I multiplied them together, so I took these two threes and added them together."}, {"video_title": "Products and exponents raised to an exponent properties   Algebra I   Khan Academy.mp3", "Sentence": "This is equal to a to the third times a to the third, and that's going to be equal to a to the three plus three power. We have the same base, and so we would add, and they're being multiplied, they're being raised to these two exponents, so it's going to be the sum of the exponents, which of course is going to be equal to a to the, that's a different color, a, it's going to be a to the sixth power. What just happened over here? Well, I took two a to the thirds, and I multiplied them together, so I took these two threes and added them together. This essentially right over here, you could view this as two times three. This right over here is two times three. That's how we got the six."}, {"video_title": "Products and exponents raised to an exponent properties   Algebra I   Khan Academy.mp3", "Sentence": "Well, I took two a to the thirds, and I multiplied them together, so I took these two threes and added them together. This essentially right over here, you could view this as two times three. This right over here is two times three. That's how we got the six. When I raise something to one exponent, and then raised it to another, that's equivalent to raising the base to the product of those two exponents. I just did it with this example right over here, but I encourage you, try other numbers to see how this works. I could do this in general."}, {"video_title": "Products and exponents raised to an exponent properties   Algebra I   Khan Academy.mp3", "Sentence": "That's how we got the six. When I raise something to one exponent, and then raised it to another, that's equivalent to raising the base to the product of those two exponents. I just did it with this example right over here, but I encourage you, try other numbers to see how this works. I could do this in general. I could say a to the b power, and then let me copy and paste that. Copy, and then I'm going to raise that to the c power. I'm going to raise that to the c power."}, {"video_title": "Products and exponents raised to an exponent properties   Algebra I   Khan Academy.mp3", "Sentence": "I could do this in general. I could say a to the b power, and then let me copy and paste that. Copy, and then I'm going to raise that to the c power. I'm going to raise that to the c power. What is that going to give me? I'm essentially going to have to take c of these, so one, two, three, I don't know how large of a number c is, so I'll just do dot, dot, dot. Dot, dot, dot."}, {"video_title": "Products and exponents raised to an exponent properties   Algebra I   Khan Academy.mp3", "Sentence": "I'm going to raise that to the c power. What is that going to give me? I'm essentially going to have to take c of these, so one, two, three, I don't know how large of a number c is, so I'll just do dot, dot, dot. Dot, dot, dot. I have c of these right over here. I have c of these, so there's c of them right over there. What is that going to be equal to?"}, {"video_title": "Products and exponents raised to an exponent properties   Algebra I   Khan Academy.mp3", "Sentence": "Dot, dot, dot. I have c of these right over here. I have c of these, so there's c of them right over there. What is that going to be equal to? That is going to be equal to a to the, well for each of these c, I'm going to have a b that I'm going to add together. Let me write this. I'm going to have a b plus b plus b plus dot, dot, dot plus b."}, {"video_title": "Products and exponents raised to an exponent properties   Algebra I   Khan Academy.mp3", "Sentence": "What is that going to be equal to? That is going to be equal to a to the, well for each of these c, I'm going to have a b that I'm going to add together. Let me write this. I'm going to have a b plus b plus b plus dot, dot, dot plus b. Now I have c of these b's, so I have c b's right over here, or you could view this as a, this is equal to a to the c times b power. C or a, you could view a to the c b power. Very useful."}, {"video_title": "Products and exponents raised to an exponent properties   Algebra I   Khan Academy.mp3", "Sentence": "I'm going to have a b plus b plus b plus dot, dot, dot plus b. Now I have c of these b's, so I have c b's right over here, or you could view this as a, this is equal to a to the c times b power. C or a, you could view a to the c b power. Very useful. If someone were to say, what is 35 to the third power, and then that raised to the seventh power, well this is obviously going to be a huge number, but we can at least simplify the expression. This is going to be equal to 35 to the product of these two exponents. It's going to be 35 to the three times seven, or 35 to the 21st power."}, {"video_title": "Positive and negative slope   Algebra I   Khan Academy.mp3", "Sentence": "So, for example, if we're looking at the XY plane here, our change in the vertical direction is gonna be a change in the Y variable divided by change in the horizontal direction is gonna be a change in the X variable. And so let's see why that is a good definition for slope. Well, I could draw something with a slope of one. A slope of one might look something like, let me, I could start it over here, and let me get my line tool out. So slope of one, as X increases by one, Y increases by one. As X increases by one, Y increases by one. So slope of one is going to look like, is going to look like this."}, {"video_title": "Positive and negative slope   Algebra I   Khan Academy.mp3", "Sentence": "A slope of one might look something like, let me, I could start it over here, and let me get my line tool out. So slope of one, as X increases by one, Y increases by one. As X increases by one, Y increases by one. So slope of one is going to look like, is going to look like this. Is going to look like, is going to look like this. Notice, as I have a change in X, however much my change in X is, so for example here, my change in X is plus two, is positive two, I'm gonna have the same change in Y. My change in Y is going to be plus two."}, {"video_title": "Positive and negative slope   Algebra I   Khan Academy.mp3", "Sentence": "So slope of one is going to look like, is going to look like this. Is going to look like, is going to look like this. Notice, as I have a change in X, however much my change in X is, so for example here, my change in X is plus two, is positive two, I'm gonna have the same change in Y. My change in Y is going to be plus two. So my change in Y divided by change in X is two divided by two is one. So for this line, I have a slope, slope is equal to one. But what would a slope of two look like?"}, {"video_title": "Positive and negative slope   Algebra I   Khan Academy.mp3", "Sentence": "My change in Y is going to be plus two. So my change in Y divided by change in X is two divided by two is one. So for this line, I have a slope, slope is equal to one. But what would a slope of two look like? Well, a slope of two should be steeper, and we could draw that. A slope of two, I could start at that same point. Actually, why don't I start at the same point, and we'll see, but you don't have to."}, {"video_title": "Positive and negative slope   Algebra I   Khan Academy.mp3", "Sentence": "But what would a slope of two look like? Well, a slope of two should be steeper, and we could draw that. A slope of two, I could start at that same point. Actually, why don't I start at the same point, and we'll see, but you don't have to. Actually, let me start at a different point. So if I start over, let's say here, a slope of two would look like, a slope of two would look like, for every one that I increase in the X direction, I'm gonna increase two in the Y direction. So it's gonna look like, it is going to look like that."}, {"video_title": "Positive and negative slope   Algebra I   Khan Academy.mp3", "Sentence": "Actually, why don't I start at the same point, and we'll see, but you don't have to. Actually, let me start at a different point. So if I start over, let's say here, a slope of two would look like, a slope of two would look like, for every one that I increase in the X direction, I'm gonna increase two in the Y direction. So it's gonna look like, it is going to look like that. This line right over here, you see it. If my change in X is one, change in X is equal to one, my change in Y, my change in Y, my change in Y is two. So change in Y over change in X is gonna be two over one."}, {"video_title": "Positive and negative slope   Algebra I   Khan Academy.mp3", "Sentence": "So it's gonna look like, it is going to look like that. This line right over here, you see it. If my change in X is one, change in X is equal to one, my change in Y, my change in Y, my change in Y is two. So change in Y over change in X is gonna be two over one. The slope here is two. And now, hopefully, you're appreciating why this definition of slope is a good one. The higher the slope, the steeper it is."}, {"video_title": "Positive and negative slope   Algebra I   Khan Academy.mp3", "Sentence": "So change in Y over change in X is gonna be two over one. The slope here is two. And now, hopefully, you're appreciating why this definition of slope is a good one. The higher the slope, the steeper it is. The faster it increases, the faster we increase in the vertical direction as we increase in the horizontal direction. Now, what would a negative slope be? So let's just think about what a line with a negative slope would mean."}, {"video_title": "Positive and negative slope   Algebra I   Khan Academy.mp3", "Sentence": "The higher the slope, the steeper it is. The faster it increases, the faster we increase in the vertical direction as we increase in the horizontal direction. Now, what would a negative slope be? So let's just think about what a line with a negative slope would mean. Well, a negative slope would mean that, well, we could take an example. If we have our change in Y over change in X, was, say, let's say it was equal to a negative one. That means that if we have a change in X of one, then in order to get negative one here, that means that our change in Y would have to be equal to negative one."}, {"video_title": "Positive and negative slope   Algebra I   Khan Academy.mp3", "Sentence": "So let's just think about what a line with a negative slope would mean. Well, a negative slope would mean that, well, we could take an example. If we have our change in Y over change in X, was, say, let's say it was equal to a negative one. That means that if we have a change in X of one, then in order to get negative one here, that means that our change in Y would have to be equal to negative one. So a line with a negative slope, a negative one slope would look like, let me see if I can draw it, would look like, a negative one slope would look like, would look like this. Notice, as X increases, as X increases by a certain amount, so our delta X here is one, Y decreases by that same amount instead of increasing. So now this is what we consider a downward sloping line."}, {"video_title": "Positive and negative slope   Algebra I   Khan Academy.mp3", "Sentence": "That means that if we have a change in X of one, then in order to get negative one here, that means that our change in Y would have to be equal to negative one. So a line with a negative slope, a negative one slope would look like, let me see if I can draw it, would look like, a negative one slope would look like, would look like this. Notice, as X increases, as X increases by a certain amount, so our delta X here is one, Y decreases by that same amount instead of increasing. So now this is what we consider a downward sloping line. So change in Y is equal to negative one. So our change in Y over change in X, change in Y over change in X is equal to negative one over one, which is equal to negative one. So the slope of this line is negative one."}, {"video_title": "Positive and negative slope   Algebra I   Khan Academy.mp3", "Sentence": "So now this is what we consider a downward sloping line. So change in Y is equal to negative one. So our change in Y over change in X, change in Y over change in X is equal to negative one over one, which is equal to negative one. So the slope of this line is negative one. Now, if you had a slope of negative two, it would decrease even faster. So a line with a slope of negative two, it could look something like this. Let me draw it."}, {"video_title": "Positive and negative slope   Algebra I   Khan Academy.mp3", "Sentence": "So the slope of this line is negative one. Now, if you had a slope of negative two, it would decrease even faster. So a line with a slope of negative two, it could look something like this. Let me draw it. So as X increases by one, Y would decrease by two. As X increases by one, Y would decrease by two. So it would look something like, let me see if I can, it would look like, it would look like that."}, {"video_title": "Positive and negative slope   Algebra I   Khan Academy.mp3", "Sentence": "Let me draw it. So as X increases by one, Y would decrease by two. As X increases by one, Y would decrease by two. So it would look something like, let me see if I can, it would look like, it would look like that. Notice, as our X increases by a certain amount, our Y decreases, decreases by twice as much. So this right over here has a slope. This has a slope of negative two."}, {"video_title": "Positive and negative slope   Algebra I   Khan Academy.mp3", "Sentence": "So it would look something like, let me see if I can, it would look like, it would look like that. Notice, as our X increases by a certain amount, our Y decreases, decreases by twice as much. So this right over here has a slope. This has a slope of negative two. So hopefully this gives you a little bit more intuition for what slope represents and how the number that we used to represent slope, how you can use that to visualize how steep a line is. A very high positive slope, as X increases, Y is going to increase fairly dramatically. If you have a negative slope, you're actually going to go from, you're actually going to go, as X increases, your Y is actually going to decrease."}, {"video_title": "Factoring difference of squares shared factors   High School Math   Khan Academy.mp3", "Sentence": "What binomial factor do they share? And like always, pause the video and see if you can work through this. All right, now let's work through this together. And the way I'm gonna do it is I'm just gonna try to factor both of them into the product of binomials and maybe some other things, and see if we have any common binomial factors. So first, let's focus on m squared minus four m minus 45. So let me write it over here. M squared minus four m minus 45."}, {"video_title": "Factoring difference of squares shared factors   High School Math   Khan Academy.mp3", "Sentence": "And the way I'm gonna do it is I'm just gonna try to factor both of them into the product of binomials and maybe some other things, and see if we have any common binomial factors. So first, let's focus on m squared minus four m minus 45. So let me write it over here. M squared minus four m minus 45. So when you're factoring a quadratic expression like this, where the coefficient on the, in this case, m squared term, on the second degree term is one, we could factor it as being equal to m plus a times m plus b, where a plus b is going to be equal to this coefficient right over here, and a times b is going to be equal to this coefficient right over here. So let's be clear. So a, let me just, another color."}, {"video_title": "Factoring difference of squares shared factors   High School Math   Khan Academy.mp3", "Sentence": "M squared minus four m minus 45. So when you're factoring a quadratic expression like this, where the coefficient on the, in this case, m squared term, on the second degree term is one, we could factor it as being equal to m plus a times m plus b, where a plus b is going to be equal to this coefficient right over here, and a times b is going to be equal to this coefficient right over here. So let's be clear. So a, let me just, another color. So a plus b needs to be equal to negative four. A plus b needs to be equal to negative four. And then a times b needs to be equal to negative 45."}, {"video_title": "Factoring difference of squares shared factors   High School Math   Khan Academy.mp3", "Sentence": "So a, let me just, another color. So a plus b needs to be equal to negative four. A plus b needs to be equal to negative four. And then a times b needs to be equal to negative 45. A times b is equal to negative 45. Now I like to focus on the a times b and think about, well, what could a and b be to get to negative 45? Well, if I'm taking the product of two things, and if the product is negative, that means that they're going to have different signs."}, {"video_title": "Factoring difference of squares shared factors   High School Math   Khan Academy.mp3", "Sentence": "And then a times b needs to be equal to negative 45. A times b is equal to negative 45. Now I like to focus on the a times b and think about, well, what could a and b be to get to negative 45? Well, if I'm taking the product of two things, and if the product is negative, that means that they're going to have different signs. And if when we add them, we get a negative number, that means that the negative one has a larger magnitude. So let's think about this a little bit. So a times b is equal to negative 45."}, {"video_title": "Factoring difference of squares shared factors   High School Math   Khan Academy.mp3", "Sentence": "Well, if I'm taking the product of two things, and if the product is negative, that means that they're going to have different signs. And if when we add them, we get a negative number, that means that the negative one has a larger magnitude. So let's think about this a little bit. So a times b is equal to negative 45. So this could be, let's try some values out. So one and 45, those are too far apart. Let's see, three and 15, those still seem pretty far apart."}, {"video_title": "Factoring difference of squares shared factors   High School Math   Khan Academy.mp3", "Sentence": "So a times b is equal to negative 45. So this could be, let's try some values out. So one and 45, those are too far apart. Let's see, three and 15, those still seem pretty far apart. Let's see, it looks like five and nine seem interesting. So if we say five times, if we were to say five times negative nine, that indeed is equal to negative 45. And five plus negative nine is indeed equal to negative four."}, {"video_title": "Factoring difference of squares shared factors   High School Math   Khan Academy.mp3", "Sentence": "Let's see, three and 15, those still seem pretty far apart. Let's see, it looks like five and nine seem interesting. So if we say five times, if we were to say five times negative nine, that indeed is equal to negative 45. And five plus negative nine is indeed equal to negative four. So a could be equal to five, and b could be equal to negative nine. And so if we were to factor this, this is going to be m plus five times m, I could say m plus negative nine, but I'll just write m minus nine. So just like that, I've been able to factor, I've been able to factor this first quadratic expression right over there as a product of two binomials."}, {"video_title": "Factoring difference of squares shared factors   High School Math   Khan Academy.mp3", "Sentence": "And five plus negative nine is indeed equal to negative four. So a could be equal to five, and b could be equal to negative nine. And so if we were to factor this, this is going to be m plus five times m, I could say m plus negative nine, but I'll just write m minus nine. So just like that, I've been able to factor, I've been able to factor this first quadratic expression right over there as a product of two binomials. So now let's try to factor the other quadratic expression. Let's try to factor six m squared minus 150. And let's see, the first thing I might wanna do is both six, both six m squared and 150, they're both divisible by six."}, {"video_title": "Factoring difference of squares shared factors   High School Math   Khan Academy.mp3", "Sentence": "So just like that, I've been able to factor, I've been able to factor this first quadratic expression right over there as a product of two binomials. So now let's try to factor the other quadratic expression. Let's try to factor six m squared minus 150. And let's see, the first thing I might wanna do is both six, both six m squared and 150, they're both divisible by six. So let me write it this way. I could write it as six, actually I'll just write six m squared minus six times, let's see, six goes into 150 25 times. So all I did is I rewrote this, and really I just wrote 150 as six times 25."}, {"video_title": "Factoring difference of squares shared factors   High School Math   Khan Academy.mp3", "Sentence": "And let's see, the first thing I might wanna do is both six, both six m squared and 150, they're both divisible by six. So let me write it this way. I could write it as six, actually I'll just write six m squared minus six times, let's see, six goes into 150 25 times. So all I did is I rewrote this, and really I just wrote 150 as six times 25. And now you can clearly see that we can factor out a six. You can view this as undistributing the six. So this is the same thing as six times m squared minus 25, which we recognize this is a difference of squares, so it's all gonna be six times m plus five times m minus five."}, {"video_title": "Factoring difference of squares shared factors   High School Math   Khan Academy.mp3", "Sentence": "So all I did is I rewrote this, and really I just wrote 150 as six times 25. And now you can clearly see that we can factor out a six. You can view this as undistributing the six. So this is the same thing as six times m squared minus 25, which we recognize this is a difference of squares, so it's all gonna be six times m plus five times m minus five. And so we've factored this out as a product of binomials and a constant factor here, six. And so what is their shared common, or what is their common binomial factor that they share? Well you see when we factor it out, they both have an m plus five."}, {"video_title": "Constructing, solving two-step inequality example   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "My brain immediately says that's greater than or equal to 120,000. If they had an audience of 45,000 in Mesa and another 33,000 in Denver, how many people attended their show in Las Vegas? So let's say Las Vegas. I'll just use L for Las Vegas. So the number of people who attended their show in Las Vegas plus the number that attended their show in Mesa, which is 45,000, plus the number of people that attended their show in Denver, which is 33,000. Those are our three cities right there. Las Vegas, Mesa, and Denver."}, {"video_title": "Constructing, solving two-step inequality example   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "I'll just use L for Las Vegas. So the number of people who attended their show in Las Vegas plus the number that attended their show in Mesa, which is 45,000, plus the number of people that attended their show in Denver, which is 33,000. Those are our three cities right there. Las Vegas, Mesa, and Denver. That has to be at least 120,000 people. Or another way of interpreting that is greater than or equal to 120,000. So to figure out how many people attended their show in Las Vegas, we just solve for L on this inequality."}, {"video_title": "Constructing, solving two-step inequality example   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Las Vegas, Mesa, and Denver. That has to be at least 120,000 people. Or another way of interpreting that is greater than or equal to 120,000. So to figure out how many people attended their show in Las Vegas, we just solve for L on this inequality. So if we simplify this left-hand side, we get the number of people in Las Vegas plus 45,000 plus 33,000. That is 78,000. It's going to be greater than or equal to 120,000."}, {"video_title": "Constructing, solving two-step inequality example   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So to figure out how many people attended their show in Las Vegas, we just solve for L on this inequality. So if we simplify this left-hand side, we get the number of people in Las Vegas plus 45,000 plus 33,000. That is 78,000. It's going to be greater than or equal to 120,000. Now to isolate the L on the left-hand side of the inequality, we can subtract 78,000 from both sides. So minus 78,000. On the left-hand side, these cancel out."}, {"video_title": "Constructing, solving two-step inequality example   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "It's going to be greater than or equal to 120,000. Now to isolate the L on the left-hand side of the inequality, we can subtract 78,000 from both sides. So minus 78,000. On the left-hand side, these cancel out. And we're just left with the number of people who attended the show in Las Vegas is going to be greater than or equal to 120,000 minus 78,000. So 120,000 minus 80,000 is 40,000. That's going to be another 2,000."}, {"video_title": "Constructing, solving two-step inequality example   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "On the left-hand side, these cancel out. And we're just left with the number of people who attended the show in Las Vegas is going to be greater than or equal to 120,000 minus 78,000. So 120,000 minus 80,000 is 40,000. That's going to be another 2,000. So the number of people who attended Las Vegas is going to be greater than or equal to 42,000 people. And we're done. That's it."}, {"video_title": "Express a small number in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "Scientific notation will be some number times some power of 10, where this number right here is going to be, let me write it this way, it's going to be greater than or equal to 1 and it's going to be less than 10. So over here, what we want to put here is that leading number is going to be, and in general you're going to look for the first non-zero digit, and this is the number that you're going to want to start off with. This is the only number you're going to want to put ahead of, or I guess to the left of the decimal point. So we could write 3.457 and it's going to be multiplied by 10 to something. Now let's think about what we're going to have to multiply it by. To go from 3.457 to this very, very small number, I mean we have had to move the decimal from 3.457 to get to this, you have to move the decimal to the left a bunch. You have to add a bunch of zeros to the left of the 3."}, {"video_title": "Express a small number in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "So we could write 3.457 and it's going to be multiplied by 10 to something. Now let's think about what we're going to have to multiply it by. To go from 3.457 to this very, very small number, I mean we have had to move the decimal from 3.457 to get to this, you have to move the decimal to the left a bunch. You have to add a bunch of zeros to the left of the 3. You have to keep moving the decimal over to the left. To do that, we're essentially making the number much, much, much smaller. So we're not going to multiply it by a positive exponent of 10."}, {"video_title": "Express a small number in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "You have to add a bunch of zeros to the left of the 3. You have to keep moving the decimal over to the left. To do that, we're essentially making the number much, much, much smaller. So we're not going to multiply it by a positive exponent of 10. We're going to multiply it times a negative exponent of 10. The equivalent is you're kind of dividing by a positive exponent of 10. So the best way to think about it, when you move an exponent 1 to the left, you're dividing by 10, which is equivalent to multiplying by 10 to the negative 1 power."}, {"video_title": "Express a small number in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "So we're not going to multiply it by a positive exponent of 10. We're going to multiply it times a negative exponent of 10. The equivalent is you're kind of dividing by a positive exponent of 10. So the best way to think about it, when you move an exponent 1 to the left, you're dividing by 10, which is equivalent to multiplying by 10 to the negative 1 power. Let me give you an example here. So if I have 1 times 10 is clearly just equal to 10. 1 times 10 to the negative 1, that's equal to 1 times 1 over 10, which is equal to 1 over 10."}, {"video_title": "Express a small number in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "So the best way to think about it, when you move an exponent 1 to the left, you're dividing by 10, which is equivalent to multiplying by 10 to the negative 1 power. Let me give you an example here. So if I have 1 times 10 is clearly just equal to 10. 1 times 10 to the negative 1, that's equal to 1 times 1 over 10, which is equal to 1 over 10. 1 times, and let me actually write a decimal, which is equal to 0, which is equal to, let me actually, I skipped a step right there. Let me add 1 times 10 to the 0 so we have something natural. So this is 1 times 10 to the 1st."}, {"video_title": "Express a small number in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "1 times 10 to the negative 1, that's equal to 1 times 1 over 10, which is equal to 1 over 10. 1 times, and let me actually write a decimal, which is equal to 0, which is equal to, let me actually, I skipped a step right there. Let me add 1 times 10 to the 0 so we have something natural. So this is 1 times 10 to the 1st. 1 times 10 to the 0 is equal to 1 times 1, which is equal to 1. 1 times 10 to the negative 1 is equal to 1 over 10, which is equal to 0.1. If I do 1 times 10 to the negative 2, 10 to the negative 2 is 1 over 10 squared, or 1 over 100."}, {"video_title": "Express a small number in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "So this is 1 times 10 to the 1st. 1 times 10 to the 0 is equal to 1 times 1, which is equal to 1. 1 times 10 to the negative 1 is equal to 1 over 10, which is equal to 0.1. If I do 1 times 10 to the negative 2, 10 to the negative 2 is 1 over 10 squared, or 1 over 100. So this is going to be 1 over 100, which is 0.01. What's happening here? When I raise it to a negative power, I raise it to a negative 1 power, I've essentially moved the decimal from to the right of the 1 to the left of the 1."}, {"video_title": "Express a small number in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "If I do 1 times 10 to the negative 2, 10 to the negative 2 is 1 over 10 squared, or 1 over 100. So this is going to be 1 over 100, which is 0.01. What's happening here? When I raise it to a negative power, I raise it to a negative 1 power, I've essentially moved the decimal from to the right of the 1 to the left of the 1. I moved it from there to there. When I raise it to the negative 2, I moved it 2 over to the left. So how many times are we going to have to move it over to the left to get this number right over here?"}, {"video_title": "Express a small number in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "When I raise it to a negative power, I raise it to a negative 1 power, I've essentially moved the decimal from to the right of the 1 to the left of the 1. I moved it from there to there. When I raise it to the negative 2, I moved it 2 over to the left. So how many times are we going to have to move it over to the left to get this number right over here? Let's think about how many 0s we have. We have to move it 1 time just to get in front of the 3, and then we have to move it that many more times to get all of the 0s in there. So we have to move it 1 time to get the 3."}, {"video_title": "Express a small number in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "So how many times are we going to have to move it over to the left to get this number right over here? Let's think about how many 0s we have. We have to move it 1 time just to get in front of the 3, and then we have to move it that many more times to get all of the 0s in there. So we have to move it 1 time to get the 3. If we started here, we're going to move 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 times. This is going to be 3.457 times 10 to the negative 10 power. Let me just rewrite it."}, {"video_title": "Express a small number in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "So we have to move it 1 time to get the 3. If we started here, we're going to move 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 times. This is going to be 3.457 times 10 to the negative 10 power. Let me just rewrite it. 3.457 times 10 to the negative 10 power. In general, what you want to do is you want to find the first non-zero number here. Remember, you want a number here that's between 1 and 10, and it can be equal to 1, but it has to be less than 10."}, {"video_title": "Express a small number in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "Let me just rewrite it. 3.457 times 10 to the negative 10 power. In general, what you want to do is you want to find the first non-zero number here. Remember, you want a number here that's between 1 and 10, and it can be equal to 1, but it has to be less than 10. 3.457 definitely fits that bill. It's between 1 and 10. Then you just want to count the leading 0s between the decimal and that number and include the number because that tells you how many times you have to shift the decimal over to actually get this number up here."}, {"video_title": "Comparing features of functions (example 2)   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "So let's figure out the maximum value for each of these, and they're defined in different ways, and then see which one is the lowest. And I'll start with the easiest. So h of x, we can just graphically look at it, visually look at it, and say, what's the maximum point? And the maximum point looks like it's right over here when x is equal to 4. And when x is equal to 4, y or h of x is equal to negative 1. So the maximum for h of x looks like it is negative 1. Now, what's the maximum for g of x?"}, {"video_title": "Comparing features of functions (example 2)   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "And the maximum point looks like it's right over here when x is equal to 4. And when x is equal to 4, y or h of x is equal to negative 1. So the maximum for h of x looks like it is negative 1. Now, what's the maximum for g of x? And they've given us some points here. And here, once again, we can just eyeball it and say, well, what's the maximum value they gave us? Well, 5 is the largest value."}, {"video_title": "Comparing features of functions (example 2)   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "Now, what's the maximum for g of x? And they've given us some points here. And here, once again, we can just eyeball it and say, well, what's the maximum value they gave us? Well, 5 is the largest value. It happens when x is equal to 0. g of 0 is 5. So the maximum value here is 5. Now, f of x, they just give us an expression to define it."}, {"video_title": "Comparing features of functions (example 2)   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "Well, 5 is the largest value. It happens when x is equal to 0. g of 0 is 5. So the maximum value here is 5. Now, f of x, they just give us an expression to define it. And so it's going to take a little bit of work to figure out what the maximum value is. The easiest way to do that for a quadratic is to complete the square. And so let's do it."}, {"video_title": "Comparing features of functions (example 2)   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "Now, f of x, they just give us an expression to define it. And so it's going to take a little bit of work to figure out what the maximum value is. The easiest way to do that for a quadratic is to complete the square. And so let's do it. So we have f of x is equal to negative x squared plus 6x minus 1. I never like having this negative here. So I'm going to factor it out."}, {"video_title": "Comparing features of functions (example 2)   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "And so let's do it. So we have f of x is equal to negative x squared plus 6x minus 1. I never like having this negative here. So I'm going to factor it out. This is the same thing as negative times x squared minus 6x and plus 1. And I'm going to write the plus 1 out here because I'm fixing to complete the square. Now, just as a review of completing the square, we essentially want to add and subtract the same number so that part of this expression is a perfect square."}, {"video_title": "Comparing features of functions (example 2)   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "So I'm going to factor it out. This is the same thing as negative times x squared minus 6x and plus 1. And I'm going to write the plus 1 out here because I'm fixing to complete the square. Now, just as a review of completing the square, we essentially want to add and subtract the same number so that part of this expression is a perfect square. And to figure out what number we want to add and subtract, we look at the coefficient on the x term. It's a negative 6. You take half of that."}, {"video_title": "Comparing features of functions (example 2)   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "Now, just as a review of completing the square, we essentially want to add and subtract the same number so that part of this expression is a perfect square. And to figure out what number we want to add and subtract, we look at the coefficient on the x term. It's a negative 6. You take half of that. That's negative 3. And you square it. Negative 3 squared is 9."}, {"video_title": "Comparing features of functions (example 2)   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "You take half of that. That's negative 3. And you square it. Negative 3 squared is 9. Now, we can't just add a 9. That would change the actual value of the expression. We have to add a 9 and subtract a 9."}, {"video_title": "Comparing features of functions (example 2)   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "Negative 3 squared is 9. Now, we can't just add a 9. That would change the actual value of the expression. We have to add a 9 and subtract a 9. And you might say, well, why are we adding and subtracting the same thing if it doesn't change the value of the expression? And the whole point is so that we can get this first part of the expression to represent a perfect square. This x squared minus 6x plus 9 is x minus 3 squared."}, {"video_title": "Comparing features of functions (example 2)   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "We have to add a 9 and subtract a 9. And you might say, well, why are we adding and subtracting the same thing if it doesn't change the value of the expression? And the whole point is so that we can get this first part of the expression to represent a perfect square. This x squared minus 6x plus 9 is x minus 3 squared. So I can rewrite that part as x minus 3 squared. And then minus 9 or negative 9 plus 1 is negative 8. So minus, let me do that in a different color so we can keep track of things."}, {"video_title": "Comparing features of functions (example 2)   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "This x squared minus 6x plus 9 is x minus 3 squared. So I can rewrite that part as x minus 3 squared. And then minus 9 or negative 9 plus 1 is negative 8. So minus, let me do that in a different color so we can keep track of things. So this part right over here is negative 8. And we still have the negative out front. And so we can rewrite this as, if we distribute the negative sign, negative x minus 3 squared plus 8."}, {"video_title": "Comparing features of functions (example 2)   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "So minus, let me do that in a different color so we can keep track of things. So this part right over here is negative 8. And we still have the negative out front. And so we can rewrite this as, if we distribute the negative sign, negative x minus 3 squared plus 8. Now, let's think about what the maximum value is. And to understand the maximum value, we have to interpret this negative x minus 3 squared. Well, x minus 3 squared, before we think about the negative, that is always going to be a positive value."}, {"video_title": "Comparing features of functions (example 2)   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "And so we can rewrite this as, if we distribute the negative sign, negative x minus 3 squared plus 8. Now, let's think about what the maximum value is. And to understand the maximum value, we have to interpret this negative x minus 3 squared. Well, x minus 3 squared, before we think about the negative, that is always going to be a positive value. But then when we make it negative, or it's always going to be non-negative. But then when we make it negative, it's always going to be non-positive. Think about it."}, {"video_title": "Comparing features of functions (example 2)   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "Well, x minus 3 squared, before we think about the negative, that is always going to be a positive value. But then when we make it negative, or it's always going to be non-negative. But then when we make it negative, it's always going to be non-positive. Think about it. If x is equal to 3, this thing is going to be 0. And you take the negative of that, it's going to be 0. x is anything else, x is anything other than 3, this part of the expression is going to be positive. But then you have a minus sign."}, {"video_title": "Comparing features of functions (example 2)   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "Think about it. If x is equal to 3, this thing is going to be 0. And you take the negative of that, it's going to be 0. x is anything else, x is anything other than 3, this part of the expression is going to be positive. But then you have a minus sign. So you're going to subtract that positive value from 8. So this actually has a maximum value when this term, when this first term right over here, is 0. The only thing that this part of the expression can do is subtract from the 8."}, {"video_title": "Comparing features of functions (example 2)   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "But then you have a minus sign. So you're going to subtract that positive value from 8. So this actually has a maximum value when this term, when this first term right over here, is 0. The only thing that this part of the expression can do is subtract from the 8. So you want this, if you want to get a maximum value, this should be equal to 0. This equals 0 when x is equal to 3. When x is equal to 3, this is 0."}, {"video_title": "Comparing features of functions (example 2)   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "The only thing that this part of the expression can do is subtract from the 8. So you want this, if you want to get a maximum value, this should be equal to 0. This equals 0 when x is equal to 3. When x is equal to 3, this is 0. And our function hits its maximum value of 8. So this has a max. Let me do that in a color that you can actually read."}, {"video_title": "Comparing features of functions (example 2)   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "When x is equal to 3, this is 0. And our function hits its maximum value of 8. So this has a max. Let me do that in a color that you can actually read. This has a max value of 8. So which has the lowest maximum value? h of x."}, {"video_title": "Perfect square factorization intro   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "And if someone asks you, hey, can you factor this into two binomials? Well, using techniques we learned in other videos, say, okay, I need to find two numbers whose product is nine and whose sum is six. And so I encourage you to think of, to pause this video and say, well, what two numbers can I, can add up to six, and if I take their product, I get nine? Well, nine only has so many factors, really one, three, and nine, and one plus nine does not equal six, and so, and negative one plus negative nine does not equal six, but three times three equals nine, and three plus three does equal six. Three times three, three plus three. And so we can factor this as x plus three times x plus three, which is, of course, the same thing as x plus three squared. And so what was it about this expression that made us recognize, or maybe now we will start to recognize as it being a perfect square?"}, {"video_title": "Perfect square factorization intro   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "Well, nine only has so many factors, really one, three, and nine, and one plus nine does not equal six, and so, and negative one plus negative nine does not equal six, but three times three equals nine, and three plus three does equal six. Three times three, three plus three. And so we can factor this as x plus three times x plus three, which is, of course, the same thing as x plus three squared. And so what was it about this expression that made us recognize, or maybe now we will start to recognize as it being a perfect square? Well, I have, of course, some variable that is being squared, which we need. I have some perfect square as the constant, and that whatever is being squared there, I have two times that as the coefficient on this first degree term here. Let's see if that is generally true, and I'll switch up the variables just to show that we can."}, {"video_title": "Perfect square factorization intro   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "And so what was it about this expression that made us recognize, or maybe now we will start to recognize as it being a perfect square? Well, I have, of course, some variable that is being squared, which we need. I have some perfect square as the constant, and that whatever is being squared there, I have two times that as the coefficient on this first degree term here. Let's see if that is generally true, and I'll switch up the variables just to show that we can. So let's say that I have a squared plus 14a plus 49. So a few interesting things are happening here. All right, I have my variable squared."}, {"video_title": "Perfect square factorization intro   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "Let's see if that is generally true, and I'll switch up the variables just to show that we can. So let's say that I have a squared plus 14a plus 49. So a few interesting things are happening here. All right, I have my variable squared. I have a perfect square constant term, that is seven squared right over here, and my coefficient on my first degree term here, that is two times the thing that's being squared. That is two times seven, or you could say it's seven plus seven. So you can immediately say, okay, if I want to factor this, this is going to be a plus seven squared."}, {"video_title": "Perfect square factorization intro   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "All right, I have my variable squared. I have a perfect square constant term, that is seven squared right over here, and my coefficient on my first degree term here, that is two times the thing that's being squared. That is two times seven, or you could say it's seven plus seven. So you can immediately say, okay, if I want to factor this, this is going to be a plus seven squared. And you can, of course, verify that by multiplying out, by figuring out what a plus seven squared is. Sometimes when you're first learning, it's like, hey, isn't that just a squared plus seven squared? No, remember, this is the same thing as a plus seven times a plus seven, and you can calculate this by using the FOIL, F-O-I-L technique."}, {"video_title": "Perfect square factorization intro   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "So you can immediately say, okay, if I want to factor this, this is going to be a plus seven squared. And you can, of course, verify that by multiplying out, by figuring out what a plus seven squared is. Sometimes when you're first learning, it's like, hey, isn't that just a squared plus seven squared? No, remember, this is the same thing as a plus seven times a plus seven, and you can calculate this by using the FOIL, F-O-I-L technique. I don't like that so much, because you're not thinking mathematically about what's happening. Really, you just have to do the distributive property twice here. First, you can multiply a plus seven times a, so a plus seven times a, and then multiply a plus seven times seven, so plus a plus seven times seven, and so this is going to be a squared plus seven a, plus, now we distribute this seven, plus seven a plus 49."}, {"video_title": "Perfect square factorization intro   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "No, remember, this is the same thing as a plus seven times a plus seven, and you can calculate this by using the FOIL, F-O-I-L technique. I don't like that so much, because you're not thinking mathematically about what's happening. Really, you just have to do the distributive property twice here. First, you can multiply a plus seven times a, so a plus seven times a, and then multiply a plus seven times seven, so plus a plus seven times seven, and so this is going to be a squared plus seven a, plus, now we distribute this seven, plus seven a plus 49. So now you see where that 14a came from. It's from the seven a plus the seven a. You see where the a squared came from, and you see where the 49 came from."}, {"video_title": "Perfect square factorization intro   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "First, you can multiply a plus seven times a, so a plus seven times a, and then multiply a plus seven times seven, so plus a plus seven times seven, and so this is going to be a squared plus seven a, plus, now we distribute this seven, plus seven a plus 49. So now you see where that 14a came from. It's from the seven a plus the seven a. You see where the a squared came from, and you see where the 49 came from. And you can speak of this in more general terms. If I wanted to just take the expression a plus b and square it, that's just a plus b times a plus b, and we do exactly what we did just here, but here I'm just doing it in very general terms with a or b, and you can think of a as either a constant number or even a variable, and so this is going to be, if we distribute this, it's going to be a plus b times that a, plus a plus b times that b. And so this is going to be a squared, now I'm just doing the distributive property again, a squared plus a b plus a b plus b squared."}, {"video_title": "Perfect square factorization intro   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "You see where the a squared came from, and you see where the 49 came from. And you can speak of this in more general terms. If I wanted to just take the expression a plus b and square it, that's just a plus b times a plus b, and we do exactly what we did just here, but here I'm just doing it in very general terms with a or b, and you can think of a as either a constant number or even a variable, and so this is going to be, if we distribute this, it's going to be a plus b times that a, plus a plus b times that b. And so this is going to be a squared, now I'm just doing the distributive property again, a squared plus a b plus a b plus b squared. So it's a squared plus two a b plus b b squared. So this is going to be the general form. So if a is the variable, which was x or a in this case, then it's just going to be whatever squared in the constant term, it's going to be two times that times the variable."}, {"video_title": "Perfect square factorization intro   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "And so this is going to be a squared, now I'm just doing the distributive property again, a squared plus a b plus a b plus b squared. So it's a squared plus two a b plus b b squared. So this is going to be the general form. So if a is the variable, which was x or a in this case, then it's just going to be whatever squared in the constant term, it's going to be two times that times the variable. And I want to show that there's some variation that you can entertain here. So if you were to see 25 plus 10x plus x squared, and someone said, hey, why don't you factor that? You could say, look, this right here is a perfect square, it's five squared, I have the variable squared right over here, and then this coefficient on our first degree term is two times five."}, {"video_title": "Perfect square factorization intro   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "So if a is the variable, which was x or a in this case, then it's just going to be whatever squared in the constant term, it's going to be two times that times the variable. And I want to show that there's some variation that you can entertain here. So if you were to see 25 plus 10x plus x squared, and someone said, hey, why don't you factor that? You could say, look, this right here is a perfect square, it's five squared, I have the variable squared right over here, and then this coefficient on our first degree term is two times five. And so you might immediately recognize this as five plus x squared. Now of course you could just rewrite this polynomial as x squared plus 10x plus 25, in which case you might say, okay, variable squared, some number squared, five squared, two times that number as a coefficient here, so that's going to be x plus five squared. And that's good, because these two things are absolutely equivalent."}, {"video_title": "Multi-step inequalities 3   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So what we really want to do is isolate the p on one side of this inequality, and preferably the left. That just makes it a little bit easier to read. It doesn't have to be, but we just want to isolate the p. So a good step to that is to get rid of this p on the right-hand side. And the best way I can think of doing that is subtracting p from the right. But of course, if we want to make sure that this inequality is always going to be true, if we do anything to the right, we also have to do that to the left. So we also have to subtract p from the left. And so the left-hand side, negative 3p minus p, that's negative 4p, and then we still have a minus 7 up here, is going to be less than p minus p. Those cancel out."}, {"video_title": "Multi-step inequalities 3   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "And the best way I can think of doing that is subtracting p from the right. But of course, if we want to make sure that this inequality is always going to be true, if we do anything to the right, we also have to do that to the left. So we also have to subtract p from the left. And so the left-hand side, negative 3p minus p, that's negative 4p, and then we still have a minus 7 up here, is going to be less than p minus p. Those cancel out. It is less than 9. Now, the next thing I'm in the mood to do is get rid of this negative 7 or this minus 7 here so that we can better isolate the p on the left-hand side. So the best way I can think of to get rid of a negative 7 is to add 7 to it."}, {"video_title": "Multi-step inequalities 3   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "And so the left-hand side, negative 3p minus p, that's negative 4p, and then we still have a minus 7 up here, is going to be less than p minus p. Those cancel out. It is less than 9. Now, the next thing I'm in the mood to do is get rid of this negative 7 or this minus 7 here so that we can better isolate the p on the left-hand side. So the best way I can think of to get rid of a negative 7 is to add 7 to it. Then it'll just cancel out to 0. So let's add 7 to both sides of this inequality. Negative 7 plus 7 cancel out."}, {"video_title": "Multi-step inequalities 3   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So the best way I can think of to get rid of a negative 7 is to add 7 to it. Then it'll just cancel out to 0. So let's add 7 to both sides of this inequality. Negative 7 plus 7 cancel out. All we're left with is negative 4p. On the right-hand side, we have 9 plus 7 is 16. And it's still less than."}, {"video_title": "Multi-step inequalities 3   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Negative 7 plus 7 cancel out. All we're left with is negative 4p. On the right-hand side, we have 9 plus 7 is 16. And it's still less than. Now, the last step to isolate the p is to get rid of this negative 4 coefficient. And the easiest way I can think of to get rid of this negative 4 coefficient is to divide both sides by negative 4. So if we divide this side by negative 4, these guys are going to cancel out."}, {"video_title": "Multi-step inequalities 3   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "And it's still less than. Now, the last step to isolate the p is to get rid of this negative 4 coefficient. And the easiest way I can think of to get rid of this negative 4 coefficient is to divide both sides by negative 4. So if we divide this side by negative 4, these guys are going to cancel out. We're just going to be left with p. We also have to do it to the right-hand side. Now, there's one thing that you really have to remember since this is an inequality. This is not an equation."}, {"video_title": "Multi-step inequalities 3   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So if we divide this side by negative 4, these guys are going to cancel out. We're just going to be left with p. We also have to do it to the right-hand side. Now, there's one thing that you really have to remember since this is an inequality. This is not an equation. If you're dealing with an inequality, and if you multiply or divide both sides of the equation by a negative number, you have to swap the inequality. So in this case, the less than becomes greater than since we're dividing by a negative number. And so negative 4 divided by negative 4, those cancel out."}, {"video_title": "Multi-step inequalities 3   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "This is not an equation. If you're dealing with an inequality, and if you multiply or divide both sides of the equation by a negative number, you have to swap the inequality. So in this case, the less than becomes greater than since we're dividing by a negative number. And so negative 4 divided by negative 4, those cancel out. We have p is greater than 16 divided by negative 4, which is negative 4. And we can plot this solution set right over here. And then we can try out some values to help us feel good about the idea of it working."}, {"video_title": "Multi-step inequalities 3   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "And so negative 4 divided by negative 4, those cancel out. We have p is greater than 16 divided by negative 4, which is negative 4. And we can plot this solution set right over here. And then we can try out some values to help us feel good about the idea of it working. So let's say this is negative 5, negative 4, negative 3, negative 2, 1, negative 1, I should say, 0. Let me write that a little bit neater. Negative 1, 0."}, {"video_title": "Multi-step inequalities 3   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "And then we can try out some values to help us feel good about the idea of it working. So let's say this is negative 5, negative 4, negative 3, negative 2, 1, negative 1, I should say, 0. Let me write that a little bit neater. Negative 1, 0. And then we can keep going to the right. And so our solution is p is not greater than or equal, so we have to exclude negative 4. p is greater than negative 4, so all the values above that. So negative 3.99999999 will work."}, {"video_title": "Multi-step inequalities 3   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Negative 1, 0. And then we can keep going to the right. And so our solution is p is not greater than or equal, so we have to exclude negative 4. p is greater than negative 4, so all the values above that. So negative 3.99999999 will work. Negative 4 will not work. And let's just try some values out to feel good that this is really the solution set. So first let's try out when p is equal to negative 3."}, {"video_title": "Multi-step inequalities 3   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So negative 3.99999999 will work. Negative 4 will not work. And let's just try some values out to feel good that this is really the solution set. So first let's try out when p is equal to negative 3. This should work. The way I've drawn it, this is in our solution set. p equals negative 3 is greater than negative 4."}, {"video_title": "Multi-step inequalities 3   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So first let's try out when p is equal to negative 3. This should work. The way I've drawn it, this is in our solution set. p equals negative 3 is greater than negative 4. So let's try that out. We have negative 3 times negative 3. The first negative 3 is this one, and then we're saying p is negative 3."}, {"video_title": "Multi-step inequalities 3   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "p equals negative 3 is greater than negative 4. So let's try that out. We have negative 3 times negative 3. The first negative 3 is this one, and then we're saying p is negative 3. Minus 7 should be less than, instead of a p, we're going to put a negative 3, should be less than negative 3 plus 9. Negative 3 times negative 3 is 9. Minus 7 should be less than negative 3 plus 9 is 6."}, {"video_title": "Multi-step inequalities 3   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "The first negative 3 is this one, and then we're saying p is negative 3. Minus 7 should be less than, instead of a p, we're going to put a negative 3, should be less than negative 3 plus 9. Negative 3 times negative 3 is 9. Minus 7 should be less than negative 3 plus 9 is 6. 9 minus 7 is 2. 2 should be less than 6, of which, of course, it is. Now let's try a value that definitely should not work."}, {"video_title": "Multi-step inequalities 3   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Minus 7 should be less than negative 3 plus 9 is 6. 9 minus 7 is 2. 2 should be less than 6, of which, of course, it is. Now let's try a value that definitely should not work. So let's try negative 5. Negative 5 is not in our solution set, so it should not work. So we have negative 3 times negative 5 minus 7."}, {"video_title": "Multi-step inequalities 3   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Now let's try a value that definitely should not work. So let's try negative 5. Negative 5 is not in our solution set, so it should not work. So we have negative 3 times negative 5 minus 7. Let's see whether it is less than negative 5 plus 9. Negative 3 times negative 5 is 15. Minus 7, it really should not be less than negative 5 plus 9."}, {"video_title": "Multi-step inequalities 3   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So we have negative 3 times negative 5 minus 7. Let's see whether it is less than negative 5 plus 9. Negative 3 times negative 5 is 15. Minus 7, it really should not be less than negative 5 plus 9. So we're just seeing if p equals negative 5 works. 15 minus 7 is 8. And so we get 8 is less than 4, which is definitely not the case."}, {"video_title": "Multi-step inequalities 3   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Minus 7, it really should not be less than negative 5 plus 9. So we're just seeing if p equals negative 5 works. 15 minus 7 is 8. And so we get 8 is less than 4, which is definitely not the case. So p equals negative 5 doesn't work, and it shouldn't work because that's not in our solution set. And now if we want to feel really good about it, we can actually try this boundary point. Negative 4 should not work, but it should satisfy the related equation."}, {"video_title": "Multi-step inequalities 3   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "And so we get 8 is less than 4, which is definitely not the case. So p equals negative 5 doesn't work, and it shouldn't work because that's not in our solution set. And now if we want to feel really good about it, we can actually try this boundary point. Negative 4 should not work, but it should satisfy the related equation. When I talk about the related equation, negative 4 should satisfy negative 3 minus 7 is equal to p plus 9. It'll satisfy this, but it won't satisfy this because when we get the same value on both sides, the same value is not less than the same value. So let's try it out."}, {"video_title": "Multi-step inequalities 3   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Negative 4 should not work, but it should satisfy the related equation. When I talk about the related equation, negative 4 should satisfy negative 3 minus 7 is equal to p plus 9. It'll satisfy this, but it won't satisfy this because when we get the same value on both sides, the same value is not less than the same value. So let's try it out. Let's see whether negative 4 at least satisfies the related equation. So if we get negative 3 times negative 4 minus 7, this should be equal to negative 4 plus 9. So this is 12 minus 7 should be equal to negative 4 plus 9 should be equal to 5."}, {"video_title": "Multi-step inequalities 3   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So let's try it out. Let's see whether negative 4 at least satisfies the related equation. So if we get negative 3 times negative 4 minus 7, this should be equal to negative 4 plus 9. So this is 12 minus 7 should be equal to negative 4 plus 9 should be equal to 5. And this, of course, is true. 5 is equal to 5. So it satisfies the related equation, but it should not satisfy this."}, {"video_title": "Multi-step inequalities 3   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So this is 12 minus 7 should be equal to negative 4 plus 9 should be equal to 5. And this, of course, is true. 5 is equal to 5. So it satisfies the related equation, but it should not satisfy this. If you put negative 4 for p here, and I encourage you to do so, actually we could do it over here instead of an equal sign. If you put it into the original inequality, let me delete all of that, it really just becomes this. The original inequality is this right over here."}, {"video_title": "Multi-step inequalities 3   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So it satisfies the related equation, but it should not satisfy this. If you put negative 4 for p here, and I encourage you to do so, actually we could do it over here instead of an equal sign. If you put it into the original inequality, let me delete all of that, it really just becomes this. The original inequality is this right over here. If you put negative 4, you have less than, less than, and then you get 5 is less than 5, which is not the case. And that's good because we did not include that in the solution set. We put an open circle."}, {"video_title": "Multi-step inequalities 3   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "The original inequality is this right over here. If you put negative 4, you have less than, less than, and then you get 5 is less than 5, which is not the case. And that's good because we did not include that in the solution set. We put an open circle. If negative 4 was included, we would fill that in. But the only reason I would include negative 4 is if this was greater than or equal. So it's good that this does not work because negative 4 is not part of our solution set."}, {"video_title": "Reflecting and scaling absolute value function.mp3", "Sentence": "What is the equation of the new graph? So pause the video and see if you can figure that out. All right, let's work through it together now. Now you might not need to draw it visually, but I will just so that we can all together visualize what is going on. So let's say that's my x-axis and that is my y-axis. My y-axis. Y equals the absolute value of x."}, {"video_title": "Reflecting and scaling absolute value function.mp3", "Sentence": "Now you might not need to draw it visually, but I will just so that we can all together visualize what is going on. So let's say that's my x-axis and that is my y-axis. My y-axis. Y equals the absolute value of x. So for non-negative values of x, y is going to be equal to x. Absolute value of zero is zero. Absolute value of one is one."}, {"video_title": "Reflecting and scaling absolute value function.mp3", "Sentence": "Y equals the absolute value of x. So for non-negative values of x, y is going to be equal to x. Absolute value of zero is zero. Absolute value of one is one. Absolute value of two is two. So it's gonna look like this. It's just gonna have a slope of one."}, {"video_title": "Reflecting and scaling absolute value function.mp3", "Sentence": "Absolute value of one is one. Absolute value of two is two. So it's gonna look like this. It's just gonna have a slope of one. And then for negative values, when you take the absolute value, you're gonna take the opposite. You're gonna get the positive. So it's gonna look like this."}, {"video_title": "Reflecting and scaling absolute value function.mp3", "Sentence": "It's just gonna have a slope of one. And then for negative values, when you take the absolute value, you're gonna take the opposite. You're gonna get the positive. So it's gonna look like this. Let me see if I can draw that a little bit cleaner. This is a hand-drawn sketch, so bear with me. But hopefully this is familiar."}, {"video_title": "Reflecting and scaling absolute value function.mp3", "Sentence": "So it's gonna look like this. Let me see if I can draw that a little bit cleaner. This is a hand-drawn sketch, so bear with me. But hopefully this is familiar. You've seen the graph of y is equal to absolute value of x before. Now let's think about the different transformations. So first they say, is reflected across the x-axis."}, {"video_title": "Reflecting and scaling absolute value function.mp3", "Sentence": "But hopefully this is familiar. You've seen the graph of y is equal to absolute value of x before. Now let's think about the different transformations. So first they say, is reflected across the x-axis. So for example, if I have some x value right over here, before I would take the absolute value of x and I would end up there. But now we wanna reflect across the x-axis. So we wanna essentially get the negative of that value associated with that corresponding x."}, {"video_title": "Reflecting and scaling absolute value function.mp3", "Sentence": "So first they say, is reflected across the x-axis. So for example, if I have some x value right over here, before I would take the absolute value of x and I would end up there. But now we wanna reflect across the x-axis. So we wanna essentially get the negative of that value associated with that corresponding x. And so for example, this x, before we would get the absolute value of x, but now we wanna flip across the x-axis and we wanna get the negative of it. So in general, what we are doing is we are getting the negative of the absolute value of x. In general, if you're flipping over the x-axis, you're getting the negative."}, {"video_title": "Reflecting and scaling absolute value function.mp3", "Sentence": "So we wanna essentially get the negative of that value associated with that corresponding x. And so for example, this x, before we would get the absolute value of x, but now we wanna flip across the x-axis and we wanna get the negative of it. So in general, what we are doing is we are getting the negative of the absolute value of x. In general, if you're flipping over the x-axis, you're getting the negative. You're scaling the expression or the function by a negative. So this is going to be y is equal to the negative of the absolute value of x. Once again, whatever absolute value of x was giving you before for a given x, we now wanna get the negative of it."}, {"video_title": "Reflecting and scaling absolute value function.mp3", "Sentence": "In general, if you're flipping over the x-axis, you're getting the negative. You're scaling the expression or the function by a negative. So this is going to be y is equal to the negative of the absolute value of x. Once again, whatever absolute value of x was giving you before for a given x, we now wanna get the negative of it. We now wanna get the negative of it. So that's what reflecting across the x-axis does for us. But then they say, scaled vertically by a factor of seven."}, {"video_title": "Reflecting and scaling absolute value function.mp3", "Sentence": "Once again, whatever absolute value of x was giving you before for a given x, we now wanna get the negative of it. We now wanna get the negative of it. So that's what reflecting across the x-axis does for us. But then they say, scaled vertically by a factor of seven. By a factor of seven. And the way I view that is if you're scaling it vertically by a factor of seven, whatever y value you got for a given x, you now wanna get seven times the y value. Seven times the y value for a given x."}, {"video_title": "Reflecting and scaling absolute value function.mp3", "Sentence": "But then they say, scaled vertically by a factor of seven. By a factor of seven. And the way I view that is if you're scaling it vertically by a factor of seven, whatever y value you got for a given x, you now wanna get seven times the y value. Seven times the y value for a given x. And so if you think about that algebraically, well, if I want seven times the y value, I have to multiply this thing by seven. So I would get y is equal to negative seven times the absolute value of x. And that's essentially what they're asking."}, {"video_title": "Reflecting and scaling absolute value function.mp3", "Sentence": "Seven times the y value for a given x. And so if you think about that algebraically, well, if I want seven times the y value, I have to multiply this thing by seven. So I would get y is equal to negative seven times the absolute value of x. And that's essentially what they're asking. What is the equation of the new graph? And so that's what it would be. The negative flips us over the x-axis and then the seven scales vertically by a factor of seven."}, {"video_title": "Reflecting and scaling absolute value function.mp3", "Sentence": "And that's essentially what they're asking. What is the equation of the new graph? And so that's what it would be. The negative flips us over the x-axis and then the seven scales vertically by a factor of seven. But just to understand what this would look like, well, you multiply zero times seven. It doesn't change anything. But whatever x this is, this was equal to negative x, but now we're gonna get to negative seven x."}, {"video_title": "Reflecting and scaling absolute value function.mp3", "Sentence": "The negative flips us over the x-axis and then the seven scales vertically by a factor of seven. But just to understand what this would look like, well, you multiply zero times seven. It doesn't change anything. But whatever x this is, this was equal to negative x, but now we're gonna get to negative seven x. So let's see, two, three, four, five, six, seven. So we'd put something around that. So our graph is now going to look like this."}, {"video_title": "Reflecting and scaling absolute value function.mp3", "Sentence": "But whatever x this is, this was equal to negative x, but now we're gonna get to negative seven x. So let's see, two, three, four, five, six, seven. So we'd put something around that. So our graph is now going to look like this. It's going to be stretched along the vertical axis. If we were scaling vertically by something that had an absolute value less than one, then it would make the graph less tall. It would make it look wider."}, {"video_title": "Reflecting and scaling absolute value function.mp3", "Sentence": "So our graph is now going to look like this. It's going to be stretched along the vertical axis. If we were scaling vertically by something that had an absolute value less than one, then it would make the graph less tall. It would make it look wider. Let me make it at least look a little bit more symmetric. So it's gonna look something like that. But the key issue, and the reason why I'm drawing it so you can see that it looks like it's being scaled vertically."}, {"video_title": "Reflecting and scaling absolute value function.mp3", "Sentence": "It would make it look wider. Let me make it at least look a little bit more symmetric. So it's gonna look something like that. But the key issue, and the reason why I'm drawing it so you can see that it looks like it's being scaled vertically. It's being stretched in the vertical direction by a factor of seven. And the way we do that algebraically is we multiply by seven. The negative here is what flipped us over the x-axis."}, {"video_title": "Multiplying binomials area model   Mathematics II   High School Math   Khan Academy (2).mp3", "Sentence": "The first way I want to express it as the product of two binomials, and then I want to express it as a trinomial. So let's think about this a little bit. So one way to say, well look, the height of this larger rectangle from here to here, we see that that distance is x, and then from here to here it's two. So the entire height right over here, the entire height right over here is going to be x plus two. So the height is x plus two, and what's the width? Well, the width is, we go from there to there is x, and then from there to there is three. So the entire width is x plus three, x plus three."}, {"video_title": "Multiplying binomials area model   Mathematics II   High School Math   Khan Academy (2).mp3", "Sentence": "So the entire height right over here, the entire height right over here is going to be x plus two. So the height is x plus two, and what's the width? Well, the width is, we go from there to there is x, and then from there to there is three. So the entire width is x plus three, x plus three. So just like that, I've expressed the area of the entire rectangle, and it's the product of two binomials. But now let's express it as a trinomial. Well, to do that, we can break down the larger area into the areas of each of these smaller rectangles."}, {"video_title": "Multiplying binomials area model   Mathematics II   High School Math   Khan Academy (2).mp3", "Sentence": "So the entire width is x plus three, x plus three. So just like that, I've expressed the area of the entire rectangle, and it's the product of two binomials. But now let's express it as a trinomial. Well, to do that, we can break down the larger area into the areas of each of these smaller rectangles. So what's the area of this purple rectangle right over here? Well, the purple rectangle, its height is x, and its width is x, so its area is x squared. Let me write that, that's x squared."}, {"video_title": "Multiplying binomials area model   Mathematics II   High School Math   Khan Academy (2).mp3", "Sentence": "Well, to do that, we can break down the larger area into the areas of each of these smaller rectangles. So what's the area of this purple rectangle right over here? Well, the purple rectangle, its height is x, and its width is x, so its area is x squared. Let me write that, that's x squared. What's the area of this yellow rectangle? Well, its height is x, same height as right over here. Its height is x, and its width is three."}, {"video_title": "Multiplying binomials area model   Mathematics II   High School Math   Khan Academy (2).mp3", "Sentence": "Let me write that, that's x squared. What's the area of this yellow rectangle? Well, its height is x, same height as right over here. Its height is x, and its width is three. So it's gonna be x times three, or three x. It'll have an area of three x. So that area is three x."}, {"video_title": "Multiplying binomials area model   Mathematics II   High School Math   Khan Academy (2).mp3", "Sentence": "Its height is x, and its width is three. So it's gonna be x times three, or three x. It'll have an area of three x. So that area is three x. If we're summing up the area of the entire thing, this would be plus three x. So this expression right over here, that's the area of this purple region plus the area of this yellow region. And then we can move on to this green region."}, {"video_title": "Multiplying binomials area model   Mathematics II   High School Math   Khan Academy (2).mp3", "Sentence": "So that area is three x. If we're summing up the area of the entire thing, this would be plus three x. So this expression right over here, that's the area of this purple region plus the area of this yellow region. And then we can move on to this green region. What's the area going to be here? Well, the height is two, and the width is x. So multiplying height times width is gonna be two times x."}, {"video_title": "Multiplying binomials area model   Mathematics II   High School Math   Khan Academy (2).mp3", "Sentence": "And then we can move on to this green region. What's the area going to be here? Well, the height is two, and the width is x. So multiplying height times width is gonna be two times x. And we can just add that, plus two times x. And then finally, this little gray box here, its height is two, we see that right over there. Its height is two, and its width is three."}, {"video_title": "Multiplying binomials area model   Mathematics II   High School Math   Khan Academy (2).mp3", "Sentence": "So multiplying height times width is gonna be two times x. And we can just add that, plus two times x. And then finally, this little gray box here, its height is two, we see that right over there. Its height is two, and its width is three. We see it right over there. So it has an area of six, two times three. So plus six, and you might say, well, this isn't a trinomial."}, {"video_title": "Multiplying binomials area model   Mathematics II   High School Math   Khan Academy (2).mp3", "Sentence": "Its height is two, and its width is three. We see it right over there. So it has an area of six, two times three. So plus six, and you might say, well, this isn't a trinomial. This has four terms right over here. But you might notice that we can add, that we can add these middle two terms. Three x plus two x. I have three x's, and I add two x's to that, I'm gonna have five x's."}, {"video_title": "Multiplying binomials area model   Mathematics II   High School Math   Khan Academy (2).mp3", "Sentence": "So plus six, and you might say, well, this isn't a trinomial. This has four terms right over here. But you might notice that we can add, that we can add these middle two terms. Three x plus two x. I have three x's, and I add two x's to that, I'm gonna have five x's. So this entire thing simplifies to x squared, x squared, plus five x, plus six, plus six. So this and this are two ways of expressing the area, so they're going to be equal. And that makes sense, because if you multiply it out, these binomials, and simplify it, you would get this trinomial."}, {"video_title": "Multiplying binomials area model   Mathematics II   High School Math   Khan Academy (2).mp3", "Sentence": "Three x plus two x. I have three x's, and I add two x's to that, I'm gonna have five x's. So this entire thing simplifies to x squared, x squared, plus five x, plus six, plus six. So this and this are two ways of expressing the area, so they're going to be equal. And that makes sense, because if you multiply it out, these binomials, and simplify it, you would get this trinomial. We could do that really fast. You multiply the x times the x. Actually, let me do it in the same colors."}, {"video_title": "Multiplying binomials area model   Mathematics II   High School Math   Khan Academy (2).mp3", "Sentence": "And that makes sense, because if you multiply it out, these binomials, and simplify it, you would get this trinomial. We could do that really fast. You multiply the x times the x. Actually, let me do it in the same colors. You multiply, you multiply the x times the x, you get the x squared. You multiply this x times the three, you get your three x. You multiply, you multiply the two times the x, you get your two x."}, {"video_title": "Multiplying binomials area model   Mathematics II   High School Math   Khan Academy (2).mp3", "Sentence": "Actually, let me do it in the same colors. You multiply, you multiply the x times the x, you get the x squared. You multiply this x times the three, you get your three x. You multiply, you multiply the two times the x, you get your two x. And then you multiply the two times the three, and you get your six. So what this, as you can see, this area model does for us is it hopefully makes a visual representation of why it makes sense to multiply binomials the way we do. And in other videos, we talk about it as applying the distribution property twice, but this gives you a more visual representation for why it actually makes sense."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "So let's just do a bunch of these problems. So here they tell us that a line has a slope of negative 5, so m is equal to negative 5, and it has a y-intercept of 6. So b is equal to 6, so this is pretty straightforward. The equation of this line is y is equal to negative 5x plus 6. That wasn't too bad. Let's do this next one over here. The line has a slope of negative 1 and contains the point 4 fifths comma 0."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "The equation of this line is y is equal to negative 5x plus 6. That wasn't too bad. Let's do this next one over here. The line has a slope of negative 1 and contains the point 4 fifths comma 0. So they're telling us the slope, slope of negative 1. So we know that m is equal to negative 1. But we're not 100% sure about where the y-intercept is just yet."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "The line has a slope of negative 1 and contains the point 4 fifths comma 0. So they're telling us the slope, slope of negative 1. So we know that m is equal to negative 1. But we're not 100% sure about where the y-intercept is just yet. So we know that this equation is going to be of the form y is equal to the slope, negative 1x plus b, where b is the y-intercept. Now, we can use this coordinate information, the fact that it contains this point, we can use that information to solve for b. The fact that the line contains this point means that the value x is equal to 4 fifths, y is equal to 0, must satisfy this equation."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "But we're not 100% sure about where the y-intercept is just yet. So we know that this equation is going to be of the form y is equal to the slope, negative 1x plus b, where b is the y-intercept. Now, we can use this coordinate information, the fact that it contains this point, we can use that information to solve for b. The fact that the line contains this point means that the value x is equal to 4 fifths, y is equal to 0, must satisfy this equation. So let's substitute those in. y is equal to 0 when x is equal to 4 fifths, so 0 is equal to negative 1 times 4 fifths plus b. Let me scroll down a little bit."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "The fact that the line contains this point means that the value x is equal to 4 fifths, y is equal to 0, must satisfy this equation. So let's substitute those in. y is equal to 0 when x is equal to 4 fifths, so 0 is equal to negative 1 times 4 fifths plus b. Let me scroll down a little bit. So let's see, we get 0 is equal to negative 4 fifths plus b. We can add 4 fifths to both sides of this equation. So we could add a 4 fifths there."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "Let me scroll down a little bit. So let's see, we get 0 is equal to negative 4 fifths plus b. We can add 4 fifths to both sides of this equation. So we could add a 4 fifths there. We could add a 4 fifths to that side as well. The whole reason I did that is so that cancels out with that, and you get b is equal to 4 fifths. So we now have the equation of the line."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "So we could add a 4 fifths there. We could add a 4 fifths to that side as well. The whole reason I did that is so that cancels out with that, and you get b is equal to 4 fifths. So we now have the equation of the line. y is equal to negative 1 times x, which we write as negative x, plus b, which is 4 fifths. Just like that. Now we have this one."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "So we now have the equation of the line. y is equal to negative 1 times x, which we write as negative x, plus b, which is 4 fifths. Just like that. Now we have this one. The line contains the points 2, 6 and 5, 0. So they haven't given us the slope or the y-intercept explicitly, but we can figure out both of them from these coordinates. The first thing we can do is figure out the slope."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "Now we have this one. The line contains the points 2, 6 and 5, 0. So they haven't given us the slope or the y-intercept explicitly, but we can figure out both of them from these coordinates. The first thing we can do is figure out the slope. So we know that the slope m is equal to change in y over change in x, which is equal to, what is the change in y? Let's start with this one right here. 6 minus 0."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "The first thing we can do is figure out the slope. So we know that the slope m is equal to change in y over change in x, which is equal to, what is the change in y? Let's start with this one right here. 6 minus 0. Let me do it this way. I want to make it color-coded. Minus 0."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "6 minus 0. Let me do it this way. I want to make it color-coded. Minus 0. So 6 minus 0. That's our change in y. And our change in x is 2 minus 5."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "Minus 0. So 6 minus 0. That's our change in y. And our change in x is 2 minus 5. And the reason why I color-coded it is I wanted to show you, when I use this y term first, I use the 6 up here, that I have to use this x term first as well. So I wanted to show you, this is the coordinate 2, 6. This is the coordinate 5, 0."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "And our change in x is 2 minus 5. And the reason why I color-coded it is I wanted to show you, when I use this y term first, I use the 6 up here, that I have to use this x term first as well. So I wanted to show you, this is the coordinate 2, 6. This is the coordinate 5, 0. I couldn't have swapped the 2 and the 5, then I would have gotten the negative of the answer. But what do we get here? This is equal to 6 minus 0 is 6."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "This is the coordinate 5, 0. I couldn't have swapped the 2 and the 5, then I would have gotten the negative of the answer. But what do we get here? This is equal to 6 minus 0 is 6. 2 minus 5 is negative 3. So this becomes negative 6 over 3, which is the same thing as negative 2. So that's our slope."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "This is equal to 6 minus 0 is 6. 2 minus 5 is negative 3. So this becomes negative 6 over 3, which is the same thing as negative 2. So that's our slope. So, so far, we know that the line must be y is equal to the slope, we'll do that in orange, negative 2 times x plus our y-intercept. Now we can do exactly what we did in the last problem. We can use one of these points to solve for b."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "So that's our slope. So, so far, we know that the line must be y is equal to the slope, we'll do that in orange, negative 2 times x plus our y-intercept. Now we can do exactly what we did in the last problem. We can use one of these points to solve for b. We can use either one. Both of these are on the line, so both of these must satisfy this equation. I'll use the 5, 0 because it's always nice when you have a 0 there, the math is a little bit easier."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "We can use one of these points to solve for b. We can use either one. Both of these are on the line, so both of these must satisfy this equation. I'll use the 5, 0 because it's always nice when you have a 0 there, the math is a little bit easier. So let's put the 5, 0 there. So y is equal to 0 when x is equal to 5. So y is equal to 0 when you have negative 2 times 5, when x is equal to 5, plus b."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "I'll use the 5, 0 because it's always nice when you have a 0 there, the math is a little bit easier. So let's put the 5, 0 there. So y is equal to 0 when x is equal to 5. So y is equal to 0 when you have negative 2 times 5, when x is equal to 5, plus b. And so you get 0 is equal to negative 10 plus b. If you add 10 to both sides of this equation, let's add 10 to both sides, these two cancel out, and you get b is equal to 10 plus 0, or 10. So you get b is equal to 10."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "So y is equal to 0 when you have negative 2 times 5, when x is equal to 5, plus b. And so you get 0 is equal to negative 10 plus b. If you add 10 to both sides of this equation, let's add 10 to both sides, these two cancel out, and you get b is equal to 10 plus 0, or 10. So you get b is equal to 10. And now we know the equation for the line. The equation is y, let me do it in a new color, y is equal to negative 2x plus b, plus 10. And we are done."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "So you get b is equal to 10. And now we know the equation for the line. The equation is y, let me do it in a new color, y is equal to negative 2x plus b, plus 10. And we are done. Let's do another one of these. Let's do another one of these. Alright, the line contains the points 3, 5 and negative 3, 0."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "And we are done. Let's do another one of these. Let's do another one of these. Alright, the line contains the points 3, 5 and negative 3, 0. Just like the last problem, we start by figuring out the slope. The slope, which we will call m, is the same thing as the rise over the run, which is the same thing as the change in y over the change in x. If you were doing this for your homework, you wouldn't have to write all this."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "Alright, the line contains the points 3, 5 and negative 3, 0. Just like the last problem, we start by figuring out the slope. The slope, which we will call m, is the same thing as the rise over the run, which is the same thing as the change in y over the change in x. If you were doing this for your homework, you wouldn't have to write all this. I just want to make sure that you understand that these are all the same things. And then what is our change in y over change in x? This is equal to, let's start with this side first, just to show you I could have picked either of these points."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "If you were doing this for your homework, you wouldn't have to write all this. I just want to make sure that you understand that these are all the same things. And then what is our change in y over change in x? This is equal to, let's start with this side first, just to show you I could have picked either of these points. So let's say it's 0, 0, minus 5, minus 5, just like that. So I'm using this coordinate first. I'm kind of viewing it as the end point."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "This is equal to, let's start with this side first, just to show you I could have picked either of these points. So let's say it's 0, 0, minus 5, minus 5, just like that. So I'm using this coordinate first. I'm kind of viewing it as the end point. 0, minus 5. And remember, when I first learned this, I would always be tempted to do the x in the numerator. No, you use the y's in the numerator."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "I'm kind of viewing it as the end point. 0, minus 5. And remember, when I first learned this, I would always be tempted to do the x in the numerator. No, you use the y's in the numerator. So that's the second of the coordinates. And then that is going to be over negative 3, negative 3, minus 3. This is the coordinate negative 3, 0."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "No, you use the y's in the numerator. So that's the second of the coordinates. And then that is going to be over negative 3, negative 3, minus 3. This is the coordinate negative 3, 0. This is the coordinate 3, 5. We're subtracting that. So what are we going to get?"}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "This is the coordinate negative 3, 0. This is the coordinate 3, 5. We're subtracting that. So what are we going to get? This is going to be equal to, I'll do it in a neutral color, this is going to be equal to, the numerator is negative 5, over negative 3, minus 3, is negative 6. So the negatives cancel out. You get 5 over 6."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "So what are we going to get? This is going to be equal to, I'll do it in a neutral color, this is going to be equal to, the numerator is negative 5, over negative 3, minus 3, is negative 6. So the negatives cancel out. You get 5 over 6. So we know that the equation is going to be of the form y is equal to 5 6ths x, plus b. And now we can substitute one of these coordinates in to solve for b. So let's do, I always like to use the one that has a 0 in it."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "You get 5 over 6. So we know that the equation is going to be of the form y is equal to 5 6ths x, plus b. And now we can substitute one of these coordinates in to solve for b. So let's do, I always like to use the one that has a 0 in it. So y is 0 when x is negative 3, when x is negative 3, plus b. So all I did is I substituted negative 3 for x, 0 for y, and I know I can do that because this is on the line. This must satisfy the equation of the line."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "So let's do, I always like to use the one that has a 0 in it. So y is 0 when x is negative 3, when x is negative 3, plus b. So all I did is I substituted negative 3 for x, 0 for y, and I know I can do that because this is on the line. This must satisfy the equation of the line. And let's solve for b. So we get 0 is equal to, well if we divide negative 3 by 3, that becomes a 1. If you divide 6 by 3, that becomes a 2."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "This must satisfy the equation of the line. And let's solve for b. So we get 0 is equal to, well if we divide negative 3 by 3, that becomes a 1. If you divide 6 by 3, that becomes a 2. So it becomes negative 5, negative 5 halves, plus b. We can add 5 halves to both sides of the equation. Plus 5 halves, plus 5 halves."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "If you divide 6 by 3, that becomes a 2. So it becomes negative 5, negative 5 halves, plus b. We can add 5 halves to both sides of the equation. Plus 5 halves, plus 5 halves. I like to change my notation just so you get familiar with both. And so the equation becomes 5 halves is equal to, that's a 0, is equal to b. b is 5 halves. So the equation of our line is, y is equal to 5 6x, plus b, which we just figured out is 5 halves."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "Plus 5 halves, plus 5 halves. I like to change my notation just so you get familiar with both. And so the equation becomes 5 halves is equal to, that's a 0, is equal to b. b is 5 halves. So the equation of our line is, y is equal to 5 6x, plus b, which we just figured out is 5 halves. Plus 5 halves. And we are done. Let's do another one."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "So the equation of our line is, y is equal to 5 6x, plus b, which we just figured out is 5 halves. Plus 5 halves. And we are done. Let's do another one. We have a graph here. Let's figure out the equation of this graph. This is actually on some level a little bit easier."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "Let's do another one. We have a graph here. Let's figure out the equation of this graph. This is actually on some level a little bit easier. What's the slope? Slope is change in y over change in x. Let's see what happens."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "This is actually on some level a little bit easier. What's the slope? Slope is change in y over change in x. Let's see what happens. When we move in x, when our change in x is 1, that is our change in x. So change in x is 1. I'm just deciding to change my x by 1."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "Let's see what happens. When we move in x, when our change in x is 1, that is our change in x. So change in x is 1. I'm just deciding to change my x by 1. Increment by 1. What is the change in y? It looks like y changes exactly by 4."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "I'm just deciding to change my x by 1. Increment by 1. What is the change in y? It looks like y changes exactly by 4. It looks like my delta y, my change in y, is equal to 4 when my delta x is equal to 1. So change in y over change in x. Change in y is 4 when change in x is 1."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "It looks like y changes exactly by 4. It looks like my delta y, my change in y, is equal to 4 when my delta x is equal to 1. So change in y over change in x. Change in y is 4 when change in x is 1. So the slope is equal to 4. And now, what's its y-intercept? Here we can just look at the graph."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "Change in y is 4 when change in x is 1. So the slope is equal to 4. And now, what's its y-intercept? Here we can just look at the graph. It looks like it intersects the y-axis at y is equal to negative 6, or at the point 0, negative 6. So we know that b is equal to negative 6. b is equal to negative 6. So we know the equation of the line."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "Here we can just look at the graph. It looks like it intersects the y-axis at y is equal to negative 6, or at the point 0, negative 6. So we know that b is equal to negative 6. b is equal to negative 6. So we know the equation of the line. The equation of the line is y is equal to the slope times x plus the y-intercept. Actually, I should write that. Or minus 6."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "So we know the equation of the line. The equation of the line is y is equal to the slope times x plus the y-intercept. Actually, I should write that. Or minus 6. That is plus negative 6. So that is the equation of our line. Let's do one more of these."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "Or minus 6. That is plus negative 6. So that is the equation of our line. Let's do one more of these. So they tell us that f of 1.5 is negative 3. f of negative 1 is 2. What is that? All this is telling you, this is just a fancy way of telling you that the point when x is 1.5, when you put 1.5 into the function, the function evaluates as negative 3."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "Let's do one more of these. So they tell us that f of 1.5 is negative 3. f of negative 1 is 2. What is that? All this is telling you, this is just a fancy way of telling you that the point when x is 1.5, when you put 1.5 into the function, the function evaluates as negative 3. So this tells us that the coordinate 1.5, negative 3 is on the line. And then this tells us that the point when x is negative 1, f of x is equal to 2. This is just a fancy way of saying that both of these two points are on the line."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "All this is telling you, this is just a fancy way of telling you that the point when x is 1.5, when you put 1.5 into the function, the function evaluates as negative 3. So this tells us that the coordinate 1.5, negative 3 is on the line. And then this tells us that the point when x is negative 1, f of x is equal to 2. This is just a fancy way of saying that both of these two points are on the line. Nothing unusual. I think the point of this problem is to get you familiar with function notation, for you to not get intimidated if you see something like this. If you evaluate the function at 1.5, you get negative 3."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "This is just a fancy way of saying that both of these two points are on the line. Nothing unusual. I think the point of this problem is to get you familiar with function notation, for you to not get intimidated if you see something like this. If you evaluate the function at 1.5, you get negative 3. So that's the coordinate y. If you imagine that y is equal to f of x. So this would be the y coordinate, would be equal to negative 3 when x is 1.5."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "If you evaluate the function at 1.5, you get negative 3. So that's the coordinate y. If you imagine that y is equal to f of x. So this would be the y coordinate, would be equal to negative 3 when x is 1.5. Let's do this multiple times. Let's figure out the slope of this line. The slope, which is change in y over change in x, is equal to, let's start with 2 minus this guy, negative 3, right, these are the y values, over negative 1 minus this guy."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "So this would be the y coordinate, would be equal to negative 3 when x is 1.5. Let's do this multiple times. Let's figure out the slope of this line. The slope, which is change in y over change in x, is equal to, let's start with 2 minus this guy, negative 3, right, these are the y values, over negative 1 minus this guy. Let me write it this way. Negative 1 minus that guy. Minus 1.5."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "The slope, which is change in y over change in x, is equal to, let's start with 2 minus this guy, negative 3, right, these are the y values, over negative 1 minus this guy. Let me write it this way. Negative 1 minus that guy. Minus 1.5. And I do the colors because I want to show you that the negative 1 and the 2 are both coming from this. That's why I used both of them first. If I used these guys first, I would have to use both the x and the y's first."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "Minus 1.5. And I do the colors because I want to show you that the negative 1 and the 2 are both coming from this. That's why I used both of them first. If I used these guys first, I would have to use both the x and the y's first. If I use the 2 first, I have to use the negative 1 first. That's why I'm color coding it. So this is going to be equal to 2 minus negative 3."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "If I used these guys first, I would have to use both the x and the y's first. If I use the 2 first, I have to use the negative 1 first. That's why I'm color coding it. So this is going to be equal to 2 minus negative 3. That's the same thing as 2 plus 3. So that is 5. 1 minus 1.5, or negative 1 minus 1.5 is negative 2.5."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "So this is going to be equal to 2 minus negative 3. That's the same thing as 2 plus 3. So that is 5. 1 minus 1.5, or negative 1 minus 1.5 is negative 2.5. And 5 divided by 2.5 is equal to 2. So the slope of this line is negative 2. Actually, I'll take a little side to show you it doesn't matter what order I do this in, as long as I use, if I use this coordinate first, then I have to use that coordinate first."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "1 minus 1.5, or negative 1 minus 1.5 is negative 2.5. And 5 divided by 2.5 is equal to 2. So the slope of this line is negative 2. Actually, I'll take a little side to show you it doesn't matter what order I do this in, as long as I use, if I use this coordinate first, then I have to use that coordinate first. Let's do it the other way. If I did it as negative 3, negative 3 minus 2, minus 2, over 1.5 minus negative 1. This should be minus the 2, over 1.5 minus the negative 1."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "Actually, I'll take a little side to show you it doesn't matter what order I do this in, as long as I use, if I use this coordinate first, then I have to use that coordinate first. Let's do it the other way. If I did it as negative 3, negative 3 minus 2, minus 2, over 1.5 minus negative 1. This should be minus the 2, over 1.5 minus the negative 1. This should give me the same answer. This is equal to what? Negative 3 minus 2 is negative 5, over 1.5 minus negative 1."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "This should be minus the 2, over 1.5 minus the negative 1. This should give me the same answer. This is equal to what? Negative 3 minus 2 is negative 5, over 1.5 minus negative 1. That's 1.5 plus 1. So it's over 2.5. So once again, this is equal to negative 2."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "Negative 3 minus 2 is negative 5, over 1.5 minus negative 1. That's 1.5 plus 1. So it's over 2.5. So once again, this is equal to negative 2. So I just want to show you, it doesn't matter which one you pick as the starting or the end point, as long as you do it, you're consistent. If this is the starting y, this is the starting x. If this is the finishing y, this has to be the finishing x."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "So once again, this is equal to negative 2. So I just want to show you, it doesn't matter which one you pick as the starting or the end point, as long as you do it, you're consistent. If this is the starting y, this is the starting x. If this is the finishing y, this has to be the finishing x. But anyway, we know that the slope is negative 2. So we know that the equation is y is equal to negative 2x, plus some y intercept. Let's use one of these coordinates."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "If this is the finishing y, this has to be the finishing x. But anyway, we know that the slope is negative 2. So we know that the equation is y is equal to negative 2x, plus some y intercept. Let's use one of these coordinates. I'll use this one, since it doesn't have a decimal in it. So we know that y is equal to 2. So y is equal to 2 when x is equal to negative 1."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "Let's use one of these coordinates. I'll use this one, since it doesn't have a decimal in it. So we know that y is equal to 2. So y is equal to 2 when x is equal to negative 1. And of course, you have your plus b. So negative 2 times negative 1 is 2, plus b. So if you subtract 2 from both sides of this equation, minus 2, minus 2, we're subtracting it from both sides of this equation, you're going to get 0 on the left-hand side is equal to b."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "So y is equal to 2 when x is equal to negative 1. And of course, you have your plus b. So negative 2 times negative 1 is 2, plus b. So if you subtract 2 from both sides of this equation, minus 2, minus 2, we're subtracting it from both sides of this equation, you're going to get 0 on the left-hand side is equal to b. So b is 0. So the equation of our line is just y is equal to negative 2x. y is equal to negative 2x."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "So if you subtract 2 from both sides of this equation, minus 2, minus 2, we're subtracting it from both sides of this equation, you're going to get 0 on the left-hand side is equal to b. So b is 0. So the equation of our line is just y is equal to negative 2x. y is equal to negative 2x. Or actually, if you wanted to write it in function notation, it would be that f of x is equal to negative 2x. I kind of just assumed that y is equal to f of x. But this is really the equation."}, {"video_title": "More examples of constructing linear equations in slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "y is equal to negative 2x. Or actually, if you wanted to write it in function notation, it would be that f of x is equal to negative 2x. I kind of just assumed that y is equal to f of x. But this is really the equation. They never mentioned y's here. So you could just write f of x is equal to 2x right here. And each of these coordinates are the coordinates of x and f of x."}, {"video_title": "U.S. national debt (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "On February 2, 2010, the U.S. Treasury estimated the national debt at 1.2278 times 10 to the 13th power. And just to get a sense of things, 10 to the 6th, you're talking of 1 times 10 to the 6th is a million, 1 times 10 to the 9th is a billion, 1 times 10 to the 12th is a trillion, so we're talking about on the order of magnitude of 10 trillion dollars. So this is about 12 trillion dollars. Then they tell us that the U.S. Census Bureau's estimate for the U.S. population was about 3.086 times 10 to the 8th power, so this is a little over 300 million people. So that's an interesting number right there, it's a population. And then they say, using these estimates, calculate the per person share of the national debt. So essentially, we want to take the entire debt and divide by the number of people."}, {"video_title": "U.S. national debt (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "Then they tell us that the U.S. Census Bureau's estimate for the U.S. population was about 3.086 times 10 to the 8th power, so this is a little over 300 million people. So that's an interesting number right there, it's a population. And then they say, using these estimates, calculate the per person share of the national debt. So essentially, we want to take the entire debt and divide by the number of people. That'll give us the per person share of the national debt. Use scientific notation to make your calculations and express your answer in both scientific and decimal notations. Alright, which means just as a regular number."}, {"video_title": "U.S. national debt (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "So essentially, we want to take the entire debt and divide by the number of people. That'll give us the per person share of the national debt. Use scientific notation to make your calculations and express your answer in both scientific and decimal notations. Alright, which means just as a regular number. Round to four decimal places while making calculations. So we want the per person debt, so we want to take the total debt and divide by the number of people. So the total debt is 1.2278 times 10 to the 13th power, and we want to divide that by the total number of people, which is 3.086 times 10 to the 8th."}, {"video_title": "U.S. national debt (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "Alright, which means just as a regular number. Round to four decimal places while making calculations. So we want the per person debt, so we want to take the total debt and divide by the number of people. So the total debt is 1.2278 times 10 to the 13th power, and we want to divide that by the total number of people, which is 3.086 times 10 to the 8th. And we could separate this into two division problems. We could say that this is equal to the division right here. These guys, 1.2278 divided by 3.06, so 1.2278 divided by 3.086, and then times 10 to the 13th divided by 10 to the 8th."}, {"video_title": "U.S. national debt (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "So the total debt is 1.2278 times 10 to the 13th power, and we want to divide that by the total number of people, which is 3.086 times 10 to the 8th. And we could separate this into two division problems. We could say that this is equal to the division right here. These guys, 1.2278 divided by 3.06, so 1.2278 divided by 3.086, and then times 10 to the 13th divided by 10 to the 8th. Now what's 10 to the 13th divided by 10 to the 8th? Let me do it over here. 10 to the 13th over 10 to the 8th."}, {"video_title": "U.S. national debt (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "These guys, 1.2278 divided by 3.06, so 1.2278 divided by 3.086, and then times 10 to the 13th divided by 10 to the 8th. Now what's 10 to the 13th divided by 10 to the 8th? Let me do it over here. 10 to the 13th over 10 to the 8th. The way I think about it, this is the exact same thing. This is equal to 10 to the 13th times 10 to the negative 8. This is an 8 right here."}, {"video_title": "U.S. national debt (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "10 to the 13th over 10 to the 8th. The way I think about it, this is the exact same thing. This is equal to 10 to the 13th times 10 to the negative 8. This is an 8 right here. If you have a 10 to the 8th in the denominator, that's like multiplying by 10 to the negative 8. So you have 13, you have the same base, 10, so 10 to the 13th times 10 to the negative 8 is going to be 10 to the 13 minus 8, which is 10 to the 5th. Or another way to think about it, if you have the base in the denominator, you subtract the exponent, so it's 13 minus 8, 10 to the 5th."}, {"video_title": "U.S. national debt (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "This is an 8 right here. If you have a 10 to the 8th in the denominator, that's like multiplying by 10 to the negative 8. So you have 13, you have the same base, 10, so 10 to the 13th times 10 to the negative 8 is going to be 10 to the 13 minus 8, which is 10 to the 5th. Or another way to think about it, if you have the base in the denominator, you subtract the exponent, so it's 13 minus 8, 10 to the 5th. So it's this blue expression times 10 to the 5th. 10 to the 13th divided by 10 to the 8th is 10 to the 5th. 10 to the 5th power."}, {"video_title": "U.S. national debt (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "Or another way to think about it, if you have the base in the denominator, you subtract the exponent, so it's 13 minus 8, 10 to the 5th. So it's this blue expression times 10 to the 5th. 10 to the 13th divided by 10 to the 8th is 10 to the 5th. 10 to the 5th power. And let's get a calculator out to calculate this right here. Let's get the calculator out. And they say round everything to four decimal places, so I'll keep that in mind."}, {"video_title": "U.S. national debt (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "10 to the 5th power. And let's get a calculator out to calculate this right here. Let's get the calculator out. And they say round everything to four decimal places, so I'll keep that in mind. So let me turn my calculator on. 1.2278 divided by 3.086 is equal to 3, let's see how many, 3979, because we want to round right there. So 3979, let me remember that."}, {"video_title": "U.S. national debt (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "And they say round everything to four decimal places, so I'll keep that in mind. So let me turn my calculator on. 1.2278 divided by 3.086 is equal to 3, let's see how many, 3979, because we want to round right there. So 3979, let me remember that. Let me just put it on the side. Put it on the side so I can still look at it. So this little dividing decimals problem results in, this is 0.3979, and of course times 10 to the 5th."}, {"video_title": "U.S. national debt (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "So 3979, let me remember that. Let me just put it on the side. Put it on the side so I can still look at it. So this little dividing decimals problem results in, this is 0.3979, and of course times 10 to the 5th. Dollars per person. Now, once again, you might be tempted to say, hey, this isn't scientific notation, I have some number times a power of 10. But notice, this number is not greater than or equal to 1."}, {"video_title": "U.S. national debt (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "So this little dividing decimals problem results in, this is 0.3979, and of course times 10 to the 5th. Dollars per person. Now, once again, you might be tempted to say, hey, this isn't scientific notation, I have some number times a power of 10. But notice, this number is not greater than or equal to 1. Remember, this number, if you want to be formal about scientific notation, has to be greater than or equal to 1 or less than 10. So what we can do here is we can multiply, if we don't want to change the number, we can multiply this number by 10 and divide this number by 10. Or another way you can think about it is, this whole thing can be rewritten as 0.3979 times 10 times 10 to the 4th."}, {"video_title": "U.S. national debt (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "But notice, this number is not greater than or equal to 1. Remember, this number, if you want to be formal about scientific notation, has to be greater than or equal to 1 or less than 10. So what we can do here is we can multiply, if we don't want to change the number, we can multiply this number by 10 and divide this number by 10. Or another way you can think about it is, this whole thing can be rewritten as 0.3979 times 10 times 10 to the 4th. What I did just now is I broke up the 10 to the 5th into a 10 and a 10 to the 4th. And I did that because I want to multiply this by 10 so I can get a 3 out front instead of a 0.3. So let's do that."}, {"video_title": "U.S. national debt (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "Or another way you can think about it is, this whole thing can be rewritten as 0.3979 times 10 times 10 to the 4th. What I did just now is I broke up the 10 to the 5th into a 10 and a 10 to the 4th. And I did that because I want to multiply this by 10 so I can get a 3 out front instead of a 0.3. So let's do that. So if you multiply this out, so essentially I took a 10 out of the 10 to the 5th, I divided it by 10 and I multiplied this other guy by 10, not changing the whole number. So then this right here will become 3.979 times 10 to the 4th. So that's how much debt there is per person in scientific notation."}, {"video_title": "U.S. national debt (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "So let's do that. So if you multiply this out, so essentially I took a 10 out of the 10 to the 5th, I divided it by 10 and I multiplied this other guy by 10, not changing the whole number. So then this right here will become 3.979 times 10 to the 4th. So that's how much debt there is per person in scientific notation. So this is debt per person in scientific notation. Now, the problem they also wanted us to express it in decimal notation, which is just kind of standard, you know, writing it as a number with our standard numeric decimal system. So what is 3.979 times 10 to the 4th?"}, {"video_title": "U.S. national debt (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "So that's how much debt there is per person in scientific notation. So this is debt per person in scientific notation. Now, the problem they also wanted us to express it in decimal notation, which is just kind of standard, you know, writing it as a number with our standard numeric decimal system. So what is 3.979 times 10 to the 4th? So let's think about it. We have 3.979 times 10 to the 4th. Or let me just do it this way."}, {"video_title": "U.S. national debt (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "So what is 3.979 times 10 to the 4th? So let's think about it. We have 3.979 times 10 to the 4th. Or let me just do it this way. Times 10 to the 4th. So if we multiply it, let's just move the decimal space. If we multiply it by 10, we're going to get 39.79."}, {"video_title": "U.S. national debt (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "Or let me just do it this way. Times 10 to the 4th. So if we multiply it, let's just move the decimal space. If we multiply it by 10, we're going to get 39.79. If we multiply it by 10 squared, we're going to get 397.9. If we multiply it by 10 to the 3rd, we're going to get 3,979. If we multiply it by 10 to the 4th, we're going to get one more zero right there."}, {"video_title": "U.S. national debt (scientific notation word problem)   Pre-Algebra   Khan Academy.mp3", "Sentence": "If we multiply it by 10, we're going to get 39.79. If we multiply it by 10 squared, we're going to get 397.9. If we multiply it by 10 to the 3rd, we're going to get 3,979. If we multiply it by 10 to the 4th, we're going to get one more zero right there. So we're essentially going to move the decimal four to the right, and we get it's 39, so I could write it like this. This is equal to $39,790. So if you think about the national debt per person, every man, woman, and child in the United States essentially owes $39,790."}, {"video_title": "Multi-step inequalities 3   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So what we really want to do is isolate the p on one side of this inequality, and preferably the left. That just makes it a little bit easier to read. It doesn't have to be, but we just want to isolate the p. So a good step to that is to get rid of this p on the right-hand side. And the best way I can think of doing that is subtracting p from the right. But of course, if we want to make sure that this inequality is always going to be true, if we do anything to the right, we also have to do that to the left. So we also have to subtract p from the left. And so the left-hand side, negative 3p minus p, that's negative 4p, and then we still have a minus 7 up here, is going to be less than p minus p. Those cancel out."}, {"video_title": "Multi-step inequalities 3   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "And the best way I can think of doing that is subtracting p from the right. But of course, if we want to make sure that this inequality is always going to be true, if we do anything to the right, we also have to do that to the left. So we also have to subtract p from the left. And so the left-hand side, negative 3p minus p, that's negative 4p, and then we still have a minus 7 up here, is going to be less than p minus p. Those cancel out. It is less than 9. Now, the next thing I'm in the mood to do is get rid of this negative 7 or this minus 7 here so that we can better isolate the p on the left-hand side. So the best way I can think of to get rid of a negative 7 is to add 7 to it."}, {"video_title": "Multi-step inequalities 3   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "And so the left-hand side, negative 3p minus p, that's negative 4p, and then we still have a minus 7 up here, is going to be less than p minus p. Those cancel out. It is less than 9. Now, the next thing I'm in the mood to do is get rid of this negative 7 or this minus 7 here so that we can better isolate the p on the left-hand side. So the best way I can think of to get rid of a negative 7 is to add 7 to it. Then it'll just cancel out to 0. So let's add 7 to both sides of this inequality. Negative 7 plus 7 cancel out."}, {"video_title": "Multi-step inequalities 3   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So the best way I can think of to get rid of a negative 7 is to add 7 to it. Then it'll just cancel out to 0. So let's add 7 to both sides of this inequality. Negative 7 plus 7 cancel out. All we're left with is negative 4p. On the right-hand side, we have 9 plus 7 is 16. And it's still less than."}, {"video_title": "Multi-step inequalities 3   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Negative 7 plus 7 cancel out. All we're left with is negative 4p. On the right-hand side, we have 9 plus 7 is 16. And it's still less than. Now, the last step to isolate the p is to get rid of this negative 4 coefficient. And the easiest way I can think of to get rid of this negative 4 coefficient is to divide both sides by negative 4. So if we divide this side by negative 4, these guys are going to cancel out."}, {"video_title": "Multi-step inequalities 3   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "And it's still less than. Now, the last step to isolate the p is to get rid of this negative 4 coefficient. And the easiest way I can think of to get rid of this negative 4 coefficient is to divide both sides by negative 4. So if we divide this side by negative 4, these guys are going to cancel out. We're just going to be left with p. We also have to do it to the right-hand side. Now, there's one thing that you really have to remember since this is an inequality. This is not an equation."}, {"video_title": "Multi-step inequalities 3   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So if we divide this side by negative 4, these guys are going to cancel out. We're just going to be left with p. We also have to do it to the right-hand side. Now, there's one thing that you really have to remember since this is an inequality. This is not an equation. If you're dealing with an inequality, and if you multiply or divide both sides of the equation by a negative number, you have to swap the inequality. So in this case, the less than becomes greater than since we're dividing by a negative number. And so negative 4 divided by negative 4, those cancel out."}, {"video_title": "Multi-step inequalities 3   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "This is not an equation. If you're dealing with an inequality, and if you multiply or divide both sides of the equation by a negative number, you have to swap the inequality. So in this case, the less than becomes greater than since we're dividing by a negative number. And so negative 4 divided by negative 4, those cancel out. We have p is greater than 16 divided by negative 4, which is negative 4. And we can plot this solution set right over here. And then we can try out some values to help us feel good about the idea of it working."}, {"video_title": "Multi-step inequalities 3   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "And so negative 4 divided by negative 4, those cancel out. We have p is greater than 16 divided by negative 4, which is negative 4. And we can plot this solution set right over here. And then we can try out some values to help us feel good about the idea of it working. So let's say this is negative 5, negative 4, negative 3, negative 2, 1, negative 1, I should say, 0. Let me write that a little bit neater. Negative 1, 0."}, {"video_title": "Multi-step inequalities 3   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "And then we can try out some values to help us feel good about the idea of it working. So let's say this is negative 5, negative 4, negative 3, negative 2, 1, negative 1, I should say, 0. Let me write that a little bit neater. Negative 1, 0. And then we can keep going to the right. And so our solution is p is not greater than or equal, so we have to exclude negative 4. p is greater than negative 4, so all the values above that. So negative 3.99999999 will work."}, {"video_title": "Multi-step inequalities 3   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Negative 1, 0. And then we can keep going to the right. And so our solution is p is not greater than or equal, so we have to exclude negative 4. p is greater than negative 4, so all the values above that. So negative 3.99999999 will work. Negative 4 will not work. And let's just try some values out to feel good that this is really the solution set. So first let's try out when p is equal to negative 3."}, {"video_title": "Multi-step inequalities 3   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So negative 3.99999999 will work. Negative 4 will not work. And let's just try some values out to feel good that this is really the solution set. So first let's try out when p is equal to negative 3. This should work. The way I've drawn it, this is in our solution set. p equals negative 3 is greater than negative 4."}, {"video_title": "Multi-step inequalities 3   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So first let's try out when p is equal to negative 3. This should work. The way I've drawn it, this is in our solution set. p equals negative 3 is greater than negative 4. So let's try that out. We have negative 3 times negative 3. The first negative 3 is this one, and then we're saying p is negative 3."}, {"video_title": "Multi-step inequalities 3   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "p equals negative 3 is greater than negative 4. So let's try that out. We have negative 3 times negative 3. The first negative 3 is this one, and then we're saying p is negative 3. Minus 7 should be less than, instead of a p, we're going to put a negative 3, should be less than negative 3 plus 9. Negative 3 times negative 3 is 9. Minus 7 should be less than negative 3 plus 9 is 6."}, {"video_title": "Multi-step inequalities 3   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "The first negative 3 is this one, and then we're saying p is negative 3. Minus 7 should be less than, instead of a p, we're going to put a negative 3, should be less than negative 3 plus 9. Negative 3 times negative 3 is 9. Minus 7 should be less than negative 3 plus 9 is 6. 9 minus 7 is 2. 2 should be less than 6, of which, of course, it is. Now let's try a value that definitely should not work."}, {"video_title": "Multi-step inequalities 3   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Minus 7 should be less than negative 3 plus 9 is 6. 9 minus 7 is 2. 2 should be less than 6, of which, of course, it is. Now let's try a value that definitely should not work. So let's try negative 5. Negative 5 is not in our solution set, so it should not work. So we have negative 3 times negative 5 minus 7."}, {"video_title": "Multi-step inequalities 3   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Now let's try a value that definitely should not work. So let's try negative 5. Negative 5 is not in our solution set, so it should not work. So we have negative 3 times negative 5 minus 7. Let's see whether it is less than negative 5 plus 9. Negative 3 times negative 5 is 15. Minus 7, it really should not be less than negative 5 plus 9."}, {"video_title": "Multi-step inequalities 3   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So we have negative 3 times negative 5 minus 7. Let's see whether it is less than negative 5 plus 9. Negative 3 times negative 5 is 15. Minus 7, it really should not be less than negative 5 plus 9. So we're just seeing if p equals negative 5 works. 15 minus 7 is 8. And so we get 8 is less than 4, which is definitely not the case."}, {"video_title": "Multi-step inequalities 3   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Minus 7, it really should not be less than negative 5 plus 9. So we're just seeing if p equals negative 5 works. 15 minus 7 is 8. And so we get 8 is less than 4, which is definitely not the case. So p equals negative 5 doesn't work, and it shouldn't work because that's not in our solution set. And now if we want to feel really good about it, we can actually try this boundary point. Negative 4 should not work, but it should satisfy the related equation."}, {"video_title": "Multi-step inequalities 3   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "And so we get 8 is less than 4, which is definitely not the case. So p equals negative 5 doesn't work, and it shouldn't work because that's not in our solution set. And now if we want to feel really good about it, we can actually try this boundary point. Negative 4 should not work, but it should satisfy the related equation. When I talk about the related equation, negative 4 should satisfy negative 3 minus 7 is equal to p plus 9. It'll satisfy this, but it won't satisfy this because when we get the same value on both sides, the same value is not less than the same value. So let's try it out."}, {"video_title": "Multi-step inequalities 3   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Negative 4 should not work, but it should satisfy the related equation. When I talk about the related equation, negative 4 should satisfy negative 3 minus 7 is equal to p plus 9. It'll satisfy this, but it won't satisfy this because when we get the same value on both sides, the same value is not less than the same value. So let's try it out. Let's see whether negative 4 at least satisfies the related equation. So if we get negative 3 times negative 4 minus 7, this should be equal to negative 4 plus 9. So this is 12 minus 7 should be equal to negative 4 plus 9 should be equal to 5."}, {"video_title": "Multi-step inequalities 3   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So let's try it out. Let's see whether negative 4 at least satisfies the related equation. So if we get negative 3 times negative 4 minus 7, this should be equal to negative 4 plus 9. So this is 12 minus 7 should be equal to negative 4 plus 9 should be equal to 5. And this, of course, is true. 5 is equal to 5. So it satisfies the related equation, but it should not satisfy this."}, {"video_title": "Multi-step inequalities 3   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So this is 12 minus 7 should be equal to negative 4 plus 9 should be equal to 5. And this, of course, is true. 5 is equal to 5. So it satisfies the related equation, but it should not satisfy this. If you put negative 4 for p here, and I encourage you to do so, actually we could do it over here instead of an equal sign. If you put it into the original inequality, let me delete all of that, it really just becomes this. The original inequality is this right over here."}, {"video_title": "Multi-step inequalities 3   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So it satisfies the related equation, but it should not satisfy this. If you put negative 4 for p here, and I encourage you to do so, actually we could do it over here instead of an equal sign. If you put it into the original inequality, let me delete all of that, it really just becomes this. The original inequality is this right over here. If you put negative 4, you have less than, less than, and then you get 5 is less than 5, which is not the case. And that's good because we did not include that in the solution set. We put an open circle."}, {"video_title": "Multi-step inequalities 3   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "The original inequality is this right over here. If you put negative 4, you have less than, less than, and then you get 5 is less than 5, which is not the case. And that's good because we did not include that in the solution set. We put an open circle. If negative 4 was included, we would fill that in. But the only reason I would include negative 4 is if this was greater than or equal. So it's good that this does not work because negative 4 is not part of our solution set."}, {"video_title": "Similarity example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "And we have these two parallel lines. AB is parallel to DE. And then we have these two essentially transversals that form these two triangles. So let's see what we can do here. So the first thing that might jump out at you is that this angle and this angle are vertical angles, so they are going to be congruent. The other thing that might jump out at you is that angle CDE is an alternate interior angle with CBA. So we have this transversal right over here."}, {"video_title": "Similarity example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "So let's see what we can do here. So the first thing that might jump out at you is that this angle and this angle are vertical angles, so they are going to be congruent. The other thing that might jump out at you is that angle CDE is an alternate interior angle with CBA. So we have this transversal right over here. And these are alternate interior angles. And they are going to be congruent. Or you could say that if you continue this transversal, you would have a corresponding angle with CDE right up here."}, {"video_title": "Similarity example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "So we have this transversal right over here. And these are alternate interior angles. And they are going to be congruent. Or you could say that if you continue this transversal, you would have a corresponding angle with CDE right up here. And this one's just vertical. Either way, this angle and this angle are going to be congruent. So we've established that we have two triangles."}, {"video_title": "Similarity example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "Or you could say that if you continue this transversal, you would have a corresponding angle with CDE right up here. And this one's just vertical. Either way, this angle and this angle are going to be congruent. So we've established that we have two triangles. And two of the corresponding angles are the same. And that by itself is enough to establish similarity. You can actually, we actually could show that this angle and this angle are also congruent by alternate interior angles."}, {"video_title": "Similarity example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "So we've established that we have two triangles. And two of the corresponding angles are the same. And that by itself is enough to establish similarity. You can actually, we actually could show that this angle and this angle are also congruent by alternate interior angles. But we don't have to. So we already know that they are similar. Actually, we could just say it just by alternate interior angles, these are also going to be congruent."}, {"video_title": "Similarity example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "You can actually, we actually could show that this angle and this angle are also congruent by alternate interior angles. But we don't have to. So we already know that they are similar. Actually, we could just say it just by alternate interior angles, these are also going to be congruent. But we already know enough to say that they are similar even before doing that. So we already know that triangle, I'll try to write it, I'll color code it so that we have the same corresponding vertices. And that's really important to know what angles and what sides correspond to what side so that you don't mess up your ratios."}, {"video_title": "Similarity example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "Actually, we could just say it just by alternate interior angles, these are also going to be congruent. But we already know enough to say that they are similar even before doing that. So we already know that triangle, I'll try to write it, I'll color code it so that we have the same corresponding vertices. And that's really important to know what angles and what sides correspond to what side so that you don't mess up your ratios. So that you do know what's corresponding to what. So we know triangle AB, triangle ABC, is similar to triangle. So A, this vertex A corresponds to vertex E over here, is similar to vertex E. And then vertex B right over here corresponds to vertex D, EDC."}, {"video_title": "Similarity example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "And that's really important to know what angles and what sides correspond to what side so that you don't mess up your ratios. So that you do know what's corresponding to what. So we know triangle AB, triangle ABC, is similar to triangle. So A, this vertex A corresponds to vertex E over here, is similar to vertex E. And then vertex B right over here corresponds to vertex D, EDC. Now what does that do for us? Well, that tells us that the ratio of corresponding sides are going to be the same. They're going to be some constant value."}, {"video_title": "Similarity example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "So A, this vertex A corresponds to vertex E over here, is similar to vertex E. And then vertex B right over here corresponds to vertex D, EDC. Now what does that do for us? Well, that tells us that the ratio of corresponding sides are going to be the same. They're going to be some constant value. So we have corresponding side. So the ratio, for example, the corresponding side for BC is going to be DC. We can see it just the way that we've written down the similarity."}, {"video_title": "Similarity example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "They're going to be some constant value. So we have corresponding side. So the ratio, for example, the corresponding side for BC is going to be DC. We can see it just the way that we've written down the similarity. This is true. Then BC is the corresponding side to DC. So we know that the length of BC over DC is going to be equal to the length of, well, we want to figure out what CE is."}, {"video_title": "Similarity example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "We can see it just the way that we've written down the similarity. This is true. Then BC is the corresponding side to DC. So we know that the length of BC over DC is going to be equal to the length of, well, we want to figure out what CE is. That's what we care about. And I'm using BC and DC because we know those values. So BC over DC is going to be equal to what's the corresponding side to CE?"}, {"video_title": "Similarity example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "So we know that the length of BC over DC is going to be equal to the length of, well, we want to figure out what CE is. That's what we care about. And I'm using BC and DC because we know those values. So BC over DC is going to be equal to what's the corresponding side to CE? The corresponding side over here is CA. It's going to be equal to CA over CE. Corresponding sides."}, {"video_title": "Similarity example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "So BC over DC is going to be equal to what's the corresponding side to CE? The corresponding side over here is CA. It's going to be equal to CA over CE. Corresponding sides. This is the last and the first, last and the first. CA over CE. And we know what BC is."}, {"video_title": "Similarity example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "Corresponding sides. This is the last and the first, last and the first. CA over CE. And we know what BC is. BC right over here is 5. We know what DC is. It is 3."}, {"video_title": "Similarity example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "And we know what BC is. BC right over here is 5. We know what DC is. It is 3. We know what CA or AC is right over here. CA is 4. And now we can just solve for CE."}, {"video_title": "Similarity example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "It is 3. We know what CA or AC is right over here. CA is 4. And now we can just solve for CE. So we can, well, there's multiple ways that you could think about this. You could cross multiply, which is really just multiplying both sides by both denominators. So you get 5 times the length of CE is equal to 3 times 4, which is just going to be equal to 12."}, {"video_title": "Similarity example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "And now we can just solve for CE. So we can, well, there's multiple ways that you could think about this. You could cross multiply, which is really just multiplying both sides by both denominators. So you get 5 times the length of CE is equal to 3 times 4, which is just going to be equal to 12. And then we get CE is equal to 12 over 5, which is the same thing as 2 and 2 fifths or 2.4. So this is going to be 2 and 2 fifths. And we're done."}, {"video_title": "Similarity example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "So you get 5 times the length of CE is equal to 3 times 4, which is just going to be equal to 12. And then we get CE is equal to 12 over 5, which is the same thing as 2 and 2 fifths or 2.4. So this is going to be 2 and 2 fifths. And we're done. We were able to use similarity to figure out this side, just knowing that the ratio between the corresponding sides are going to be the same. Now let's do this problem right over here. Let me draw a little line here to show that this is a different problem now."}, {"video_title": "Similarity example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "And we're done. We were able to use similarity to figure out this side, just knowing that the ratio between the corresponding sides are going to be the same. Now let's do this problem right over here. Let me draw a little line here to show that this is a different problem now. This is a different problem. So in this problem, we need to figure out what DE is. And we all, once again, have these two parallel lines like this."}, {"video_title": "Similarity example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "Let me draw a little line here to show that this is a different problem now. This is a different problem. So in this problem, we need to figure out what DE is. And we all, once again, have these two parallel lines like this. And so we know corresponding angles are congruent. So we know that angle is going to be congruent to that angle, because you could view this as a transversal. We also know that this angle right over here is going to be congruent to that angle right over there."}, {"video_title": "Similarity example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "And we all, once again, have these two parallel lines like this. And so we know corresponding angles are congruent. So we know that angle is going to be congruent to that angle, because you could view this as a transversal. We also know that this angle right over here is going to be congruent to that angle right over there. Once again, corresponding angles for transversal. And also in both triangles. So I'm looking at triangle CBD and triangle CAE."}, {"video_title": "Similarity example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "We also know that this angle right over here is going to be congruent to that angle right over there. Once again, corresponding angles for transversal. And also in both triangles. So I'm looking at triangle CBD and triangle CAE. They both share this angle up here. Once again, we could have stopped at two angles. But we've actually shown that all three angles of these two triangles, all three of the corresponding angles, are congruent to each other."}, {"video_title": "Similarity example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "So I'm looking at triangle CBD and triangle CAE. They both share this angle up here. Once again, we could have stopped at two angles. But we've actually shown that all three angles of these two triangles, all three of the corresponding angles, are congruent to each other. So we now know, and once again, this is an important thing to do, is to make sure that you write it in the right order when you write your similarity. We now know that triangle CBD is similar. Not congruent."}, {"video_title": "Similarity example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "But we've actually shown that all three angles of these two triangles, all three of the corresponding angles, are congruent to each other. So we now know, and once again, this is an important thing to do, is to make sure that you write it in the right order when you write your similarity. We now know that triangle CBD is similar. Not congruent. It is similar to triangle CCAE, which means that the ratio of corresponding sides are going to be constant. So we know, for example, that the ratio of CB to CA, let's write this down. We know that the ratio of CB over CA is going to be equal to the ratio of CD over CE."}, {"video_title": "Similarity example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "Not congruent. It is similar to triangle CCAE, which means that the ratio of corresponding sides are going to be constant. So we know, for example, that the ratio of CB to CA, let's write this down. We know that the ratio of CB over CA is going to be equal to the ratio of CD over CE. And we know what CB is. CB over here is 5. We know what CA is."}, {"video_title": "Similarity example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "We know that the ratio of CB over CA is going to be equal to the ratio of CD over CE. And we know what CB is. CB over here is 5. We know what CA is. And we have to be careful here. It's not 3. CA, this entire side, is going to be 5 plus 3."}, {"video_title": "Similarity example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "We know what CA is. And we have to be careful here. It's not 3. CA, this entire side, is going to be 5 plus 3. So this is going to be 8. And we know what CD is. CD is going to be 4."}, {"video_title": "Similarity example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "CA, this entire side, is going to be 5 plus 3. So this is going to be 8. And we know what CD is. CD is going to be 4. And so once again, we can cross multiply. We have 5 times CE is equal to 8 times 4. 8 times 4 is 32."}, {"video_title": "Similarity example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "CD is going to be 4. And so once again, we can cross multiply. We have 5 times CE is equal to 8 times 4. 8 times 4 is 32. And so CE is equal to 32 over 5. Or this is another way to think about it. That's 6 and 2 fifths."}, {"video_title": "Similarity example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "8 times 4 is 32. And so CE is equal to 32 over 5. Or this is another way to think about it. That's 6 and 2 fifths. Now we're not done, because they didn't ask for what CE is. They're asking for just this part right over here. They're asking for DE."}, {"video_title": "Similarity example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "That's 6 and 2 fifths. Now we're not done, because they didn't ask for what CE is. They're asking for just this part right over here. They're asking for DE. So we know that this entire length, CE right over here, this is 6 and 2 fifths. And so DE right over here, what we actually have to figure out, it's going to be this entire length, 6 and 2 fifths, minus 4, minus CD right over here. So it's going to be 2 and 2 fifths."}, {"video_title": "Similarity example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "They're asking for DE. So we know that this entire length, CE right over here, this is 6 and 2 fifths. And so DE right over here, what we actually have to figure out, it's going to be this entire length, 6 and 2 fifths, minus 4, minus CD right over here. So it's going to be 2 and 2 fifths. 6 and 2 fifths minus 4 and 2 fifths is 2 and 2 fifths. So we're done. DE is 2 and 2 fifths."}, {"video_title": "Basic linear equation word problem   7th grade   Khan Academy.mp3", "Sentence": "What is the minimum number of computers she needs to sell in a month to make a profit? So I'll let you think about that for a second. Well, let's think about what we have to figure out. We have to figure out the minimum number of computers she needs to sell. So let's set that to a variable or set a variable to represent that. So let's let x equal the number of computers she sells. Number of computers sold."}, {"video_title": "Basic linear equation word problem   7th grade   Khan Academy.mp3", "Sentence": "We have to figure out the minimum number of computers she needs to sell. So let's set that to a variable or set a variable to represent that. So let's let x equal the number of computers she sells. Number of computers sold. Now let's think about how much net profit she will make in a month. And that's what we're thinking about. How many computers, the minimum number she needs to sell in order to make a net profit."}, {"video_title": "Basic linear equation word problem   7th grade   Khan Academy.mp3", "Sentence": "Number of computers sold. Now let's think about how much net profit she will make in a month. And that's what we're thinking about. How many computers, the minimum number she needs to sell in order to make a net profit. So I'll write her profit. Her profit is going to be how much money she brings in from selling the computers. And she makes $27 on every computer she sells."}, {"video_title": "Basic linear equation word problem   7th grade   Khan Academy.mp3", "Sentence": "How many computers, the minimum number she needs to sell in order to make a net profit. So I'll write her profit. Her profit is going to be how much money she brings in from selling the computers. And she makes $27 on every computer she sells. So her profit is going to be $27 times the number of computers she sells. She gets $27 per computer times the number she sells. But we're not done yet."}, {"video_title": "Basic linear equation word problem   7th grade   Khan Academy.mp3", "Sentence": "And she makes $27 on every computer she sells. So her profit is going to be $27 times the number of computers she sells. She gets $27 per computer times the number she sells. But we're not done yet. She still has expenses of $10,000 per month. So we're going to have to subtract out the $10,000. What we care about is making a profit."}, {"video_title": "Basic linear equation word problem   7th grade   Khan Academy.mp3", "Sentence": "But we're not done yet. She still has expenses of $10,000 per month. So we're going to have to subtract out the $10,000. What we care about is making a profit. We want this number right over here to be greater than 0. So let's just think about what number of computers would get us to 0 if maybe she needs to sell a little bit more than that. So let's see what gets her to break even."}, {"video_title": "Basic linear equation word problem   7th grade   Khan Academy.mp3", "Sentence": "What we care about is making a profit. We want this number right over here to be greater than 0. So let's just think about what number of computers would get us to 0 if maybe she needs to sell a little bit more than that. So let's see what gets her to break even. So break even, that's 0 profit, neither positive or negative, is equal to 27 times, and I'll do it all in one color now, 27x minus 10,000. Well, we've seen equations like this before. We can add 10,000 to both sides."}, {"video_title": "Basic linear equation word problem   7th grade   Khan Academy.mp3", "Sentence": "So let's see what gets her to break even. So break even, that's 0 profit, neither positive or negative, is equal to 27 times, and I'll do it all in one color now, 27x minus 10,000. Well, we've seen equations like this before. We can add 10,000 to both sides. So let's do both sides. Add 10,000 to both sides so it's no longer on the right-hand side. And we are left with 10,000 is equal to 27x."}, {"video_title": "Basic linear equation word problem   7th grade   Khan Academy.mp3", "Sentence": "We can add 10,000 to both sides. So let's do both sides. Add 10,000 to both sides so it's no longer on the right-hand side. And we are left with 10,000 is equal to 27x. And then to solve for x, we just have to divide both sides by 27. Let's do that. Divide both sides by 27."}, {"video_title": "Basic linear equation word problem   7th grade   Khan Academy.mp3", "Sentence": "And we are left with 10,000 is equal to 27x. And then to solve for x, we just have to divide both sides by 27. Let's do that. Divide both sides by 27. On our right-hand side, we have x. So let me just write this down. So we have x on our right-hand side is going to be equal to 10,000 over 27."}, {"video_title": "Basic linear equation word problem   7th grade   Khan Academy.mp3", "Sentence": "Divide both sides by 27. On our right-hand side, we have x. So let me just write this down. So we have x on our right-hand side is going to be equal to 10,000 over 27. I've switched the right and the left-hand sides here. Now, what is this going to be? Well, we can do a little bit of long division to handle that."}, {"video_title": "Basic linear equation word problem   7th grade   Khan Academy.mp3", "Sentence": "So we have x on our right-hand side is going to be equal to 10,000 over 27. I've switched the right and the left-hand sides here. Now, what is this going to be? Well, we can do a little bit of long division to handle that. So 27 goes into 10,000. So 27 doesn't go into 1, doesn't go into 10. It goes into 100 three times."}, {"video_title": "Basic linear equation word problem   7th grade   Khan Academy.mp3", "Sentence": "Well, we can do a little bit of long division to handle that. So 27 goes into 10,000. So 27 doesn't go into 1, doesn't go into 10. It goes into 100 three times. 3 times 27 is what, 81. 100 minus 81 is 19. Then we can bring down a 0."}, {"video_title": "Basic linear equation word problem   7th grade   Khan Academy.mp3", "Sentence": "It goes into 100 three times. 3 times 27 is what, 81. 100 minus 81 is 19. Then we can bring down a 0. 27 goes into 190. It looks like it'll go into it about 6 times. Let's see if that's right."}, {"video_title": "Basic linear equation word problem   7th grade   Khan Academy.mp3", "Sentence": "Then we can bring down a 0. 27 goes into 190. It looks like it'll go into it about 6 times. Let's see if that's right. 6 times 7 is 42. 6 times 2 is 12 plus 4 is 16. Let's see, 90 minus 62 is actually 28."}, {"video_title": "Basic linear equation word problem   7th grade   Khan Academy.mp3", "Sentence": "Let's see if that's right. 6 times 7 is 42. 6 times 2 is 12 plus 4 is 16. Let's see, 90 minus 62 is actually 28. Oh, yes. Let me turn this back. So it goes 7 times."}, {"video_title": "Basic linear equation word problem   7th grade   Khan Academy.mp3", "Sentence": "Let's see, 90 minus 62 is actually 28. Oh, yes. Let me turn this back. So it goes 7 times. 7 times 7 is 49. 7 times 2 is 14 plus 4 is 18. There you go."}, {"video_title": "Basic linear equation word problem   7th grade   Khan Academy.mp3", "Sentence": "So it goes 7 times. 7 times 7 is 49. 7 times 2 is 14 plus 4 is 18. There you go. Look at that. So 190 minus 189, we get 1. Let's bring down another 0."}, {"video_title": "Basic linear equation word problem   7th grade   Khan Academy.mp3", "Sentence": "There you go. Look at that. So 190 minus 189, we get 1. Let's bring down another 0. We have a 0 right there. 27 goes into 10 how many times? Well, it doesn't go into 10 at all."}, {"video_title": "Basic linear equation word problem   7th grade   Khan Academy.mp3", "Sentence": "Let's bring down another 0. We have a 0 right there. 27 goes into 10 how many times? Well, it doesn't go into 10 at all. So we put a 0 right there. 0 times 27 is 0. Then we subtract."}, {"video_title": "Basic linear equation word problem   7th grade   Khan Academy.mp3", "Sentence": "Well, it doesn't go into 10 at all. So we put a 0 right there. 0 times 27 is 0. Then we subtract. And then we get 10 again. And now we're in the decimal range, or we're going to start getting decimal values. But we could bring down another 0."}, {"video_title": "Basic linear equation word problem   7th grade   Khan Academy.mp3", "Sentence": "Then we subtract. And then we get 10 again. And now we're in the decimal range, or we're going to start getting decimal values. But we could bring down another 0. We could get 27 goes into 100 three times. So our x value is going to be approximately 370.3. And then we're going to keep going on and on and on and on."}, {"video_title": "Basic linear equation word problem   7th grade   Khan Academy.mp3", "Sentence": "But we could bring down another 0. We could get 27 goes into 100 three times. So our x value is going to be approximately 370.3. And then we're going to keep going on and on and on and on. But this is enough information for us to answer our question. What is the minimum number of computers she needs to sell in a month to make a profit? Well, she can't sell a decimal number of computers, a third of a computer."}, {"video_title": "Basic linear equation word problem   7th grade   Khan Academy.mp3", "Sentence": "And then we're going to keep going on and on and on and on. But this is enough information for us to answer our question. What is the minimum number of computers she needs to sell in a month to make a profit? Well, she can't sell a decimal number of computers, a third of a computer. She could either sell 370 computers. If she sells 370 computers, then she's not going to get to break even, because that's less than the quantity she needs for break even. So she needs to sell 371 computers in a month to make a profit."}, {"video_title": "Checking ordered pair solutions to equations example 1   Algebra I   Khan Academy.mp3", "Sentence": "What we have to remind ourselves is when we're given an ordered pair, the first number is the x coordinate and the second number is the y coordinate, or the y value. So when they tell us the ordered pair negative four comma four, they're saying, hey look, if x is equal to negative four and y is equal to positive four, does that satisfy this equation? And what we can do is we can just try that out. So we have negative three, and everywhere where we see an x, everywhere where we see an x, we can replace it with negative four. So it's negative three times negative four minus, minus, and everywhere we see, and everywhere we see a, and everywhere we see a y, we can replace it with positive four. We replace it with positive four. So negative three times x minus y, which is four, needs to be equal to six."}, {"video_title": "Checking ordered pair solutions to equations example 1   Algebra I   Khan Academy.mp3", "Sentence": "So we have negative three, and everywhere where we see an x, everywhere where we see an x, we can replace it with negative four. So it's negative three times negative four minus, minus, and everywhere we see, and everywhere we see a, and everywhere we see a y, we can replace it with positive four. We replace it with positive four. So negative three times x minus y, which is four, needs to be equal to six. Needs to be equal to six. Now is this indeed the case? Negative three times negative four is positive 12."}, {"video_title": "Checking ordered pair solutions to equations example 1   Algebra I   Khan Academy.mp3", "Sentence": "So negative three times x minus y, which is four, needs to be equal to six. Needs to be equal to six. Now is this indeed the case? Negative three times negative four is positive 12. Positive 12 minus four, positive 12 minus four is equal to eight. It's not equal to six. Is not equal, is not equal to six."}, {"video_title": "Checking ordered pair solutions to equations example 1   Algebra I   Khan Academy.mp3", "Sentence": "Negative three times negative four is positive 12. Positive 12 minus four, positive 12 minus four is equal to eight. It's not equal to six. Is not equal, is not equal to six. So this one does not work out. So let's see, negative three comma three. So we can do the same thing here."}, {"video_title": "Checking ordered pair solutions to equations example 1   Algebra I   Khan Academy.mp3", "Sentence": "Is not equal, is not equal to six. So this one does not work out. So let's see, negative three comma three. So we can do the same thing here. Let's see what happens when x is equal to negative three and y is equal to positive three. So we substitute back in. We get negative three, negative three times x, which now we're gonna try out x being equal to negative three minus y, minus y, y is positive three here, minus y, let me do that y color, that blue, minus y now needs to be equal to, now needs to be equal, just like before, it needs to be equal to six."}, {"video_title": "Checking ordered pair solutions to equations example 1   Algebra I   Khan Academy.mp3", "Sentence": "So we can do the same thing here. Let's see what happens when x is equal to negative three and y is equal to positive three. So we substitute back in. We get negative three, negative three times x, which now we're gonna try out x being equal to negative three minus y, minus y, y is positive three here, minus y, let me do that y color, that blue, minus y now needs to be equal to, now needs to be equal, just like before, it needs to be equal to six. So negative three times negative three, that's going to be positive nine. Nine minus three is indeed equal to six. Nine minus three is indeed equal to six."}, {"video_title": "Checking ordered pair solutions to equations example 1   Algebra I   Khan Academy.mp3", "Sentence": "We get negative three, negative three times x, which now we're gonna try out x being equal to negative three minus y, minus y, y is positive three here, minus y, let me do that y color, that blue, minus y now needs to be equal to, now needs to be equal, just like before, it needs to be equal to six. So negative three times negative three, that's going to be positive nine. Nine minus three is indeed equal to six. Nine minus three is indeed equal to six. Nine minus three is six, that is equal to six. This works out. So negative three comma three is an ordered pair that is a solution to this equation."}, {"video_title": "Adding fractions with different signs   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "It's pretty straightforward. We have two positive numbers. Let me draw a number line. And I'll try to focus in. So we're going to start at 3 1 8ths. So let's make this 0. So you have 1, 2, 3, and then you have 4."}, {"video_title": "Adding fractions with different signs   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "And I'll try to focus in. So we're going to start at 3 1 8ths. So let's make this 0. So you have 1, 2, 3, and then you have 4. 3 1 8ths is going to be right about there. So let me just draw its absolute value. So this 3 1 8ths is going to be 3 1 8ths to the right of 0."}, {"video_title": "Adding fractions with different signs   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "So you have 1, 2, 3, and then you have 4. 3 1 8ths is going to be right about there. So let me just draw its absolute value. So this 3 1 8ths is going to be 3 1 8ths to the right of 0. So it's going to be exactly that distance from 0 to the right. So this right here, the length of this arrow, you could view it as 3 1 8ths. Now, whenever I like to deal with fractions, especially when they have different denominators and all of that, I like to deal with them as improper fractions."}, {"video_title": "Adding fractions with different signs   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "So this 3 1 8ths is going to be 3 1 8ths to the right of 0. So it's going to be exactly that distance from 0 to the right. So this right here, the length of this arrow, you could view it as 3 1 8ths. Now, whenever I like to deal with fractions, especially when they have different denominators and all of that, I like to deal with them as improper fractions. It makes the addition and the subtraction and actually the multiplication and the division a lot easier. So 3 1 8ths is the same thing as 8 times 3 is 24, plus 1 is 25 over 8. So this is 25 over 8, which is the same thing as 3 1 8ths."}, {"video_title": "Adding fractions with different signs   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "Now, whenever I like to deal with fractions, especially when they have different denominators and all of that, I like to deal with them as improper fractions. It makes the addition and the subtraction and actually the multiplication and the division a lot easier. So 3 1 8ths is the same thing as 8 times 3 is 24, plus 1 is 25 over 8. So this is 25 over 8, which is the same thing as 3 1 8ths. Another way to think about it, 3 is 24 over 8. And you add 1 8th to that, so you get 25 over 8. So this is our starting point."}, {"video_title": "Adding fractions with different signs   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "So this is 25 over 8, which is the same thing as 3 1 8ths. Another way to think about it, 3 is 24 over 8. And you add 1 8th to that, so you get 25 over 8. So this is our starting point. Now, to that, we're going to add 3 4ths. So we're going to move another 3 4ths. We are going to move another 3 4ths."}, {"video_title": "Adding fractions with different signs   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "So this is our starting point. Now, to that, we're going to add 3 4ths. So we're going to move another 3 4ths. We are going to move another 3 4ths. It's hard drawing these arrows. We're going to move another 3 4ths to the right. So this right here, the length of this that we're moving to the right is 3 4ths, so plus 3 4ths."}, {"video_title": "Adding fractions with different signs   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "We are going to move another 3 4ths. It's hard drawing these arrows. We're going to move another 3 4ths to the right. So this right here, the length of this that we're moving to the right is 3 4ths, so plus 3 4ths. Now, where does this put us? Well, both of these are positive integers, so we can just add them. We just have to find a like denominator."}, {"video_title": "Adding fractions with different signs   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "So this right here, the length of this that we're moving to the right is 3 4ths, so plus 3 4ths. Now, where does this put us? Well, both of these are positive integers, so we can just add them. We just have to find a like denominator. So we have 25 over 8. We have 25 over 8 plus 3 4ths. That's the same thing as we need to find a common denominator here."}, {"video_title": "Adding fractions with different signs   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "We just have to find a like denominator. So we have 25 over 8. We have 25 over 8 plus 3 4ths. That's the same thing as we need to find a common denominator here. The common denominator, or the least common multiple of 4 and 8 is 8. So it's going to be something over 8. To get from 4 to 8, we multiply by 2."}, {"video_title": "Adding fractions with different signs   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "That's the same thing as we need to find a common denominator here. The common denominator, or the least common multiple of 4 and 8 is 8. So it's going to be something over 8. To get from 4 to 8, we multiply by 2. So we have to multiply 3 by 2 as well. So you get 6. So 3 4ths is the same thing as 6 8ths."}, {"video_title": "Adding fractions with different signs   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "To get from 4 to 8, we multiply by 2. So we have to multiply 3 by 2 as well. So you get 6. So 3 4ths is the same thing as 6 8ths. If we have 25 8ths and we're adding 6 8ths to that, that gives us 25 plus 6 is 31 over 8. So this number right over here is 31 over 8. And it makes sense, because 32 over 8 would be 4."}, {"video_title": "Adding fractions with different signs   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "So 3 4ths is the same thing as 6 8ths. If we have 25 8ths and we're adding 6 8ths to that, that gives us 25 plus 6 is 31 over 8. So this number right over here is 31 over 8. And it makes sense, because 32 over 8 would be 4. So it should be a little bit less than 4. So this number right over here is 31 over 8. Or the length of this arrow, the absolute value of that number is 31 over 8."}, {"video_title": "Adding fractions with different signs   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "And it makes sense, because 32 over 8 would be 4. So it should be a little bit less than 4. So this number right over here is 31 over 8. Or the length of this arrow, the absolute value of that number is 31 over 8. A little bit less than 4. If you wanted to write that as a mixed number, it would be 3 and 7 8ths. So that's that right over here."}, {"video_title": "Adding fractions with different signs   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "Or the length of this arrow, the absolute value of that number is 31 over 8. A little bit less than 4. If you wanted to write that as a mixed number, it would be 3 and 7 8ths. So that's that right over here. This is 31 over 8. That's that part right over there. Now to that, we want to add a negative 2 and 1 6."}, {"video_title": "Adding fractions with different signs   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "So that's that right over here. This is 31 over 8. That's that part right over there. Now to that, we want to add a negative 2 and 1 6. So we're going to add a negative number. So think about what negative 2 and 1 6 is going to be like. Let me do this in a new color."}, {"video_title": "Adding fractions with different signs   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "Now to that, we want to add a negative 2 and 1 6. So we're going to add a negative number. So think about what negative 2 and 1 6 is going to be like. Let me do this in a new color. Do it in pink. Negative 2 and 1 6. So we're going to subtract, or I guess we're going to add a negative 1."}, {"video_title": "Adding fractions with different signs   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "Let me do this in a new color. Do it in pink. Negative 2 and 1 6. So we're going to subtract, or I guess we're going to add a negative 1. We're going to add a negative 2 and then a negative 1 6. Let me draw. So negative 2 and 1 6, we could literally draw like this."}, {"video_title": "Adding fractions with different signs   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "So we're going to subtract, or I guess we're going to add a negative 1. We're going to add a negative 2 and then a negative 1 6. Let me draw. So negative 2 and 1 6, we could literally draw like this. Negative 2 and 1 6, we can draw with an arrow that looks something like that. So this is negative 2 and 1 6. Now there's a couple of ways to think about it."}, {"video_title": "Adding fractions with different signs   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "So negative 2 and 1 6, we could literally draw like this. Negative 2 and 1 6, we can draw with an arrow that looks something like that. So this is negative 2 and 1 6. Now there's a couple of ways to think about it. We could just say, hey look, when you add this arrow, this thing that's moving to the left, we could put it over here and you would get straight to negative 2 and 1 6. But we're adding this negative 2 and 1 6. It's the same thing as subtracting a positive 2 and 1 6."}, {"video_title": "Adding fractions with different signs   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "Now there's a couple of ways to think about it. We could just say, hey look, when you add this arrow, this thing that's moving to the left, we could put it over here and you would get straight to negative 2 and 1 6. But we're adding this negative 2 and 1 6. It's the same thing as subtracting a positive 2 and 1 6. We're moving 2 and 1 6 to the left. And we're going to end up with a number whose absolute value is going to look something like that. And it's actually going to be to the right."}, {"video_title": "Adding fractions with different signs   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "It's the same thing as subtracting a positive 2 and 1 6. We're moving 2 and 1 6 to the left. And we're going to end up with a number whose absolute value is going to look something like that. And it's actually going to be to the right. So it's not going to only be its absolute value. It's going to be what's absolute value is going to be the number, since it's going to be a positive number. So let's just think about what it is."}, {"video_title": "Adding fractions with different signs   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "And it's actually going to be to the right. So it's not going to only be its absolute value. It's going to be what's absolute value is going to be the number, since it's going to be a positive number. So let's just think about what it is. This value right here, which is going to be the answer to our problem, is just going to be the difference of 31 over 8 and 2 and 1 6. And it's the positive difference, because we're dealing with a positive number. So we just take 31 over 8."}, {"video_title": "Adding fractions with different signs   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "So let's just think about what it is. This value right here, which is going to be the answer to our problem, is just going to be the difference of 31 over 8 and 2 and 1 6. And it's the positive difference, because we're dealing with a positive number. So we just take 31 over 8. And from that, we will subtract 2 and 1 6. So let's do this. So this orange value is going to be 31 over 8 minus 2 and 1 6."}, {"video_title": "Adding fractions with different signs   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "So we just take 31 over 8. And from that, we will subtract 2 and 1 6. So let's do this. So this orange value is going to be 31 over 8 minus 2 and 1 6. So 2 and 1 6 is the same thing as 6 times 2 is 12 plus 1 is 13 minus 13 over 6. And this is equal to, once again, we need to get a common denominator over here. And it looks like 24 will be the common denominator."}, {"video_title": "Adding fractions with different signs   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "So this orange value is going to be 31 over 8 minus 2 and 1 6. So 2 and 1 6 is the same thing as 6 times 2 is 12 plus 1 is 13 minus 13 over 6. And this is equal to, once again, we need to get a common denominator over here. And it looks like 24 will be the common denominator. Let me make it very clear. This is the 31 over 8. And this is the 2 and 1 6."}, {"video_title": "Adding fractions with different signs   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "And it looks like 24 will be the common denominator. Let me make it very clear. This is the 31 over 8. And this is the 2 and 1 6. This right here is the 2 and 1 6. So 31 over 8 over 24, you have to multiply by 3 to get to the 24 over here. So we multiply by 3 on the 31."}, {"video_title": "Adding fractions with different signs   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "And this is the 2 and 1 6. This right here is the 2 and 1 6. So 31 over 8 over 24, you have to multiply by 3 to get to the 24 over here. So we multiply by 3 on the 31. That gives us 93. And then to go from 6 to 24, you have to multiply by 4. Let me do that in another color."}, {"video_title": "Adding fractions with different signs   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "So we multiply by 3 on the 31. That gives us 93. And then to go from 6 to 24, you have to multiply by 4. Let me do that in another color. You have to multiply it by 4. So we have to multiply by 4 up here as well. So 4 times 13."}, {"video_title": "Adding fractions with different signs   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "Let me do that in another color. You have to multiply it by 4. So we have to multiply by 4 up here as well. So 4 times 13. Let's see. 4 times 10 is 40. 4 times 3 is 12."}, {"video_title": "Adding fractions with different signs   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "So 4 times 13. Let's see. 4 times 10 is 40. 4 times 3 is 12. So that's 52. So this is going to be equal to 93 minus 52 over 24. And that is, so 93 minus 52."}, {"video_title": "Adding fractions with different signs   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "4 times 3 is 12. So that's 52. So this is going to be equal to 93 minus 52 over 24. And that is, so 93 minus 52. 3 minus 2 is 1. 9 minus 5 is 4. So it is 41 over 24."}, {"video_title": "Adding fractions with different signs   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "And that is, so 93 minus 52. 3 minus 2 is 1. 9 minus 5 is 4. So it is 41 over 24. Positive. And you can see that here just by looking at the number line. This right here is 41 over 24."}, {"video_title": "Adding fractions with different signs   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "So it is 41 over 24. Positive. And you can see that here just by looking at the number line. This right here is 41 over 24. And it should be a little bit less than 2, because 2 would be 48 over 24. So this would be 48 over 24. So it makes sense that we're a little bit less than that."}, {"video_title": "Solving system with elimination   Algebra   Khan Academy.mp3", "Sentence": "And the second equation, negative x plus three y is equal to 11. Now what we're gonna do is find an x and y pair that satisfies both of these equations. That's what solving the system actually means. As you might already have seen, there's a bunch of x and y pairs that satisfy this first equation. In fact, if you were to graph them, they would form a line. And there's a bunch of other x and y pairs that satisfy this other equation, the second equation. And if you were to graph them, it would form a line."}, {"video_title": "Solving system with elimination   Algebra   Khan Academy.mp3", "Sentence": "As you might already have seen, there's a bunch of x and y pairs that satisfy this first equation. In fact, if you were to graph them, they would form a line. And there's a bunch of other x and y pairs that satisfy this other equation, the second equation. And if you were to graph them, it would form a line. And so if you find the x and y pair that satisfy both, that would be the intersection of the lines. So let's do that. So actually, I'm just gonna rewrite the first equation over here."}, {"video_title": "Solving system with elimination   Algebra   Khan Academy.mp3", "Sentence": "And if you were to graph them, it would form a line. And so if you find the x and y pair that satisfy both, that would be the intersection of the lines. So let's do that. So actually, I'm just gonna rewrite the first equation over here. So I'm gonna write x minus four y is equal to negative 18. So we've already seen in algebra that as long as we do the same thing to both sides of the equation, we can maintain our equality. So what if we were to add, and our goal here is to eliminate one of the variables."}, {"video_title": "Solving system with elimination   Algebra   Khan Academy.mp3", "Sentence": "So actually, I'm just gonna rewrite the first equation over here. So I'm gonna write x minus four y is equal to negative 18. So we've already seen in algebra that as long as we do the same thing to both sides of the equation, we can maintain our equality. So what if we were to add, and our goal here is to eliminate one of the variables. So we have one equation with one unknown. So what if we were to add this negative x plus three y to the left-hand side here? So negative x plus three y."}, {"video_title": "Solving system with elimination   Algebra   Khan Academy.mp3", "Sentence": "So what if we were to add, and our goal here is to eliminate one of the variables. So we have one equation with one unknown. So what if we were to add this negative x plus three y to the left-hand side here? So negative x plus three y. Well, that looks pretty good because an x and a negative x are going to cancel out. And we are going to be left with negative four y plus three y. Well, that's just going to be negative y."}, {"video_title": "Solving system with elimination   Algebra   Khan Academy.mp3", "Sentence": "So negative x plus three y. Well, that looks pretty good because an x and a negative x are going to cancel out. And we are going to be left with negative four y plus three y. Well, that's just going to be negative y. So by adding the left-hand side of this bottom equation to the left-hand side of the top equation, we were able to cancel out the x's. We had x and we had a negative x. That was very nice for us."}, {"video_title": "Solving system with elimination   Algebra   Khan Academy.mp3", "Sentence": "Well, that's just going to be negative y. So by adding the left-hand side of this bottom equation to the left-hand side of the top equation, we were able to cancel out the x's. We had x and we had a negative x. That was very nice for us. So what do we do on the right-hand side? We've already said that we have to add the same thing to both sides of an equation. We might be tempted, we might be tempted to just say, well, if I have to add the same thing to both sides, well, maybe I have to add a negative x plus three y to that side."}, {"video_title": "Solving system with elimination   Algebra   Khan Academy.mp3", "Sentence": "That was very nice for us. So what do we do on the right-hand side? We've already said that we have to add the same thing to both sides of an equation. We might be tempted, we might be tempted to just say, well, if I have to add the same thing to both sides, well, maybe I have to add a negative x plus three y to that side. But that's not going to help us much. We're going to have negative 18 minus x plus three y. We would have introduced an x on the right-hand side of the equation."}, {"video_title": "Solving system with elimination   Algebra   Khan Academy.mp3", "Sentence": "We might be tempted, we might be tempted to just say, well, if I have to add the same thing to both sides, well, maybe I have to add a negative x plus three y to that side. But that's not going to help us much. We're going to have negative 18 minus x plus three y. We would have introduced an x on the right-hand side of the equation. But what if we could add something that's equivalent to negative x plus three y that does not introduce the x variable? Well, we know that the number 11 is equivalent to negative x plus three y. How do we know that?"}, {"video_title": "Solving system with elimination   Algebra   Khan Academy.mp3", "Sentence": "We would have introduced an x on the right-hand side of the equation. But what if we could add something that's equivalent to negative x plus three y that does not introduce the x variable? Well, we know that the number 11 is equivalent to negative x plus three y. How do we know that? Well, that second equation tells us that. So once again, all I'm doing is I'm adding the same thing to both sides of that top equation. On the left, I'm expressing it as negative x plus three y."}, {"video_title": "Solving system with elimination   Algebra   Khan Academy.mp3", "Sentence": "How do we know that? Well, that second equation tells us that. So once again, all I'm doing is I'm adding the same thing to both sides of that top equation. On the left, I'm expressing it as negative x plus three y. But the second equation tells us that negative x plus three y is going to be equal to 11. It's introducing that second constraint. And so let's add 11 to the right-hand side, which is, once again, I know I keep repeating it, it's the same thing as negative x plus three y."}, {"video_title": "Solving system with elimination   Algebra   Khan Academy.mp3", "Sentence": "On the left, I'm expressing it as negative x plus three y. But the second equation tells us that negative x plus three y is going to be equal to 11. It's introducing that second constraint. And so let's add 11 to the right-hand side, which is, once again, I know I keep repeating it, it's the same thing as negative x plus three y. So negative 18 plus 11 is negative seven. And since we added the same thing to both sides, the equality still holds. And we get negative y is equal to negative seven."}, {"video_title": "Solving system with elimination   Algebra   Khan Academy.mp3", "Sentence": "And so let's add 11 to the right-hand side, which is, once again, I know I keep repeating it, it's the same thing as negative x plus three y. So negative 18 plus 11 is negative seven. And since we added the same thing to both sides, the equality still holds. And we get negative y is equal to negative seven. Or divide both sides by negative one or multiply both sides by negative one. So multiply both sides by negative one. We get y is equal to seven."}, {"video_title": "Solving system with elimination   Algebra   Khan Academy.mp3", "Sentence": "And we get negative y is equal to negative seven. Or divide both sides by negative one or multiply both sides by negative one. So multiply both sides by negative one. We get y is equal to seven. So we have the y coordinate of the xy pair that satisfies both of these. Now how do we find the x? Well, we can just substitute this y equals seven to either one of these."}, {"video_title": "Solving system with elimination   Algebra   Khan Academy.mp3", "Sentence": "We get y is equal to seven. So we have the y coordinate of the xy pair that satisfies both of these. Now how do we find the x? Well, we can just substitute this y equals seven to either one of these. When y equals seven, we should get the same x regardless of which equation we use. So let's use the top equation. So we know that x minus four times, instead of writing y, I'm gonna write four times seven, because we're gonna figure out what is x when y is seven."}, {"video_title": "Solving system with elimination   Algebra   Khan Academy.mp3", "Sentence": "Well, we can just substitute this y equals seven to either one of these. When y equals seven, we should get the same x regardless of which equation we use. So let's use the top equation. So we know that x minus four times, instead of writing y, I'm gonna write four times seven, because we're gonna figure out what is x when y is seven. That is going to be equal to negative 18. And so let's see, negative four, or four times seven, that is 28. So let's see, I could, to solve for x, I could add 28 to both sides."}, {"video_title": "Solving system with elimination   Algebra   Khan Academy.mp3", "Sentence": "So we know that x minus four times, instead of writing y, I'm gonna write four times seven, because we're gonna figure out what is x when y is seven. That is going to be equal to negative 18. And so let's see, negative four, or four times seven, that is 28. So let's see, I could, to solve for x, I could add 28 to both sides. So add 28 to both sides. On the left-hand side, negative 28, positive 28, those cancel out, I'm just left with an x. And on the right-hand side, I get negative eight, negative 18 plus 28 is 10."}, {"video_title": "Solving system with elimination   Algebra   Khan Academy.mp3", "Sentence": "So let's see, I could, to solve for x, I could add 28 to both sides. So add 28 to both sides. On the left-hand side, negative 28, positive 28, those cancel out, I'm just left with an x. And on the right-hand side, I get negative eight, negative 18 plus 28 is 10. So there you have it. I have the xy pair that satisfies both. X equals 10, y equals seven."}, {"video_title": "Solving system with elimination   Algebra   Khan Academy.mp3", "Sentence": "And on the right-hand side, I get negative eight, negative 18 plus 28 is 10. So there you have it. I have the xy pair that satisfies both. X equals 10, y equals seven. I could write it here. So I could write it as coordinates. I could write it as 10 comma seven."}, {"video_title": "Solving system with elimination   Algebra   Khan Academy.mp3", "Sentence": "X equals 10, y equals seven. I could write it here. So I could write it as coordinates. I could write it as 10 comma seven. And notice, what I just did here, I encourage you to substitute y equals seven here, and you will also get x equals 10. Either way, you would have come to x equals 10. And to visualize what is going on here, let's visualize it really fast."}, {"video_title": "Solving system with elimination   Algebra   Khan Academy.mp3", "Sentence": "I could write it as 10 comma seven. And notice, what I just did here, I encourage you to substitute y equals seven here, and you will also get x equals 10. Either way, you would have come to x equals 10. And to visualize what is going on here, let's visualize it really fast. Let me draw some coordinate axes. Whoops, I meant to draw a straighter line than that. All right, there you go."}, {"video_title": "Solving system with elimination   Algebra   Khan Academy.mp3", "Sentence": "And to visualize what is going on here, let's visualize it really fast. Let me draw some coordinate axes. Whoops, I meant to draw a straighter line than that. All right, there you go. So let's say that is our y-axis, and that is, whoops, that is our x-axis. And then, let's see, the top equation is gonna look something like this. It's gonna look something like this."}, {"video_title": "Solving system with elimination   Algebra   Khan Academy.mp3", "Sentence": "All right, there you go. So let's say that is our y-axis, and that is, whoops, that is our x-axis. And then, let's see, the top equation is gonna look something like this. It's gonna look something like this. And then that bottom equation is gonna look something, something like, let me draw it a little bit nicer than that. It's gonna look something like this, something like that. Let me draw that bottom one here so you see the point of intersection."}, {"video_title": "Solving system with elimination   Algebra   Khan Academy.mp3", "Sentence": "It's gonna look something like this. And then that bottom equation is gonna look something, something like, let me draw it a little bit nicer than that. It's gonna look something like this, something like that. Let me draw that bottom one here so you see the point of intersection. And so the point of intersection right over here, that is an xy pair that satisfies both of these equations. And that, we just saw, it happens when x is equal to 10 and y is equal to seven. Once again, this white line, that's all the x and y pairs that satisfy the top equation."}, {"video_title": "Solving system with elimination   Algebra   Khan Academy.mp3", "Sentence": "Let me draw that bottom one here so you see the point of intersection. And so the point of intersection right over here, that is an xy pair that satisfies both of these equations. And that, we just saw, it happens when x is equal to 10 and y is equal to seven. Once again, this white line, that's all the x and y pairs that satisfy the top equation. This orange line, that's all the x and y pairs that satisfy the orange equation. And where they intersect, that point is on both lines. It satisfies both equations."}, {"video_title": "Solving a quadratic equation by factoring   Algebra II   Khan Academy.mp3", "Sentence": "Now, if this is the first time that you've seen this type of what's essentially a quadratic equation, you might be tempted to try to solve for s using traditional algebraic means, but the best way to solve this, especially when it's explicitly equal to 0, is to factor the left-hand side. And then think about the fact that those binomials that you factor into, that they have to be equal to 0. So let's just do that. So how can we factor this? We've seen it several ways. I'll show you kind of the standard way we've been doing it, by grouping. And then there's a little bit of a shortcut when you have a 1 as a coefficient over here."}, {"video_title": "Solving a quadratic equation by factoring   Algebra II   Khan Academy.mp3", "Sentence": "So how can we factor this? We've seen it several ways. I'll show you kind of the standard way we've been doing it, by grouping. And then there's a little bit of a shortcut when you have a 1 as a coefficient over here. So when you do something by grouping, when you factor by grouping, you think about two numbers whose sum is going to be equal to negative 2. So you think about two numbers whose sum, a plus b, is equal to negative 2, and whose product is going to be equal to negative 35. Since their product is a negative number, one has to be positive, one has to be negative."}, {"video_title": "Solving a quadratic equation by factoring   Algebra II   Khan Academy.mp3", "Sentence": "And then there's a little bit of a shortcut when you have a 1 as a coefficient over here. So when you do something by grouping, when you factor by grouping, you think about two numbers whose sum is going to be equal to negative 2. So you think about two numbers whose sum, a plus b, is equal to negative 2, and whose product is going to be equal to negative 35. Since their product is a negative number, one has to be positive, one has to be negative. And so if you think about it, ones that are about two apart, you have 5 and negative 7. I think that'll work. 5 plus negative 7 is equal to negative 2."}, {"video_title": "Solving a quadratic equation by factoring   Algebra II   Khan Academy.mp3", "Sentence": "Since their product is a negative number, one has to be positive, one has to be negative. And so if you think about it, ones that are about two apart, you have 5 and negative 7. I think that'll work. 5 plus negative 7 is equal to negative 2. So to factor by grouping, you split this middle term into a let me write it this way, we have s squared, and then this middle term right here, I'll do it in pink, this middle term right there, I can write it as plus 5s minus 7s, and then we have the minus 35. And of course, all of that is equal to 0. Now we call it factoring by grouping because we group it."}, {"video_title": "Solving a quadratic equation by factoring   Algebra II   Khan Academy.mp3", "Sentence": "5 plus negative 7 is equal to negative 2. So to factor by grouping, you split this middle term into a let me write it this way, we have s squared, and then this middle term right here, I'll do it in pink, this middle term right there, I can write it as plus 5s minus 7s, and then we have the minus 35. And of course, all of that is equal to 0. Now we call it factoring by grouping because we group it. So we can group these first two terms. And these first two terms, they have a common factor of s. So let's factor that out. You have s times s plus 5."}, {"video_title": "Solving a quadratic equation by factoring   Algebra II   Khan Academy.mp3", "Sentence": "Now we call it factoring by grouping because we group it. So we can group these first two terms. And these first two terms, they have a common factor of s. So let's factor that out. You have s times s plus 5. That's the same thing as s squared plus 5s. Now in these second two terms right here, you have a common factor of negative 7. So let's factor that out."}, {"video_title": "Solving a quadratic equation by factoring   Algebra II   Khan Academy.mp3", "Sentence": "You have s times s plus 5. That's the same thing as s squared plus 5s. Now in these second two terms right here, you have a common factor of negative 7. So let's factor that out. So you have negative 7 times s plus 5. And of course, all of that is equal to 0. Now we have two terms here where both of them have s plus 5 as a factor."}, {"video_title": "Solving a quadratic equation by factoring   Algebra II   Khan Academy.mp3", "Sentence": "So let's factor that out. So you have negative 7 times s plus 5. And of course, all of that is equal to 0. Now we have two terms here where both of them have s plus 5 as a factor. Both of them have this s plus 5 as a factor. So we can factor that out. So let's do that."}, {"video_title": "Solving a quadratic equation by factoring   Algebra II   Khan Academy.mp3", "Sentence": "Now we have two terms here where both of them have s plus 5 as a factor. Both of them have this s plus 5 as a factor. So we can factor that out. So let's do that. So you have s plus 5 times this s times this s right here. s plus 5 times s would give you this term. And then you have minus that 7 right there."}, {"video_title": "Solving a quadratic equation by factoring   Algebra II   Khan Academy.mp3", "Sentence": "So let's do that. So you have s plus 5 times this s times this s right here. s plus 5 times s would give you this term. And then you have minus that 7 right there. I undistributed the s plus 5. And then this is going to be equal to 0. Now that we've factored it, we just have to think a little bit about what happens when you take the product of two numbers."}, {"video_title": "Solving a quadratic equation by factoring   Algebra II   Khan Academy.mp3", "Sentence": "And then you have minus that 7 right there. I undistributed the s plus 5. And then this is going to be equal to 0. Now that we've factored it, we just have to think a little bit about what happens when you take the product of two numbers. I mean, s plus 5 is a number. s minus 7 is another number. And we're saying that the product of those two numbers is equal to 0."}, {"video_title": "Solving a quadratic equation by factoring   Algebra II   Khan Academy.mp3", "Sentence": "Now that we've factored it, we just have to think a little bit about what happens when you take the product of two numbers. I mean, s plus 5 is a number. s minus 7 is another number. And we're saying that the product of those two numbers is equal to 0. If I were told you that I had two numbers, if I told you that I had the numbers a times b and that they equal to 0, what do we know about either a or b or both of them? Well, at least one of them has to be equal to 0, or both of them have to be equal to 0. So the fact that this number times that number is equal to 0 tells us that either s plus 5 is equal to 0, or s minus 7 is equal to 0."}, {"video_title": "Solving a quadratic equation by factoring   Algebra II   Khan Academy.mp3", "Sentence": "And we're saying that the product of those two numbers is equal to 0. If I were told you that I had two numbers, if I told you that I had the numbers a times b and that they equal to 0, what do we know about either a or b or both of them? Well, at least one of them has to be equal to 0, or both of them have to be equal to 0. So the fact that this number times that number is equal to 0 tells us that either s plus 5 is equal to 0, or s minus 7 is equal to 0. And so you have these two equations. Actually, we could say and or. It could be or and either way."}, {"video_title": "Solving a quadratic equation by factoring   Algebra II   Khan Academy.mp3", "Sentence": "So the fact that this number times that number is equal to 0 tells us that either s plus 5 is equal to 0, or s minus 7 is equal to 0. And so you have these two equations. Actually, we could say and or. It could be or and either way. Both of them could be equal to 0. So let's see how we can solve for this. Well, we could just subtract 5 from both sides of this equation right there."}, {"video_title": "Solving a quadratic equation by factoring   Algebra II   Khan Academy.mp3", "Sentence": "It could be or and either way. Both of them could be equal to 0. So let's see how we can solve for this. Well, we could just subtract 5 from both sides of this equation right there. And so you get on the left-hand side, you have s is equal to negative 5. That is one solution to the equation. Or you have, let's see, you can add 7 to both sides of that equation."}, {"video_title": "Solving a quadratic equation by factoring   Algebra II   Khan Academy.mp3", "Sentence": "Well, we could just subtract 5 from both sides of this equation right there. And so you get on the left-hand side, you have s is equal to negative 5. That is one solution to the equation. Or you have, let's see, you can add 7 to both sides of that equation. And you get s is equal to 7. So if s is equal to negative 5 or s is equal to 7, then we have satisfied this equation. We can even verify it."}, {"video_title": "Solving a quadratic equation by factoring   Algebra II   Khan Academy.mp3", "Sentence": "Or you have, let's see, you can add 7 to both sides of that equation. And you get s is equal to 7. So if s is equal to negative 5 or s is equal to 7, then we have satisfied this equation. We can even verify it. If you make s equal to negative 5, you have positive 25 plus 10, which is 35, minus 35. That does equal 0. If you have 7, 49 minus 14 minus 35 does equal 0."}, {"video_title": "Solving a quadratic equation by factoring   Algebra II   Khan Academy.mp3", "Sentence": "We can even verify it. If you make s equal to negative 5, you have positive 25 plus 10, which is 35, minus 35. That does equal 0. If you have 7, 49 minus 14 minus 35 does equal 0. So we've solved for s. Now I mentioned there's an easier way to do it. And when you have something like this, where you have 1 as the leading coefficient, you don't have to do this two-step factoring. Let me just show you an example."}, {"video_title": "Solving a quadratic equation by factoring   Algebra II   Khan Academy.mp3", "Sentence": "If you have 7, 49 minus 14 minus 35 does equal 0. So we've solved for s. Now I mentioned there's an easier way to do it. And when you have something like this, where you have 1 as the leading coefficient, you don't have to do this two-step factoring. Let me just show you an example. If I just have x plus a times x plus b, what does that equal to? x times x is x squared. x times b is bx."}, {"video_title": "Solving a quadratic equation by factoring   Algebra II   Khan Academy.mp3", "Sentence": "Let me just show you an example. If I just have x plus a times x plus b, what does that equal to? x times x is x squared. x times b is bx. a times x is plus ax. a times b is ab. So you get x squared plus, these two can be added, plus a plus bx plus ab."}, {"video_title": "Solving a quadratic equation by factoring   Algebra II   Khan Academy.mp3", "Sentence": "x times b is bx. a times x is plus ax. a times b is ab. So you get x squared plus, these two can be added, plus a plus bx plus ab. And that's the pattern that we have right here. We have 1 as a leading coefficient here. We have 1 as a leading coefficient here."}, {"video_title": "Solving a quadratic equation by factoring   Algebra II   Khan Academy.mp3", "Sentence": "So you get x squared plus, these two can be added, plus a plus bx plus ab. And that's the pattern that we have right here. We have 1 as a leading coefficient here. We have 1 as a leading coefficient here. So once we have our two numbers that add up to negative 2, that's our a plus b. And we have our product that gets to negative 35. Then we can straight just factor it into the product of those two things."}, {"video_title": "Solving a quadratic equation by factoring   Algebra II   Khan Academy.mp3", "Sentence": "We have 1 as a leading coefficient here. So once we have our two numbers that add up to negative 2, that's our a plus b. And we have our product that gets to negative 35. Then we can straight just factor it into the product of those two things. Or the product of the binomials where those will be the a's and the b's. So we figured out it's 5 and negative 7. 5 plus negative 7 is negative 2."}, {"video_title": "Solving a quadratic equation by factoring   Algebra II   Khan Academy.mp3", "Sentence": "Then we can straight just factor it into the product of those two things. Or the product of the binomials where those will be the a's and the b's. So we figured out it's 5 and negative 7. 5 plus negative 7 is negative 2. 5 times negative 7 is negative 35. So we could have just straight factored at this point to, well actually this was a case of s. So we could have factored straight to the case of s plus 5 times s minus 7. We could have done that straight away and we've gotten to that right there."}, {"video_title": "Introduction to slope   Algebra I   Khan Academy.mp3", "Sentence": "As we start to graph lines, we might notice that there are differences between lines. For example, this pink or this magenta line here, it looks steeper than this blue line. And what we'll see is this notion of steepness, how steep a line is, how quickly does it increase or how quickly does it decrease, is a really useful idea in mathematics. So ideally, we'd be able to assign a number to each of these lines or to any line that describes how steep it is, how quickly does it increase or decrease. So what's a reasonable way to do that? What's a reasonable way to assign a number to these lines that describe their steepness? Well, one way to think about it could say, well, how much does a line increase for, how much does a line increase in the vertical direction for a given increase in the horizontal direction?"}, {"video_title": "Introduction to slope   Algebra I   Khan Academy.mp3", "Sentence": "So ideally, we'd be able to assign a number to each of these lines or to any line that describes how steep it is, how quickly does it increase or decrease. So what's a reasonable way to do that? What's a reasonable way to assign a number to these lines that describe their steepness? Well, one way to think about it could say, well, how much does a line increase for, how much does a line increase in the vertical direction for a given increase in the horizontal direction? So let's write this down. So let's say if we, if an increase in vertical for a given increase in horizontal, for a given increase in horizontal. So how could this give us a value?"}, {"video_title": "Introduction to slope   Algebra I   Khan Academy.mp3", "Sentence": "Well, one way to think about it could say, well, how much does a line increase for, how much does a line increase in the vertical direction for a given increase in the horizontal direction? So let's write this down. So let's say if we, if an increase in vertical for a given increase in horizontal, for a given increase in horizontal. So how could this give us a value? Well, let's look at that magenta line again. Now, let's just start at an arbitrary point to that magenta line, but I'll start at a point where it's gonna be easy for me to figure out what point we're at. So if we were to start right here, and if I were to increase in the horizontal direction by one, so I move one to the right, to get back on the line, how much do I have to increase in the vertical direction?"}, {"video_title": "Introduction to slope   Algebra I   Khan Academy.mp3", "Sentence": "So how could this give us a value? Well, let's look at that magenta line again. Now, let's just start at an arbitrary point to that magenta line, but I'll start at a point where it's gonna be easy for me to figure out what point we're at. So if we were to start right here, and if I were to increase in the horizontal direction by one, so I move one to the right, to get back on the line, how much do I have to increase in the vertical direction? Well, I have to increase in the vertical direction by two, by two. So at least for this magenta line, it looks like our increase in vertical is two. Whenever we have an increase in one in the horizontal direction."}, {"video_title": "Introduction to slope   Algebra I   Khan Academy.mp3", "Sentence": "So if we were to start right here, and if I were to increase in the horizontal direction by one, so I move one to the right, to get back on the line, how much do I have to increase in the vertical direction? Well, I have to increase in the vertical direction by two, by two. So at least for this magenta line, it looks like our increase in vertical is two. Whenever we have an increase in one in the horizontal direction. And we could, let's see, does that apply? Let's see, does that still work? If I were to go, if I were to start here, instead of increasing the horizontal direction by one, if I were to increase in the horizontal direction, let's increase by three."}, {"video_title": "Introduction to slope   Algebra I   Khan Academy.mp3", "Sentence": "Whenever we have an increase in one in the horizontal direction. And we could, let's see, does that apply? Let's see, does that still work? If I were to go, if I were to start here, instead of increasing the horizontal direction by one, if I were to increase in the horizontal direction, let's increase by three. So now, I've gone plus three in the horizontal direction. Then to get back on the line, how much do I have to increase in the vertical direction? I have to increase by one, two, three, four, five, six."}, {"video_title": "Introduction to slope   Algebra I   Khan Academy.mp3", "Sentence": "If I were to go, if I were to start here, instead of increasing the horizontal direction by one, if I were to increase in the horizontal direction, let's increase by three. So now, I've gone plus three in the horizontal direction. Then to get back on the line, how much do I have to increase in the vertical direction? I have to increase by one, two, three, four, five, six. I have to increase by six, so plus six. So when I increase by three in the horizontal direction, I increase by six in the vertical. We're just saying, hey, let's just measure how much we increase in vertical for a given increase in horizontal."}, {"video_title": "Introduction to slope   Algebra I   Khan Academy.mp3", "Sentence": "I have to increase by one, two, three, four, five, six. I have to increase by six, so plus six. So when I increase by three in the horizontal direction, I increase by six in the vertical. We're just saying, hey, let's just measure how much we increase in vertical for a given increase in horizontal. Well, two over one is just two, and that's the same thing as six over three. So no matter where I start on this line, no matter where I start on this line, if I take and if I increase in the horizontal direction by a given amount, I'm gonna increase twice as much, twice as much in the vertical direction, twice as much in the vertical direction. So this notion of increase in vertical divided by increase in horizontal, this is what mathematicians use to describe the steepness of lines, and this is called the slope."}, {"video_title": "Introduction to slope   Algebra I   Khan Academy.mp3", "Sentence": "We're just saying, hey, let's just measure how much we increase in vertical for a given increase in horizontal. Well, two over one is just two, and that's the same thing as six over three. So no matter where I start on this line, no matter where I start on this line, if I take and if I increase in the horizontal direction by a given amount, I'm gonna increase twice as much, twice as much in the vertical direction, twice as much in the vertical direction. So this notion of increase in vertical divided by increase in horizontal, this is what mathematicians use to describe the steepness of lines, and this is called the slope. So this is called the slope of a line. And you're probably familiar with the notion of the word slope being used for, say, a ski slope, and that's because a ski slope has a certain inclination. It could have a steep slope or a shallow slope."}, {"video_title": "Introduction to slope   Algebra I   Khan Academy.mp3", "Sentence": "So this notion of increase in vertical divided by increase in horizontal, this is what mathematicians use to describe the steepness of lines, and this is called the slope. So this is called the slope of a line. And you're probably familiar with the notion of the word slope being used for, say, a ski slope, and that's because a ski slope has a certain inclination. It could have a steep slope or a shallow slope. So slope is a measure for how steep something is, and the convention is we measure the increase in vertical for a given increase in horizontal. So two over one is equal to six over three is equal to two. This is equal to the slope of this magenta line."}, {"video_title": "Introduction to slope   Algebra I   Khan Academy.mp3", "Sentence": "It could have a steep slope or a shallow slope. So slope is a measure for how steep something is, and the convention is we measure the increase in vertical for a given increase in horizontal. So two over one is equal to six over three is equal to two. This is equal to the slope of this magenta line. So slope, so let me write this down. So this slope right over here, the slope of that line is going to be equal to two. And one way to interpret that, if you, for whatever amount you increase in the horizontal direction, you're going to increase twice as much in the vertical direction."}, {"video_title": "Introduction to slope   Algebra I   Khan Academy.mp3", "Sentence": "This is equal to the slope of this magenta line. So slope, so let me write this down. So this slope right over here, the slope of that line is going to be equal to two. And one way to interpret that, if you, for whatever amount you increase in the horizontal direction, you're going to increase twice as much in the vertical direction. Now what about this blue line here? What would be the slope of the blue line? Well, let me rewrite another way that you'll typically see the definition of slope, and this is just the convention that mathematicians have defined for slope, but it's a valuable one."}, {"video_title": "Introduction to slope   Algebra I   Khan Academy.mp3", "Sentence": "And one way to interpret that, if you, for whatever amount you increase in the horizontal direction, you're going to increase twice as much in the vertical direction. Now what about this blue line here? What would be the slope of the blue line? Well, let me rewrite another way that you'll typically see the definition of slope, and this is just the convention that mathematicians have defined for slope, but it's a valuable one. What is our change in vertical for a given change in horizontal? And I'll introduce a new notation for you. So change in vertical, and in this coordinate it's going to be, the vertical is our y-coordinate, divided by our change in horizontal."}, {"video_title": "Introduction to slope   Algebra I   Khan Academy.mp3", "Sentence": "Well, let me rewrite another way that you'll typically see the definition of slope, and this is just the convention that mathematicians have defined for slope, but it's a valuable one. What is our change in vertical for a given change in horizontal? And I'll introduce a new notation for you. So change in vertical, and in this coordinate it's going to be, the vertical is our y-coordinate, divided by our change in horizontal. And x is our horizontal coordinate in this coordinate plane right over here. And you might say, wait, wait, wait, you said change in, but then you drew this triangle. Well, this triangle, this is the Greek letter delta."}, {"video_title": "Introduction to slope   Algebra I   Khan Academy.mp3", "Sentence": "So change in vertical, and in this coordinate it's going to be, the vertical is our y-coordinate, divided by our change in horizontal. And x is our horizontal coordinate in this coordinate plane right over here. And you might say, wait, wait, wait, you said change in, but then you drew this triangle. Well, this triangle, this is the Greek letter delta. This is the Greek letter delta, and it's a math symbol used to represent change in. So that's delta, delta. And it literally means change in y, change in y, divided by change in x."}, {"video_title": "Introduction to slope   Algebra I   Khan Academy.mp3", "Sentence": "Well, this triangle, this is the Greek letter delta. This is the Greek letter delta, and it's a math symbol used to represent change in. So that's delta, delta. And it literally means change in y, change in y, divided by change in x. Change in x. So if we want to find the slope of the blue line, we just have to say, well, how much does y change for a given change in x? So the slope of the blue line."}, {"video_title": "Introduction to slope   Algebra I   Khan Academy.mp3", "Sentence": "And it literally means change in y, change in y, divided by change in x. Change in x. So if we want to find the slope of the blue line, we just have to say, well, how much does y change for a given change in x? So the slope of the blue line. So let's see, let's see, actually let me do this, let me do it this way. Let me have a, when if I, let's just start at some point here, and let's say if my x changes by two, so my delta x is equal to positive two, what's my delta y going to be? What's going to be my change in y?"}, {"video_title": "Introduction to slope   Algebra I   Khan Academy.mp3", "Sentence": "So the slope of the blue line. So let's see, let's see, actually let me do this, let me do it this way. Let me have a, when if I, let's just start at some point here, and let's say if my x changes by two, so my delta x is equal to positive two, what's my delta y going to be? What's going to be my change in y? Well, if I go to the right by two, to get back on the line, I have to increase my y by two. So my change in y is also going to be plus two. So the slope of this blue line, the slope of the blue line, which is change in y over change in x, we just saw that when our change in x is positive two, our change in y is also positive two."}, {"video_title": "Introduction to slope   Algebra I   Khan Academy.mp3", "Sentence": "What's going to be my change in y? Well, if I go to the right by two, to get back on the line, I have to increase my y by two. So my change in y is also going to be plus two. So the slope of this blue line, the slope of the blue line, which is change in y over change in x, we just saw that when our change in x is positive two, our change in y is also positive two. So our slope is two divided by two, which is equal to one. Which tells us, however much we increase in x, we're going to increase that same amount in y. And you see that, you increase one in x, you increase one in y."}, {"video_title": "Introduction to slope   Algebra I   Khan Academy.mp3", "Sentence": "So the slope of this blue line, the slope of the blue line, which is change in y over change in x, we just saw that when our change in x is positive two, our change in y is also positive two. So our slope is two divided by two, which is equal to one. Which tells us, however much we increase in x, we're going to increase that same amount in y. And you see that, you increase one in x, you increase one in y. Increase one in x, increase one in y. From any point on the line, that's going to be true. You increase three in x, you're going to increase three in y."}, {"video_title": "Introduction to slope   Algebra I   Khan Academy.mp3", "Sentence": "And you see that, you increase one in x, you increase one in y. Increase one in x, increase one in y. From any point on the line, that's going to be true. You increase three in x, you're going to increase three in y. It's actually true the other way. If you decrease one in x, you're going to decrease one in y. If you decrease two in x, you're going to decrease two in y."}, {"video_title": "Introduction to slope   Algebra I   Khan Academy.mp3", "Sentence": "You increase three in x, you're going to increase three in y. It's actually true the other way. If you decrease one in x, you're going to decrease one in y. If you decrease two in x, you're going to decrease two in y. And that makes sense from the math of it as well. Because if your change in x is negative two, that's what we did right over here, our change in x was negative two, we went two back, then your change in y is going to be negative two as well. Your change in y is going to be negative two, and negative two divided by negative two is positive one, which is your slope again."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "We're going to have to massage the equations a little bit in order to prepare them for elimination. So let's say that we have an equation 5x minus 10y is equal to 15. And we have another equation. 3x minus 2y is equal to 3. And I said we want to do this using elimination. Once again, we could use substitution. We could graph both of these lines and figure out where they intersect."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "3x minus 2y is equal to 3. And I said we want to do this using elimination. Once again, we could use substitution. We could graph both of these lines and figure out where they intersect. But we're going to use elimination. But the first thing you might say, hey, Sal, with elimination, you are subtracting the left-hand side of one equation from another or adding the two and then adding the two right-hand sides. And I could do that because it's essentially adding the same thing to both sides of the equation."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "We could graph both of these lines and figure out where they intersect. But we're going to use elimination. But the first thing you might say, hey, Sal, with elimination, you are subtracting the left-hand side of one equation from another or adding the two and then adding the two right-hand sides. And I could do that because it's essentially adding the same thing to both sides of the equation. But here, it's not obvious that that would be of any help. If we added these two left-hand sides, you would get 8x minus 12y. That wouldn't eliminate any variables."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "And I could do that because it's essentially adding the same thing to both sides of the equation. But here, it's not obvious that that would be of any help. If we added these two left-hand sides, you would get 8x minus 12y. That wouldn't eliminate any variables. And on the right-hand side, you would just be left with a number. And if you subtracted, that wouldn't eliminate any variables. So how is elimination going to help here?"}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "That wouldn't eliminate any variables. And on the right-hand side, you would just be left with a number. And if you subtracted, that wouldn't eliminate any variables. So how is elimination going to help here? And the answer is we can multiply both of these equations in such a way that maybe we can get one of these terms to cancel out with one of the others. So let's say we want, and you could really pick which term you want to cancel out. Let's say we want to cancel out the y terms."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "So how is elimination going to help here? And the answer is we can multiply both of these equations in such a way that maybe we can get one of these terms to cancel out with one of the others. So let's say we want, and you could really pick which term you want to cancel out. Let's say we want to cancel out the y terms. So I'll just rewrite this 5x minus 10y here. 5x minus 10y is equal to 15. Now, is there anything that I can multiply this green equation by so that this negative 2y term becomes a term that will cancel out with the negative 10y?"}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "Let's say we want to cancel out the y terms. So I'll just rewrite this 5x minus 10y here. 5x minus 10y is equal to 15. Now, is there anything that I can multiply this green equation by so that this negative 2y term becomes a term that will cancel out with the negative 10y? So I essentially want to make this negative 2y into a positive 10y, right? Because if this was a positive 10y, it'll cancel out when I add the left-hand side to this equation. So what could I multiply this equation by?"}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "Now, is there anything that I can multiply this green equation by so that this negative 2y term becomes a term that will cancel out with the negative 10y? So I essentially want to make this negative 2y into a positive 10y, right? Because if this was a positive 10y, it'll cancel out when I add the left-hand side to this equation. So what could I multiply this equation by? Well, if I multiply it by negative 5, negative 5 times negative 2 right here would be positive 10. So let's do that. Let's multiply this equation times 5, times negative 5."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "So what could I multiply this equation by? Well, if I multiply it by negative 5, negative 5 times negative 2 right here would be positive 10. So let's do that. Let's multiply this equation times 5, times negative 5. So you multiply the left-hand side by negative 5 and multiply the right-hand side by negative 5. And what do you get? Remember, we're not fundamentally changing the equation."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "Let's multiply this equation times 5, times negative 5. So you multiply the left-hand side by negative 5 and multiply the right-hand side by negative 5. And what do you get? Remember, we're not fundamentally changing the equation. We're not changing the information in the equation. We're doing the same thing to both sides of it. So the left-hand side of the equation becomes negative 5 times 3x is negative 15x."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "Remember, we're not fundamentally changing the equation. We're not changing the information in the equation. We're doing the same thing to both sides of it. So the left-hand side of the equation becomes negative 5 times 3x is negative 15x. And the negative 5 times negative 2y is plus 10y is equal to 3 times negative 5 is negative 15. And now we're ready to do our elimination. If we add this to the left-hand side of the yellow equation and we add the negative 15 to the right-hand side of the yellow equation, we are adding the same thing to both sides of the equation because this is equal to that."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "So the left-hand side of the equation becomes negative 5 times 3x is negative 15x. And the negative 5 times negative 2y is plus 10y is equal to 3 times negative 5 is negative 15. And now we're ready to do our elimination. If we add this to the left-hand side of the yellow equation and we add the negative 15 to the right-hand side of the yellow equation, we are adding the same thing to both sides of the equation because this is equal to that. So let's do that. So 5x minus 15y, we have this little negative sign there. We don't want to lose that."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "If we add this to the left-hand side of the yellow equation and we add the negative 15 to the right-hand side of the yellow equation, we are adding the same thing to both sides of the equation because this is equal to that. So let's do that. So 5x minus 15y, we have this little negative sign there. We don't want to lose that. That's negative 10x. The y's cancel out. Negative 10y plus 10y, that's 0y."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "We don't want to lose that. That's negative 10x. The y's cancel out. Negative 10y plus 10y, that's 0y. That was the whole point behind multiplying this by negative 5. Is going to be equal to 15 minus 15 is 0. So negative 10x is equal to 0."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "Negative 10y plus 10y, that's 0y. That was the whole point behind multiplying this by negative 5. Is going to be equal to 15 minus 15 is 0. So negative 10x is equal to 0. Divide both sides by negative 10. And you get x is equal to 0. And now we can substitute back into either of these equations to figure out what y must be equal to."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "So negative 10x is equal to 0. Divide both sides by negative 10. And you get x is equal to 0. And now we can substitute back into either of these equations to figure out what y must be equal to. Let's substitute into the top equation. So we get 5 times 0 minus 10y is equal to 15. Or negative 10y is equal to 15."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "And now we can substitute back into either of these equations to figure out what y must be equal to. Let's substitute into the top equation. So we get 5 times 0 minus 10y is equal to 15. Or negative 10y is equal to 15. Let me write that. Negative 10y is equal to 15. Divide both sides by negative 10."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "Or negative 10y is equal to 15. Let me write that. Negative 10y is equal to 15. Divide both sides by negative 10. And we are left with y is equal to 15 over 10 is negative 3 over 2. So if you were to graph it, the point of intersection would be the point 0, negative 3 halves. And you can verify that it also satisfies this equation."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "Divide both sides by negative 10. And we are left with y is equal to 15 over 10 is negative 3 over 2. So if you were to graph it, the point of intersection would be the point 0, negative 3 halves. And you can verify that it also satisfies this equation. The original equation over here was 3x minus 2y is equal to 3. 3 times 0 is equal to 3. 3 times 0, which is 0, minus 2 times negative 3 halves is 0."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "And you can verify that it also satisfies this equation. The original equation over here was 3x minus 2y is equal to 3. 3 times 0 is equal to 3. 3 times 0, which is 0, minus 2 times negative 3 halves is 0. This is positive 3. These cancel out. These become positive."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "3 times 0, which is 0, minus 2 times negative 3 halves is 0. This is positive 3. These cancel out. These become positive. Plus positive 3 is equal to 3. So this does indeed satisfy both equations. Let's do another one of these where we have to multiply and to massage the equations."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "These become positive. Plus positive 3 is equal to 3. So this does indeed satisfy both equations. Let's do another one of these where we have to multiply and to massage the equations. And then we can eliminate one of the variables. Let's say we have 5x plus 7y is equal to 15. And we have 7x minus 3y is equal to 5."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "Let's do another one of these where we have to multiply and to massage the equations. And then we can eliminate one of the variables. Let's say we have 5x plus 7y is equal to 15. And we have 7x minus 3y is equal to 5. Now once again, if you just added or subtracted both the left-hand sides, you're not eliminating any variables. These aren't in any way kind of have the same coefficient or the negative of their coefficient. So let's pick a variable to eliminate."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "And we have 7x minus 3y is equal to 5. Now once again, if you just added or subtracted both the left-hand sides, you're not eliminating any variables. These aren't in any way kind of have the same coefficient or the negative of their coefficient. So let's pick a variable to eliminate. Let's say we want to eliminate the x's this time. And you could literally pick on one of the variables or another. Doesn't matter."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "So let's pick a variable to eliminate. Let's say we want to eliminate the x's this time. And you could literally pick on one of the variables or another. Doesn't matter. You could say, oh, let's eliminate the y's first. But I'm going to choose to eliminate the x's first. And so what I need to do is massage one or both of these equations in a way that these guys have the same coefficients or their coefficients are the negatives of each other."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "Doesn't matter. You could say, oh, let's eliminate the y's first. But I'm going to choose to eliminate the x's first. And so what I need to do is massage one or both of these equations in a way that these guys have the same coefficients or their coefficients are the negatives of each other. So that when I add the left-hand sides, they're going to eliminate each other. Now, there's nothing obvious. I could multiply this by a fraction to make it equal to negative 5."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "And so what I need to do is massage one or both of these equations in a way that these guys have the same coefficients or their coefficients are the negatives of each other. So that when I add the left-hand sides, they're going to eliminate each other. Now, there's nothing obvious. I could multiply this by a fraction to make it equal to negative 5. Or I could multiply this by a fraction to make it equal to negative 7. But even a more fun thing to do is I can try to get both of them to be their least common multiple. I could get both of these to 35."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "I could multiply this by a fraction to make it equal to negative 5. Or I could multiply this by a fraction to make it equal to negative 7. But even a more fun thing to do is I can try to get both of them to be their least common multiple. I could get both of these to 35. And the way I can do it is by multiplying by each other. So I can multiply this top equation by 7. And I'm picking 7 so that this becomes a 35."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "I could get both of these to 35. And the way I can do it is by multiplying by each other. So I can multiply this top equation by 7. And I'm picking 7 so that this becomes a 35. And I could multiply this bottom equation by negative 5. And the reason why I'm doing that is so that this becomes a negative 35. Remember, my point is I want to eliminate the x's."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "And I'm picking 7 so that this becomes a 35. And I could multiply this bottom equation by negative 5. And the reason why I'm doing that is so that this becomes a negative 35. Remember, my point is I want to eliminate the x's. So if I make this a 35 and if I make this a negative 35, then I'm going to be all set. I can add the two, the left-hand and the right-hand sides of the equations. So this top equation, when you multiply by 7, it becomes."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "Remember, my point is I want to eliminate the x's. So if I make this a 35 and if I make this a negative 35, then I'm going to be all set. I can add the two, the left-hand and the right-hand sides of the equations. So this top equation, when you multiply by 7, it becomes. Let me scroll up a little bit. When you multiply by 7, it becomes 35x plus 49y is equal to 70 plus 35 is equal to 105. 15 at 70 plus 35 is equal to 105."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "So this top equation, when you multiply by 7, it becomes. Let me scroll up a little bit. When you multiply by 7, it becomes 35x plus 49y is equal to 70 plus 35 is equal to 105. 15 at 70 plus 35 is equal to 105. That's what the top equation becomes. This bottom equation becomes negative 5 times 7x is negative 35x. Negative 5 times negative 3y is plus 15y."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "15 at 70 plus 35 is equal to 105. That's what the top equation becomes. This bottom equation becomes negative 5 times 7x is negative 35x. Negative 5 times negative 3y is plus 15y. The negatives cancel out. And then 5, this isn't a minus 5. This is times negative 5."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "Negative 5 times negative 3y is plus 15y. The negatives cancel out. And then 5, this isn't a minus 5. This is times negative 5. 5 times negative 5 is equal to negative 25. Now we can start with this top equation and add the same thing to both sides, where that same thing is negative 25, which is also equal to this expression. So let's add the left-hand sides and the right-hand sides because we're really adding the same thing to both sides of the equation."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "This is times negative 5. 5 times negative 5 is equal to negative 25. Now we can start with this top equation and add the same thing to both sides, where that same thing is negative 25, which is also equal to this expression. So let's add the left-hand sides and the right-hand sides because we're really adding the same thing to both sides of the equation. So the left-hand side, the x's cancel out. 35x minus 35x. That was the whole point."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "So let's add the left-hand sides and the right-hand sides because we're really adding the same thing to both sides of the equation. So the left-hand side, the x's cancel out. 35x minus 35x. That was the whole point. They cancel out. And on the y's, you get 49y plus 15y. That is 64y is equal to 105 minus 25 is equal to 80."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "That was the whole point. They cancel out. And on the y's, you get 49y plus 15y. That is 64y is equal to 105 minus 25 is equal to 80. Divide both sides by 64. And you get y is equal to 80 over 64. And let's see, if you divide the numerator and the denominator by 8, actually, if I do 16, 16 would be better."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "That is 64y is equal to 105 minus 25 is equal to 80. Divide both sides by 64. And you get y is equal to 80 over 64. And let's see, if you divide the numerator and the denominator by 8, actually, if I do 16, 16 would be better. Well, let's do 8 first just because we know our 8 times tables. So that becomes 10 over 8. And you can divide this by 2, and you get 5 over 4."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "And let's see, if you divide the numerator and the denominator by 8, actually, if I do 16, 16 would be better. Well, let's do 8 first just because we know our 8 times tables. So that becomes 10 over 8. And you can divide this by 2, and you get 5 over 4. If you divided just straight up by 16, you would have gone straight to 5 over 4. So y is equal to 5 over 4. Let's figure out what x is."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "And you can divide this by 2, and you get 5 over 4. If you divided just straight up by 16, you would have gone straight to 5 over 4. So y is equal to 5 over 4. Let's figure out what x is. So we can substitute either into one of these equations or into one of the original equations. Let's substitute into the second of the original equations where we had 7x minus 3y is equal to 5. That was the original version of the second equation that we later transformed into this."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "Let's figure out what x is. So we can substitute either into one of these equations or into one of the original equations. Let's substitute into the second of the original equations where we had 7x minus 3y is equal to 5. That was the original version of the second equation that we later transformed into this. So we get 7x minus 3 times y times 5 over 4 is equal to 5. Or 7x minus 15 over 4 is equal to 5. Let's add 15 over 4."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "That was the original version of the second equation that we later transformed into this. So we get 7x minus 3 times y times 5 over 4 is equal to 5. Or 7x minus 15 over 4 is equal to 5. Let's add 15 over 4. Oh, sorry, I didn't do that right. This would be 7x minus 3 times 4. Oh, sorry, that was right."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "Let's add 15 over 4. Oh, sorry, I didn't do that right. This would be 7x minus 3 times 4. Oh, sorry, that was right. What am I doing? 3 times 15 over 4 is equal to 5. Let's add 15 over 4 to both sides."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "Oh, sorry, that was right. What am I doing? 3 times 15 over 4 is equal to 5. Let's add 15 over 4 to both sides. So plus 15 over 4. Plus 15 over 4. And what do we get?"}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "Let's add 15 over 4 to both sides. So plus 15 over 4. Plus 15 over 4. And what do we get? The left-hand side just becomes a 7x. These guys cancel out. And that's going to be equal to 5 is the same thing as 20 over 4."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "And what do we get? The left-hand side just becomes a 7x. These guys cancel out. And that's going to be equal to 5 is the same thing as 20 over 4. 20 over 4 plus 15 over 4. Or we get that, let me scroll down a little bit, 7x is equal to 35 over 4. We can multiply both sides by 1 7th."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "And that's going to be equal to 5 is the same thing as 20 over 4. 20 over 4 plus 15 over 4. Or we get that, let me scroll down a little bit, 7x is equal to 35 over 4. We can multiply both sides by 1 7th. Or we could divide both sides by 7. Same thing. Let's multiply both sides by 1 7th."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "We can multiply both sides by 1 7th. Or we could divide both sides by 7. Same thing. Let's multiply both sides by 1 7th. The same thing as dividing by 7. So these cancel out. And you're left with x is equal to."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "Let's multiply both sides by 1 7th. The same thing as dividing by 7. So these cancel out. And you're left with x is equal to. Here, if you divide 35 by 7, you get 5. You divide 7 by 7, you get 1. So x is equal to 5 4ths as well."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "And you're left with x is equal to. Here, if you divide 35 by 7, you get 5. You divide 7 by 7, you get 1. So x is equal to 5 4ths as well. So the point of intersection of this right here is both x and y are going to be equal to 5 4ths. So if you looked at it as a graph, it would be 5 4ths comma 5 4ths. Now let's verify that this satisfies the top equation."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "So x is equal to 5 4ths as well. So the point of intersection of this right here is both x and y are going to be equal to 5 4ths. So if you looked at it as a graph, it would be 5 4ths comma 5 4ths. Now let's verify that this satisfies the top equation. If you take 5 times 5 over 4 plus 7 times 5 over 4, what do you get? It should be equal to 15. So this is equal to 25 over 4 plus 35 over 4, which is equal to 60 over 4, which is indeed equal to 15."}, {"video_title": "Systems of equations with elimination (and manipulation)   High School Math   Khan Academy.mp3", "Sentence": "Now let's verify that this satisfies the top equation. If you take 5 times 5 over 4 plus 7 times 5 over 4, what do you get? It should be equal to 15. So this is equal to 25 over 4 plus 35 over 4, which is equal to 60 over 4, which is indeed equal to 15. So it does definitely satisfy that top equation. And you can check out this bottom equation for yourself. But it should, because we actually use this bottom equation to figure out that x is equal to 5 4ths."}, {"video_title": "Finding the slope of a line from its graph   Algebra I   Khan Academy.mp3", "Sentence": "And the best way to view it, slope is equal to change in y, change in y over change in x. Over change in x. And for a line, this will always be constant. And sometimes you might see it written like this. You might see this triangle. That's a capital delta. That means change in y over change in x."}, {"video_title": "Finding the slope of a line from its graph   Algebra I   Khan Academy.mp3", "Sentence": "And sometimes you might see it written like this. You might see this triangle. That's a capital delta. That means change in y over change in x. That's just a fancy way of saying change in y over change in x. So let's see what this change in y is for any change in x. So let's start at some point that seems pretty reasonable to read from this table right here, from this graph."}, {"video_title": "Finding the slope of a line from its graph   Algebra I   Khan Academy.mp3", "Sentence": "That means change in y over change in x. That's just a fancy way of saying change in y over change in x. So let's see what this change in y is for any change in x. So let's start at some point that seems pretty reasonable to read from this table right here, from this graph. So let's see, we're starting here. Let me do it in a more vibrant color. So let's say we start at that point right there."}, {"video_title": "Finding the slope of a line from its graph   Algebra I   Khan Academy.mp3", "Sentence": "So let's start at some point that seems pretty reasonable to read from this table right here, from this graph. So let's see, we're starting here. Let me do it in a more vibrant color. So let's say we start at that point right there. And we want to go to another point that's pretty straightforward to read. So we can move to that point right there. We can literally pick any two points on this line."}, {"video_title": "Finding the slope of a line from its graph   Algebra I   Khan Academy.mp3", "Sentence": "So let's say we start at that point right there. And we want to go to another point that's pretty straightforward to read. So we can move to that point right there. We can literally pick any two points on this line. I'm just picking ones that are at nice integer coordinates so it's easy to read. So what is the change in y and what is the change in x? So first let's look at the change in x."}, {"video_title": "Finding the slope of a line from its graph   Algebra I   Khan Academy.mp3", "Sentence": "We can literally pick any two points on this line. I'm just picking ones that are at nice integer coordinates so it's easy to read. So what is the change in y and what is the change in x? So first let's look at the change in x. So if we go from there to there, what is the change in x? My change in x is equal to what? Well, I could just count it out."}, {"video_title": "Finding the slope of a line from its graph   Algebra I   Khan Academy.mp3", "Sentence": "So first let's look at the change in x. So if we go from there to there, what is the change in x? My change in x is equal to what? Well, I could just count it out. I went one step, two steps, three steps. My change in x is 3. And you could even see it from the x values."}, {"video_title": "Finding the slope of a line from its graph   Algebra I   Khan Academy.mp3", "Sentence": "Well, I could just count it out. I went one step, two steps, three steps. My change in x is 3. And you could even see it from the x values. If I go from negative 3 to 0, I went up by 3. So my change in x is 3. So let me write this."}, {"video_title": "Finding the slope of a line from its graph   Algebra I   Khan Academy.mp3", "Sentence": "And you could even see it from the x values. If I go from negative 3 to 0, I went up by 3. So my change in x is 3. So let me write this. Change in x delta x is equal to 3. And what's my change in y? Well, my change in y, I'm going from negative 3 up to negative 1."}, {"video_title": "Finding the slope of a line from its graph   Algebra I   Khan Academy.mp3", "Sentence": "So let me write this. Change in x delta x is equal to 3. And what's my change in y? Well, my change in y, I'm going from negative 3 up to negative 1. Or you could just say 1, 2. So my change in y is equal to positive 2. So let me write that down."}, {"video_title": "Finding the slope of a line from its graph   Algebra I   Khan Academy.mp3", "Sentence": "Well, my change in y, I'm going from negative 3 up to negative 1. Or you could just say 1, 2. So my change in y is equal to positive 2. So let me write that down. Change in y is equal to 2. So what is my change in y for a change in x? Well, when my change in x was 3, my change in y is 2."}, {"video_title": "Finding the slope of a line from its graph   Algebra I   Khan Academy.mp3", "Sentence": "So let me write that down. Change in y is equal to 2. So what is my change in y for a change in x? Well, when my change in x was 3, my change in y is 2. So this is my slope. And one thing I want to do, I want to show you that I could have really picked any two points here. Let's say I didn't pick those two points."}, {"video_title": "Finding the slope of a line from its graph   Algebra I   Khan Academy.mp3", "Sentence": "Well, when my change in x was 3, my change in y is 2. So this is my slope. And one thing I want to do, I want to show you that I could have really picked any two points here. Let's say I didn't pick those two points. Let me pick some other points. And I'll even go in a different direction. And I want to show you that you're going to get the same answer."}, {"video_title": "Finding the slope of a line from its graph   Algebra I   Khan Academy.mp3", "Sentence": "Let's say I didn't pick those two points. Let me pick some other points. And I'll even go in a different direction. And I want to show you that you're going to get the same answer. Let's say I viewed this as my starting point. And I want to go all the way over there. So what is my, well, let's think about the change in y first."}, {"video_title": "Finding the slope of a line from its graph   Algebra I   Khan Academy.mp3", "Sentence": "And I want to show you that you're going to get the same answer. Let's say I viewed this as my starting point. And I want to go all the way over there. So what is my, well, let's think about the change in y first. So the change in y, I am going down by how many units? 1, 2, 3, 4 units. So my change in y in this example is negative 4."}, {"video_title": "Finding the slope of a line from its graph   Algebra I   Khan Academy.mp3", "Sentence": "So what is my, well, let's think about the change in y first. So the change in y, I am going down by how many units? 1, 2, 3, 4 units. So my change in y in this example is negative 4. I went from 1 to negative 3. That's negative 4. That's my change in y."}, {"video_title": "Finding the slope of a line from its graph   Algebra I   Khan Academy.mp3", "Sentence": "So my change in y in this example is negative 4. I went from 1 to negative 3. That's negative 4. That's my change in y. Change in y is equal to negative 4. Now what is my change in x? Well, I'm going from this point, or from this x value, all the way back like this."}, {"video_title": "Finding the slope of a line from its graph   Algebra I   Khan Academy.mp3", "Sentence": "That's my change in y. Change in y is equal to negative 4. Now what is my change in x? Well, I'm going from this point, or from this x value, all the way back like this. So I'm going to the left. So it's going to be a negative change in x. And I went 1, 2, 3, 4, 5, 6 units back."}, {"video_title": "Finding the slope of a line from its graph   Algebra I   Khan Academy.mp3", "Sentence": "Well, I'm going from this point, or from this x value, all the way back like this. So I'm going to the left. So it's going to be a negative change in x. And I went 1, 2, 3, 4, 5, 6 units back. So my change in x is equal to negative 6. Change in x is equal to negative 6. And you can even see I started at an x is equal to 3."}, {"video_title": "Finding the slope of a line from its graph   Algebra I   Khan Academy.mp3", "Sentence": "And I went 1, 2, 3, 4, 5, 6 units back. So my change in x is equal to negative 6. Change in x is equal to negative 6. And you can even see I started at an x is equal to 3. And I went all the way to x is equal to a negative 3. That's a change of negative 6. I went 6 to the left, or a change of negative 6."}, {"video_title": "Finding the slope of a line from its graph   Algebra I   Khan Academy.mp3", "Sentence": "And you can even see I started at an x is equal to 3. And I went all the way to x is equal to a negative 3. That's a change of negative 6. I went 6 to the left, or a change of negative 6. So what is my change in y over change in x? My change in y over change in x is equal to negative 4 over negative 6. The negatives cancel out."}, {"video_title": "Finding the slope of a line from its graph   Algebra I   Khan Academy.mp3", "Sentence": "I went 6 to the left, or a change of negative 6. So what is my change in y over change in x? My change in y over change in x is equal to negative 4 over negative 6. The negatives cancel out. And what's 4 over 6? Well, that's just 2 over 3. So it's the same value."}, {"video_title": "Finding the slope of a line from its graph   Algebra I   Khan Academy.mp3", "Sentence": "The negatives cancel out. And what's 4 over 6? Well, that's just 2 over 3. So it's the same value. You just have to be consistent. If this is my start point, I went down 4. And then I went back 6."}, {"video_title": "Finding the slope of a line from its graph   Algebra I   Khan Academy.mp3", "Sentence": "So it's the same value. You just have to be consistent. If this is my start point, I went down 4. And then I went back 6. Negative 4 over negative 6. If I viewed this as my starting point, I could say that I went up 4. So it would be a change in y would be 4."}, {"video_title": "Finding the slope of a line from its graph   Algebra I   Khan Academy.mp3", "Sentence": "And then I went back 6. Negative 4 over negative 6. If I viewed this as my starting point, I could say that I went up 4. So it would be a change in y would be 4. And then my change in x would be 6. And either way, once again, change in y over change in x is going to be 4 over 6, 2 thirds. So no matter which point you choose, as long as you kind of think about it in a consistent way, you're going to get the same value for slope."}, {"video_title": "Systems of equations with graphing exact & approximate solutions   High School Math   Khan Academy.mp3", "Sentence": "And then I have another equation that involves x and y, so it's gonna define another line. It says graph the system of equations and find its solution. So we're gonna try to find it visually. So let's graph this first one. And to graph this line, I have a little graphing tool here. Notice I can, if I can figure out two points, I can move those points around and it's going to define our line for us. So let's pick, I'm gonna pick two x values and figure out the corresponding y values and then graph the line."}, {"video_title": "Systems of equations with graphing exact & approximate solutions   High School Math   Khan Academy.mp3", "Sentence": "So let's graph this first one. And to graph this line, I have a little graphing tool here. Notice I can, if I can figure out two points, I can move those points around and it's going to define our line for us. So let's pick, I'm gonna pick two x values and figure out the corresponding y values and then graph the line. So let's see how I could do this. So let's see, an easy one is what happens when x is equal to zero? Well, if x is equal to zero, everything I just shaded goes away, and we're left with negative three y is equal to nine."}, {"video_title": "Systems of equations with graphing exact & approximate solutions   High School Math   Khan Academy.mp3", "Sentence": "So let's pick, I'm gonna pick two x values and figure out the corresponding y values and then graph the line. So let's see how I could do this. So let's see, an easy one is what happens when x is equal to zero? Well, if x is equal to zero, everything I just shaded goes away, and we're left with negative three y is equal to nine. So negative three y equals nine, y would be negative three. So when x is equal to zero, y would be negative three. So let me graph that."}, {"video_title": "Systems of equations with graphing exact & approximate solutions   High School Math   Khan Academy.mp3", "Sentence": "Well, if x is equal to zero, everything I just shaded goes away, and we're left with negative three y is equal to nine. So negative three y equals nine, y would be negative three. So when x is equal to zero, y would be negative three. So let me graph that. When x is equal to zero, y, x is zero, y is negative three. Now another easy point, actually, instead of trying another x value, let's think about when y is equal to zero, because these equations are written in standard form, so it's easy to just test. Well, what are the x and y intercepts?"}, {"video_title": "Systems of equations with graphing exact & approximate solutions   High School Math   Khan Academy.mp3", "Sentence": "So let me graph that. When x is equal to zero, y, x is zero, y is negative three. Now another easy point, actually, instead of trying another x value, let's think about when y is equal to zero, because these equations are written in standard form, so it's easy to just test. Well, what are the x and y intercepts? So when y is equal to zero, when y is equal to zero, this term goes away and you have negative x is equal to nine, or x would be equal to negative nine. So when y is zero, x is negative nine. So when y is zero, x is negative nine, or when x is negative nine, y is zero."}, {"video_title": "Systems of equations with graphing exact & approximate solutions   High School Math   Khan Academy.mp3", "Sentence": "Well, what are the x and y intercepts? So when y is equal to zero, when y is equal to zero, this term goes away and you have negative x is equal to nine, or x would be equal to negative nine. So when y is zero, x is negative nine. So when y is zero, x is negative nine, or when x is negative nine, y is zero. So I've just plotted this first equation. So now let's do the second one. We'll do the same thing."}, {"video_title": "Systems of equations with graphing exact & approximate solutions   High School Math   Khan Academy.mp3", "Sentence": "So when y is zero, x is negative nine, or when x is negative nine, y is zero. So I've just plotted this first equation. So now let's do the second one. We'll do the same thing. What happens when x is equal to zero? When x is equal to zero, so this is going to be our y intercept now. When x is equal to zero, negative six y is equal to negative six."}, {"video_title": "Systems of equations with graphing exact & approximate solutions   High School Math   Khan Academy.mp3", "Sentence": "We'll do the same thing. What happens when x is equal to zero? When x is equal to zero, so this is going to be our y intercept now. When x is equal to zero, negative six y is equal to negative six. Well, y would have to be equal to one. So when x is zero, y is equal to one. So when x is zero, x is zero, y is equal to one."}, {"video_title": "Systems of equations with graphing exact & approximate solutions   High School Math   Khan Academy.mp3", "Sentence": "When x is equal to zero, negative six y is equal to negative six. Well, y would have to be equal to one. So when x is zero, y is equal to one. So when x is zero, x is zero, y is equal to one. Get one more point here. When y is zero, when this term is, y being zero would make this entire term zero, then six x is equal to negative six, or x is equal to negative one. So when y is zero, x is negative one, or when x is negative one, y is zero."}, {"video_title": "Systems of equations with graphing exact & approximate solutions   High School Math   Khan Academy.mp3", "Sentence": "So when x is zero, x is zero, y is equal to one. Get one more point here. When y is zero, when this term is, y being zero would make this entire term zero, then six x is equal to negative six, or x is equal to negative one. So when y is zero, x is negative one, or when x is negative one, y is zero. When x is negative one, y is zero. And so just like that, I've plotted the two lines, and the solution to the system are the x and y values that satisfy both equations. And if they satisfy both equations, that means they sit on both lines."}, {"video_title": "Systems of equations with graphing exact & approximate solutions   High School Math   Khan Academy.mp3", "Sentence": "So when y is zero, x is negative one, or when x is negative one, y is zero. When x is negative one, y is zero. And so just like that, I've plotted the two lines, and the solution to the system are the x and y values that satisfy both equations. And if they satisfy both equations, that means they sit on both lines. And so in order to be on both lines, they're going to be at the point of intersection. And I see this point of intersection right over here. It looks pretty clear that this is the point x is equal to negative three, and y is equal to negative two."}, {"video_title": "Systems of equations with graphing exact & approximate solutions   High School Math   Khan Academy.mp3", "Sentence": "And if they satisfy both equations, that means they sit on both lines. And so in order to be on both lines, they're going to be at the point of intersection. And I see this point of intersection right over here. It looks pretty clear that this is the point x is equal to negative three, and y is equal to negative two. So it's the point negative three comma negative two. So let me write that down. Negative three comma negative two."}, {"video_title": "Systems of equations with graphing exact & approximate solutions   High School Math   Khan Academy.mp3", "Sentence": "It looks pretty clear that this is the point x is equal to negative three, and y is equal to negative two. So it's the point negative three comma negative two. So let me write that down. Negative three comma negative two. And then I could check my answer. Got it right. Let's do another one of these, maybe of a different type."}, {"video_title": "Systems of equations with graphing exact & approximate solutions   High School Math   Khan Academy.mp3", "Sentence": "Negative three comma negative two. And then I could check my answer. Got it right. Let's do another one of these, maybe of a different type. So over here it says a system of two linear equations is graphed below. Approximate the solution of the system. All right, so here I just have to just look at this carefully and think about where this point is."}, {"video_title": "Systems of equations with graphing exact & approximate solutions   High School Math   Khan Academy.mp3", "Sentence": "Let's do another one of these, maybe of a different type. So over here it says a system of two linear equations is graphed below. Approximate the solution of the system. All right, so here I just have to just look at this carefully and think about where this point is. So let's think about first its x value. So its x value, let's see, it's about right there in terms of its x value. It looks like, so this is negative one, negative, this is negative two, so negative 1.5 is gonna be right over here."}, {"video_title": "Systems of equations with graphing exact & approximate solutions   High School Math   Khan Academy.mp3", "Sentence": "All right, so here I just have to just look at this carefully and think about where this point is. So let's think about first its x value. So its x value, let's see, it's about right there in terms of its x value. It looks like, so this is negative one, negative, this is negative two, so negative 1.5 is gonna be right over here. It's a little bit to the left of negative 1.5, so it's even more negative. I would say negative 1.6. So when I'm approximating it, negative 1.6, hopefully it has a little leeway in how it checks the answer."}, {"video_title": "Systems of equations with graphing exact & approximate solutions   High School Math   Khan Academy.mp3", "Sentence": "It looks like, so this is negative one, negative, this is negative two, so negative 1.5 is gonna be right over here. It's a little bit to the left of negative 1.5, so it's even more negative. I would say negative 1.6. So when I'm approximating it, negative 1.6, hopefully it has a little leeway in how it checks the answer. What about the y value? So if I look at the y value here, it looks like it's a little less than 1.5. 1.5 would be halfway between one and two."}, {"video_title": "Systems of equations with graphing exact & approximate solutions   High School Math   Khan Academy.mp3", "Sentence": "So when I'm approximating it, negative 1.6, hopefully it has a little leeway in how it checks the answer. What about the y value? So if I look at the y value here, it looks like it's a little less than 1.5. 1.5 would be halfway between one and two. It looks like it's a little less than halfway between one and two. So I don't know, I'd give it 1.4. Positive 1.4."}, {"video_title": "Systems of equations with graphing exact & approximate solutions   High School Math   Khan Academy.mp3", "Sentence": "1.5 would be halfway between one and two. It looks like it's a little less than halfway between one and two. So I don't know, I'd give it 1.4. Positive 1.4. And let's check the answer, see how we're doing. Yep, we got it right. Let's actually just do one more for good measure."}, {"video_title": "Systems of equations with graphing exact & approximate solutions   High School Math   Khan Academy.mp3", "Sentence": "Positive 1.4. And let's check the answer, see how we're doing. Yep, we got it right. Let's actually just do one more for good measure. So this is another system. They've just written the equations in more of our slope-intercept form. So let's see, y is equal to negative seven x plus three."}, {"video_title": "Systems of equations with graphing exact & approximate solutions   High School Math   Khan Academy.mp3", "Sentence": "Let's actually just do one more for good measure. So this is another system. They've just written the equations in more of our slope-intercept form. So let's see, y is equal to negative seven x plus three. When x is equal to zero, we have our y-intercept. Y is equal to three. So when x is equal to zero, y is equal to three."}, {"video_title": "Systems of equations with graphing exact & approximate solutions   High School Math   Khan Academy.mp3", "Sentence": "So let's see, y is equal to negative seven x plus three. When x is equal to zero, we have our y-intercept. Y is equal to three. So when x is equal to zero, y is equal to three. And then we see that our slope is negative seven. When you increase x by one, you decrease y by seven. So when you increase x by one, you decrease y by one, two, three, four, five, six, and seven."}, {"video_title": "Systems of equations with graphing exact & approximate solutions   High School Math   Khan Academy.mp3", "Sentence": "So when x is equal to zero, y is equal to three. And then we see that our slope is negative seven. When you increase x by one, you decrease y by seven. So when you increase x by one, you decrease y by one, two, three, four, five, six, and seven. When x goes from zero to one, y went from three to negative four. It went down by seven. So that's that first one."}, {"video_title": "Systems of equations with graphing exact & approximate solutions   High School Math   Khan Academy.mp3", "Sentence": "So when you increase x by one, you decrease y by one, two, three, four, five, six, and seven. When x goes from zero to one, y went from three to negative four. It went down by seven. So that's that first one. Now the second one, our y-intercept. When x is equal to zero, y is negative three. So let me graph that."}, {"video_title": "Systems of equations with graphing exact & approximate solutions   High School Math   Khan Academy.mp3", "Sentence": "So that's that first one. Now the second one, our y-intercept. When x is equal to zero, y is negative three. So let me graph that. When x is zero, y is equal to negative three. And then its slope is negative one. When x increases by one, y decreases by one."}, {"video_title": "Systems of equations with graphing exact & approximate solutions   High School Math   Khan Academy.mp3", "Sentence": "So let me graph that. When x is zero, y is equal to negative three. And then its slope is negative one. When x increases by one, y decreases by one. So the slope here is negative one. So when x increases by one, y decreases by one. And there you have it."}, {"video_title": "Systems of equations with graphing exact & approximate solutions   High School Math   Khan Academy.mp3", "Sentence": "When x increases by one, y decreases by one. So the slope here is negative one. So when x increases by one, y decreases by one. And there you have it. You have your point of intersection. You have the xy pair that satisfies both equations. That is the point of intersection."}, {"video_title": "Systems of equations with graphing exact & approximate solutions   High School Math   Khan Academy.mp3", "Sentence": "And there you have it. You have your point of intersection. You have the xy pair that satisfies both equations. That is the point of intersection. It's gonna sit on both lines, which is why it's the point of intersection. And that's the point x equals one, y is equal to negative four. So you have x equals one, and y is equal to negative four."}, {"video_title": "Example 3 Subtracting polynomials   Algebra I   Khan Academy.mp3", "Sentence": "So when you subtract an entire expression, this is the exact same thing as having 16x plus 14, and then you're adding the opposite of this whole thing, or you're adding negative 1, you're adding negative 1 times 3x squared plus x minus 9. Or another way to think about it is you can distribute this negative sign along all of those terms. That's essentially what we're about to do here. We're just adding the negative of this entire thing. We're adding the opposite of it. So this first part, I'm not going to change it. That's still just 16x plus 14."}, {"video_title": "Example 3 Subtracting polynomials   Algebra I   Khan Academy.mp3", "Sentence": "We're just adding the negative of this entire thing. We're adding the opposite of it. So this first part, I'm not going to change it. That's still just 16x plus 14. But now I'm going to distribute the negative sign here. So negative 1 times 3x squared is negative 3x squared. Negative 1 times positive x is negative x, because that's positive 1x."}, {"video_title": "Example 3 Subtracting polynomials   Algebra I   Khan Academy.mp3", "Sentence": "That's still just 16x plus 14. But now I'm going to distribute the negative sign here. So negative 1 times 3x squared is negative 3x squared. Negative 1 times positive x is negative x, because that's positive 1x. Negative 1 times negative 9, remember you have to consider this negative right over there. That is part of the term. Negative 1 times negative 9 is positive 9."}, {"video_title": "Example 3 Subtracting polynomials   Algebra I   Khan Academy.mp3", "Sentence": "Negative 1 times positive x is negative x, because that's positive 1x. Negative 1 times negative 9, remember you have to consider this negative right over there. That is part of the term. Negative 1 times negative 9 is positive 9. Negative times a negative is a positive. So then we have positive 9. And now we just have to combine like terms."}, {"video_title": "Example 3 Subtracting polynomials   Algebra I   Khan Academy.mp3", "Sentence": "Negative 1 times negative 9 is positive 9. Negative times a negative is a positive. So then we have positive 9. And now we just have to combine like terms. So what's our highest degree term here? I like to write it in that order. We have only one x squared term, second degree term."}, {"video_title": "Example 3 Subtracting polynomials   Algebra I   Khan Academy.mp3", "Sentence": "And now we just have to combine like terms. So what's our highest degree term here? I like to write it in that order. We have only one x squared term, second degree term. We only have one of those, so let me write it over here. Negative 3x squared. And then what do we have in terms of first degree terms?"}, {"video_title": "Example 3 Subtracting polynomials   Algebra I   Khan Academy.mp3", "Sentence": "We have only one x squared term, second degree term. We only have one of those, so let me write it over here. Negative 3x squared. And then what do we have in terms of first degree terms? Just an x, x to the first power. We have a 16x, and then from that we're going to subtract an x. Subtract 1x. So 16x minus 1x is 15x."}, {"video_title": "Example 3 Subtracting polynomials   Algebra I   Khan Academy.mp3", "Sentence": "And then what do we have in terms of first degree terms? Just an x, x to the first power. We have a 16x, and then from that we're going to subtract an x. Subtract 1x. So 16x minus 1x is 15x. If you have 16 of something and you subtract one of them away, you're going to have 15 of that something. And then finally, you have 14. You can view that as 14 times x to the 0, or just 14."}, {"video_title": "Example 3 Subtracting polynomials   Algebra I   Khan Academy.mp3", "Sentence": "So 16x minus 1x is 15x. If you have 16 of something and you subtract one of them away, you're going to have 15 of that something. And then finally, you have 14. You can view that as 14 times x to the 0, or just 14. 14 plus 9, they're both constant terms, or they're both being multiplied by x to the 0. 14 plus 9 is 23. And we're done."}, {"video_title": "Powers of zero   Exponents, radicals, and scientific notation   Pre-algebra   Khan Academy (2).mp3", "Sentence": "And just as a little bit of a reminder, let's start with a non-zero number, just to remind ourselves what exponentiation is all about. So if I were to take two to the first power, one way to think about this is we always start with a one, and then we multiply this base that many times times that one. So here, we're only gonna have one, two. So it's gonna be one times two, which is of course equal to two. If I were to say what is two to the second power, well, that's going to be equal to one times, and now I'm gonna have two twos, so times two times two, which is equal to four. And you could keep going like that. Now, the reason why I have this one here, and we've done this before, is to justify, and there's many other good reasons why two to the zero power should be equal to one."}, {"video_title": "Powers of zero   Exponents, radicals, and scientific notation   Pre-algebra   Khan Academy (2).mp3", "Sentence": "So it's gonna be one times two, which is of course equal to two. If I were to say what is two to the second power, well, that's going to be equal to one times, and now I'm gonna have two twos, so times two times two, which is equal to four. And you could keep going like that. Now, the reason why I have this one here, and we've done this before, is to justify, and there's many other good reasons why two to the zero power should be equal to one. But you could see if we use the same exact idea here, you start with a one, and then you multiply it by two zero times. Well, that's just going to end up with a one. So, so far, I've told you this video's about powers of zero, but I've been doing powers of two, so let's focus on zero now."}, {"video_title": "Powers of zero   Exponents, radicals, and scientific notation   Pre-algebra   Khan Academy (2).mp3", "Sentence": "Now, the reason why I have this one here, and we've done this before, is to justify, and there's many other good reasons why two to the zero power should be equal to one. But you could see if we use the same exact idea here, you start with a one, and then you multiply it by two zero times. Well, that's just going to end up with a one. So, so far, I've told you this video's about powers of zero, but I've been doing powers of two, so let's focus on zero now. So what do you think zero to the first power is going to be? Pause this video and try to figure that out. Well, you do the exact same idea."}, {"video_title": "Powers of zero   Exponents, radicals, and scientific notation   Pre-algebra   Khan Academy (2).mp3", "Sentence": "So, so far, I've told you this video's about powers of zero, but I've been doing powers of two, so let's focus on zero now. So what do you think zero to the first power is going to be? Pause this video and try to figure that out. Well, you do the exact same idea. You start with a one, and then multiply it by zero one time. So times zero, and this is going to be equal to zero. What do you think zero to the second power is going to be equal to?"}, {"video_title": "Powers of zero   Exponents, radicals, and scientific notation   Pre-algebra   Khan Academy (2).mp3", "Sentence": "Well, you do the exact same idea. You start with a one, and then multiply it by zero one time. So times zero, and this is going to be equal to zero. What do you think zero to the second power is going to be equal to? Pause this video and think about that. Well, it's gonna be one times zero twice. So times zero times zero, and I think you see where this is going."}, {"video_title": "Powers of zero   Exponents, radicals, and scientific notation   Pre-algebra   Khan Academy (2).mp3", "Sentence": "What do you think zero to the second power is going to be equal to? Pause this video and think about that. Well, it's gonna be one times zero twice. So times zero times zero, and I think you see where this is going. This is also going to be equal to zero. What do you think zero to some arbitrary positive, positive integer is going to be? Well, it's going to be equal to one times zero, that positive integer number of times."}, {"video_title": "Powers of zero   Exponents, radicals, and scientific notation   Pre-algebra   Khan Academy (2).mp3", "Sentence": "So times zero times zero, and I think you see where this is going. This is also going to be equal to zero. What do you think zero to some arbitrary positive, positive integer is going to be? Well, it's going to be equal to one times zero, that positive integer number of times. So once again, it's going to be equal to zero. And in general, you can extend that zero to any positive value exponent is going to give you zero. So that's pretty straightforward."}, {"video_title": "Powers of zero   Exponents, radicals, and scientific notation   Pre-algebra   Khan Academy (2).mp3", "Sentence": "Well, it's going to be equal to one times zero, that positive integer number of times. So once again, it's going to be equal to zero. And in general, you can extend that zero to any positive value exponent is going to give you zero. So that's pretty straightforward. But there is an interesting edge case here. What do you think zero to the zeroth power should be? Pause this video and think about that."}, {"video_title": "Powers of zero   Exponents, radicals, and scientific notation   Pre-algebra   Khan Academy (2).mp3", "Sentence": "So that's pretty straightforward. But there is an interesting edge case here. What do you think zero to the zeroth power should be? Pause this video and think about that. Well, this is actually contested. Different people will tell you different things. If you use the intuition behind exponentiation that we've been using in this video, you would say, all right, I would start with a one and then multiply it by zero zero times."}, {"video_title": "Powers of zero   Exponents, radicals, and scientific notation   Pre-algebra   Khan Academy (2).mp3", "Sentence": "Pause this video and think about that. Well, this is actually contested. Different people will tell you different things. If you use the intuition behind exponentiation that we've been using in this video, you would say, all right, I would start with a one and then multiply it by zero zero times. Or in other words, I just wouldn't multiply it by zero, in which case I'm just left with the one. The zero to the zeroth power should be equal to one. Other folks would say, hey, no, I'm with a zero and that's the zeroth power, maybe it should be a zero."}, {"video_title": "Analyzing solutions to linear systems graphically 1   Algebra I   Khan Academy.mp3", "Sentence": "We're told to look at the coordinate grid above. I put it on the side here. Identify one system of two lines that has a single solution, then identify one system of two lines that does not have a solution. So let's do the first part first. So a single solution. And they say identify one system. But we can see here, actually, there's actually going to be two systems that have a single solution."}, {"video_title": "Analyzing solutions to linear systems graphically 1   Algebra I   Khan Academy.mp3", "Sentence": "So let's do the first part first. So a single solution. And they say identify one system. But we can see here, actually, there's actually going to be two systems that have a single solution. And when we talk about a single solution, we're talking about a single x and y value that will satisfy both equations in the system. So if we look right here at the points of intersection, this point right there, that satisfies this equation, y is equal to 0.1x plus 1. And it also satisfies this blue line, but the graph that that line represents, y is equal to 4x plus 10."}, {"video_title": "Analyzing solutions to linear systems graphically 1   Algebra I   Khan Academy.mp3", "Sentence": "But we can see here, actually, there's actually going to be two systems that have a single solution. And when we talk about a single solution, we're talking about a single x and y value that will satisfy both equations in the system. So if we look right here at the points of intersection, this point right there, that satisfies this equation, y is equal to 0.1x plus 1. And it also satisfies this blue line, but the graph that that line represents, y is equal to 4x plus 10. So this dot right here, that point represents the solution to both of these. Or I guess another way to think about it, it represents an x and y value that satisfy both of these constraints. So one system that has one solution is a system that has y is equal to 0.1x plus 1."}, {"video_title": "Analyzing solutions to linear systems graphically 1   Algebra I   Khan Academy.mp3", "Sentence": "And it also satisfies this blue line, but the graph that that line represents, y is equal to 4x plus 10. So this dot right here, that point represents the solution to both of these. Or I guess another way to think about it, it represents an x and y value that satisfy both of these constraints. So one system that has one solution is a system that has y is equal to 0.1x plus 1. And then this blue line right here, which is y is equal to 4x plus 10. Now they only want us to identify one system of two lines that has a single solution. We've already done that."}, {"video_title": "Analyzing solutions to linear systems graphically 1   Algebra I   Khan Academy.mp3", "Sentence": "So one system that has one solution is a system that has y is equal to 0.1x plus 1. And then this blue line right here, which is y is equal to 4x plus 10. Now they only want us to identify one system of two lines that has a single solution. We've already done that. But just so you see it, there's actually another system here. So this is one system right here. Or another system would be the green line and this red line."}, {"video_title": "Analyzing solutions to linear systems graphically 1   Algebra I   Khan Academy.mp3", "Sentence": "We've already done that. But just so you see it, there's actually another system here. So this is one system right here. Or another system would be the green line and this red line. This point of intersection right here, once again, that represents an x and y value that satisfies both the equation y is equal to 0.1x plus 1. So y is equal to 0.1x plus 1. And this point right here satisfies the equation y is equal to 4x minus 6."}, {"video_title": "Analyzing solutions to linear systems graphically 1   Algebra I   Khan Academy.mp3", "Sentence": "Or another system would be the green line and this red line. This point of intersection right here, once again, that represents an x and y value that satisfies both the equation y is equal to 0.1x plus 1. So y is equal to 0.1x plus 1. And this point right here satisfies the equation y is equal to 4x minus 6. So if you look at this system, there's one solution because there's one point of intersection of these two equations or these two lines. And this system also has one solution because it has one point of intersection. Now the second part of this problem, they say identify one system of two lines that does not have a single solution or does not have a solution."}, {"video_title": "Analyzing solutions to linear systems graphically 1   Algebra I   Khan Academy.mp3", "Sentence": "And this point right here satisfies the equation y is equal to 4x minus 6. So if you look at this system, there's one solution because there's one point of intersection of these two equations or these two lines. And this system also has one solution because it has one point of intersection. Now the second part of this problem, they say identify one system of two lines that does not have a single solution or does not have a solution. So no solution. So in order for there to be no solution, that means that the two constraints don't overlap. That there's no point that is common to both equations or there's no pair of xy values that's common to both equations."}, {"video_title": "Analyzing solutions to linear systems graphically 1   Algebra I   Khan Academy.mp3", "Sentence": "Now the second part of this problem, they say identify one system of two lines that does not have a single solution or does not have a solution. So no solution. So in order for there to be no solution, that means that the two constraints don't overlap. That there's no point that is common to both equations or there's no pair of xy values that's common to both equations. And that's the case of the two parallel lines here, this blue line and this green line. Because they never intersect, there's no coordinate on the coordinate plane that satisfies both equations. So there's no x and y that satisfy both."}, {"video_title": "Analyzing solutions to linear systems graphically 1   Algebra I   Khan Academy.mp3", "Sentence": "That there's no point that is common to both equations or there's no pair of xy values that's common to both equations. And that's the case of the two parallel lines here, this blue line and this green line. Because they never intersect, there's no coordinate on the coordinate plane that satisfies both equations. So there's no x and y that satisfy both. So the second part of the question, a system that has no solution is y is equal to 4x plus 10. And then the other one is y is equal to 4x minus 6. And notice, they have the exact same slope."}, {"video_title": "Analyzing solutions to linear systems graphically 1   Algebra I   Khan Academy.mp3", "Sentence": "So there's no x and y that satisfy both. So the second part of the question, a system that has no solution is y is equal to 4x plus 10. And then the other one is y is equal to 4x minus 6. And notice, they have the exact same slope. They have the exact same slope. And they're two different lines. They have different intercepts."}, {"video_title": "Factoring polynomials common binomial factor   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "We're asked to factor the polynomial below as the product of two binomials. And we have n times n minus one plus three times n minus one. So I encourage you to pause this video and see if you can figure this out. Well, the key is to realizing that both of these terms have n minus one as a factor. Let me just rewrite the whole thing so we can work on it down here. So this is n times n minus one plus three times n minus one. And notice, both of them have an n minus one, have an n minus one as a factor."}, {"video_title": "Factoring polynomials common binomial factor   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "Well, the key is to realizing that both of these terms have n minus one as a factor. Let me just rewrite the whole thing so we can work on it down here. So this is n times n minus one plus three times n minus one. And notice, both of them have an n minus one, have an n minus one as a factor. So what we could do is factor out the n minus one, or you could view it as undistributing the n minus one. And if we do that, we're gonna factor out the n minus one, and what are we going to have left over? Well, if you take out the n minus one here, if you undistribute it out, you're just going to be left with, you are just going to be left with that n. So you're gonna have an n there."}, {"video_title": "Factoring polynomials common binomial factor   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "And notice, both of them have an n minus one, have an n minus one as a factor. So what we could do is factor out the n minus one, or you could view it as undistributing the n minus one. And if we do that, we're gonna factor out the n minus one, and what are we going to have left over? Well, if you take out the n minus one here, if you undistribute it out, you're just going to be left with, you are just going to be left with that n. So you're gonna have an n there. And then for this second term, you factor this n minus one out, you're just going to be left with this positive three, plus three. And just like that, we are done. We have factored the polynomial below as a product of two binomials."}, {"video_title": "Factoring polynomials common binomial factor   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "Well, if you take out the n minus one here, if you undistribute it out, you're just going to be left with, you are just going to be left with that n. So you're gonna have an n there. And then for this second term, you factor this n minus one out, you're just going to be left with this positive three, plus three. And just like that, we are done. We have factored the polynomial below as a product of two binomials. So this is the same thing as n minus one times n plus three. And once again, you can check this. You can take this n minus one and distribute it."}, {"video_title": "Powers of zero   Exponents, radicals, and scientific notation   Pre-algebra   Khan Academy.mp3", "Sentence": "In this video, we're going to talk about powers of zero. And just as a little bit of a reminder, let's start with a non-zero number, just to remind ourselves what exponentiation is all about. So if I were to take two to the first power, one way to think about this is we always start with a one, and then we multiply this base that many times times that one. So here, we're only gonna have one, two. So it's gonna be one times two, which is of course equal to two. If I were to say what is two to the second power, well, that's going to be equal to one times, and now I'm gonna have two twos, so times two times two, which is equal to four. And you could keep going like that."}, {"video_title": "Powers of zero   Exponents, radicals, and scientific notation   Pre-algebra   Khan Academy.mp3", "Sentence": "So here, we're only gonna have one, two. So it's gonna be one times two, which is of course equal to two. If I were to say what is two to the second power, well, that's going to be equal to one times, and now I'm gonna have two twos, so times two times two, which is equal to four. And you could keep going like that. Now, the reason why I have this one here, and we've done this before, is to justify, and there's many other good reasons why two to the zero power should be equal to one. But you could see if we use the same exact idea here, you start with a one, and then you multiply it by two zero times. Well, that's just going to end up with a one."}, {"video_title": "Powers of zero   Exponents, radicals, and scientific notation   Pre-algebra   Khan Academy.mp3", "Sentence": "And you could keep going like that. Now, the reason why I have this one here, and we've done this before, is to justify, and there's many other good reasons why two to the zero power should be equal to one. But you could see if we use the same exact idea here, you start with a one, and then you multiply it by two zero times. Well, that's just going to end up with a one. So, so far, I've told you this video's about powers of zero, but I've been doing powers of two, so let's focus on zero now. So what do you think zero to the first power is going to be? Pause this video and try to figure that out."}, {"video_title": "Powers of zero   Exponents, radicals, and scientific notation   Pre-algebra   Khan Academy.mp3", "Sentence": "Well, that's just going to end up with a one. So, so far, I've told you this video's about powers of zero, but I've been doing powers of two, so let's focus on zero now. So what do you think zero to the first power is going to be? Pause this video and try to figure that out. Well, you do the exact same idea. You start with a one, and then multiply it by zero one time. So times zero, and this is going to be equal to zero."}, {"video_title": "Powers of zero   Exponents, radicals, and scientific notation   Pre-algebra   Khan Academy.mp3", "Sentence": "Pause this video and try to figure that out. Well, you do the exact same idea. You start with a one, and then multiply it by zero one time. So times zero, and this is going to be equal to zero. What do you think zero to the second power is going to be equal to? Pause this video and think about that. Well, it's gonna be one times zero twice."}, {"video_title": "Powers of zero   Exponents, radicals, and scientific notation   Pre-algebra   Khan Academy.mp3", "Sentence": "So times zero, and this is going to be equal to zero. What do you think zero to the second power is going to be equal to? Pause this video and think about that. Well, it's gonna be one times zero twice. So times zero times zero, and I think you see where this is going. This is also going to be equal to zero. What do you think zero to some arbitrary positive, positive integer is going to be?"}, {"video_title": "Powers of zero   Exponents, radicals, and scientific notation   Pre-algebra   Khan Academy.mp3", "Sentence": "Well, it's gonna be one times zero twice. So times zero times zero, and I think you see where this is going. This is also going to be equal to zero. What do you think zero to some arbitrary positive, positive integer is going to be? Well, it's going to be equal to one times zero, that positive integer number of times. So once again, it's going to be equal to zero. And in general, you can extend that zero to any positive value exponent is going to give you zero."}, {"video_title": "Powers of zero   Exponents, radicals, and scientific notation   Pre-algebra   Khan Academy.mp3", "Sentence": "What do you think zero to some arbitrary positive, positive integer is going to be? Well, it's going to be equal to one times zero, that positive integer number of times. So once again, it's going to be equal to zero. And in general, you can extend that zero to any positive value exponent is going to give you zero. So that's pretty straightforward. But there is an interesting edge case here. What do you think zero to the zeroth power should be?"}, {"video_title": "Powers of zero   Exponents, radicals, and scientific notation   Pre-algebra   Khan Academy.mp3", "Sentence": "And in general, you can extend that zero to any positive value exponent is going to give you zero. So that's pretty straightforward. But there is an interesting edge case here. What do you think zero to the zeroth power should be? Pause this video and think about that. Well, this is actually contested. Different people will tell you different things."}, {"video_title": "Powers of zero   Exponents, radicals, and scientific notation   Pre-algebra   Khan Academy.mp3", "Sentence": "What do you think zero to the zeroth power should be? Pause this video and think about that. Well, this is actually contested. Different people will tell you different things. If you use the intuition behind exponentiation that we've been using in this video, you would say, all right, I would start with a one and then multiply it by zero zero times. Or in other words, I just wouldn't multiply it by zero, in which case I'm just left with the one. The zero to the zeroth power should be equal to one."}, {"video_title": "Powers of zero   Exponents, radicals, and scientific notation   Pre-algebra   Khan Academy.mp3", "Sentence": "Different people will tell you different things. If you use the intuition behind exponentiation that we've been using in this video, you would say, all right, I would start with a one and then multiply it by zero zero times. Or in other words, I just wouldn't multiply it by zero, in which case I'm just left with the one. The zero to the zeroth power should be equal to one. Other folks would say, hey, no, I'm with a zero and that's the zeroth power, maybe it should be a zero. And that's why a lot of folks leave it undefined. Most of the time, you're going to see zero to the zeroth power either being undefined or that it is equal to one."}, {"video_title": "Example 4 Factor a polynomial with two variables by taking a common factor   Khan Academy.mp3", "Sentence": "So let me rewrite it. So we have 4x to the fourth y, and we have minus 8x to the third y, and then we have minus 2x squared. So in the other videos, we looked at it in terms of breaking it down to its simplest parts, but I think we have enough practice now to be able to do a little bit more of it in our head. So what is the largest number that divides into all of these? When I say number, I'm actually talking about the actual, I guess, coefficients. We have a 4, an 8, a 2. We don't have to worry about the negative signs just yet."}, {"video_title": "Example 4 Factor a polynomial with two variables by taking a common factor   Khan Academy.mp3", "Sentence": "So what is the largest number that divides into all of these? When I say number, I'm actually talking about the actual, I guess, coefficients. We have a 4, an 8, a 2. We don't have to worry about the negative signs just yet. And we say, well, the largest common factor of 2, 8, and 4 is 2. 2 goes into all of them, and obviously that's the largest number that can go into 2. So that is the largest number that's going to be part of the greatest common factor."}, {"video_title": "Example 4 Factor a polynomial with two variables by taking a common factor   Khan Academy.mp3", "Sentence": "We don't have to worry about the negative signs just yet. And we say, well, the largest common factor of 2, 8, and 4 is 2. 2 goes into all of them, and obviously that's the largest number that can go into 2. So that is the largest number that's going to be part of the greatest common factor. So let's write that down. So it's going to be 2. And then what's the greatest, I guess, factor, what's the greatest degree of x that's divisible into all three of these?"}, {"video_title": "Example 4 Factor a polynomial with two variables by taking a common factor   Khan Academy.mp3", "Sentence": "So that is the largest number that's going to be part of the greatest common factor. So let's write that down. So it's going to be 2. And then what's the greatest, I guess, factor, what's the greatest degree of x that's divisible into all three of these? Well, x squared goes into all three of these, and obviously that's the greatest degree of x that can be divided into this last term. So x squared is going to be the greatest common x degree in all of them, 2x squared. And then what's the largest degree of y that's divisible into all of them?"}, {"video_title": "Example 4 Factor a polynomial with two variables by taking a common factor   Khan Academy.mp3", "Sentence": "And then what's the greatest, I guess, factor, what's the greatest degree of x that's divisible into all three of these? Well, x squared goes into all three of these, and obviously that's the greatest degree of x that can be divided into this last term. So x squared is going to be the greatest common x degree in all of them, 2x squared. And then what's the largest degree of y that's divisible into all of them? Well, these two guys are divisible by y, but this guy isn't. So there is no degree of y that's divisible into all of them. So the greatest common factor of all three of these guys right here is 2x squared."}, {"video_title": "Example 4 Factor a polynomial with two variables by taking a common factor   Khan Academy.mp3", "Sentence": "And then what's the largest degree of y that's divisible into all of them? Well, these two guys are divisible by y, but this guy isn't. So there is no degree of y that's divisible into all of them. So the greatest common factor of all three of these guys right here is 2x squared. So what we can do now is we can think about each of these terms as a product of 2x squared and something else. And to figure out something else, we can literally undistribute the 2x squared and say this is the same thing. Or even before we undistribute the 2x squared, we could say, look, 4x to the fourth y is the same thing as 2x squared times 4x to the fourth y over 2x squared."}, {"video_title": "Example 4 Factor a polynomial with two variables by taking a common factor   Khan Academy.mp3", "Sentence": "So the greatest common factor of all three of these guys right here is 2x squared. So what we can do now is we can think about each of these terms as a product of 2x squared and something else. And to figure out something else, we can literally undistribute the 2x squared and say this is the same thing. Or even before we undistribute the 2x squared, we could say, look, 4x to the fourth y is the same thing as 2x squared times 4x to the fourth y over 2x squared. Right? If you just multiply this out, you'd get 4xy. Similarly, you could say that 8x to the third y, I'll put the negative out front, you could say that 8x to the third y is the same thing as 2x squared, our greatest common factor, times 8x to the third y over 2x squared."}, {"video_title": "Example 4 Factor a polynomial with two variables by taking a common factor   Khan Academy.mp3", "Sentence": "Or even before we undistribute the 2x squared, we could say, look, 4x to the fourth y is the same thing as 2x squared times 4x to the fourth y over 2x squared. Right? If you just multiply this out, you'd get 4xy. Similarly, you could say that 8x to the third y, I'll put the negative out front, you could say that 8x to the third y is the same thing as 2x squared, our greatest common factor, times 8x to the third y over 2x squared. And then finally, 2x squared is the same thing as if we factor out 2x squared. So we have that negative sign out front. We have this negative sign, 2x squared."}, {"video_title": "Example 4 Factor a polynomial with two variables by taking a common factor   Khan Academy.mp3", "Sentence": "Similarly, you could say that 8x to the third y, I'll put the negative out front, you could say that 8x to the third y is the same thing as 2x squared, our greatest common factor, times 8x to the third y over 2x squared. And then finally, 2x squared is the same thing as if we factor out 2x squared. So we have that negative sign out front. We have this negative sign, 2x squared. If we factor out 2x squared, same thing as 2x squared times 2x squared over 2x squared. This is almost silly what I'm doing here, but I'm just showing you that I'm just multiplying and dividing both of these terms by 2x squared. Multiplying and dividing."}, {"video_title": "Example 4 Factor a polynomial with two variables by taking a common factor   Khan Academy.mp3", "Sentence": "We have this negative sign, 2x squared. If we factor out 2x squared, same thing as 2x squared times 2x squared over 2x squared. This is almost silly what I'm doing here, but I'm just showing you that I'm just multiplying and dividing both of these terms by 2x squared. Multiplying and dividing. Here it's trivially simple. This just simplifies to 2x squared right there, or this 2x squared times 1. That simplifies to 1."}, {"video_title": "Example 4 Factor a polynomial with two variables by taking a common factor   Khan Academy.mp3", "Sentence": "Multiplying and dividing. Here it's trivially simple. This just simplifies to 2x squared right there, or this 2x squared times 1. That simplifies to 1. Maybe I should write it below. That simplifies to 1. But what do these simplify to?"}, {"video_title": "Example 4 Factor a polynomial with two variables by taking a common factor   Khan Academy.mp3", "Sentence": "That simplifies to 1. Maybe I should write it below. That simplifies to 1. But what do these simplify to? This first term over here, this simplifies to 2x squared times, now you get 4 divided by 2 is 2. X to the fourth divided by x squared is x squared. And then y divided by 1 is just going to be a y."}, {"video_title": "Example 4 Factor a polynomial with two variables by taking a common factor   Khan Academy.mp3", "Sentence": "But what do these simplify to? This first term over here, this simplifies to 2x squared times, now you get 4 divided by 2 is 2. X to the fourth divided by x squared is x squared. And then y divided by 1 is just going to be a y. So it's 2x squared times 2x squared y. And then you have minus 2x squared times 8 divided by 2 is 4. X to the third divided by x squared is x."}, {"video_title": "Example 4 Factor a polynomial with two variables by taking a common factor   Khan Academy.mp3", "Sentence": "And then y divided by 1 is just going to be a y. So it's 2x squared times 2x squared y. And then you have minus 2x squared times 8 divided by 2 is 4. X to the third divided by x squared is x. And y divided by 1, you can imagine, is just y. And then finally, of course, you have minus 2x squared times this right here simplifies to 1. Times 1."}, {"video_title": "Example 4 Factor a polynomial with two variables by taking a common factor   Khan Academy.mp3", "Sentence": "X to the third divided by x squared is x. And y divided by 1, you can imagine, is just y. And then finally, of course, you have minus 2x squared times this right here simplifies to 1. Times 1. Now, if you were to undistribute 2x squared out of the expression, you'd essentially get 2x squared times this term minus this term minus this term. If you distribute this out, if you take that out of each of the terms, you're going to get 2x squared times this 2x squared y minus 4xy. And then you have minus 1."}, {"video_title": "Example 4 Factor a polynomial with two variables by taking a common factor   Khan Academy.mp3", "Sentence": "Times 1. Now, if you were to undistribute 2x squared out of the expression, you'd essentially get 2x squared times this term minus this term minus this term. If you distribute this out, if you take that out of each of the terms, you're going to get 2x squared times this 2x squared y minus 4xy. And then you have minus 1. And we're done. We've factored the problem. Now, it looks like we did a lot of steps."}, {"video_title": "Example 4 Factor a polynomial with two variables by taking a common factor   Khan Academy.mp3", "Sentence": "And then you have minus 1. And we're done. We've factored the problem. Now, it looks like we did a lot of steps. And the reason why I kind of went through great pains to show you exactly what we're doing is so you know exactly what we're doing. In the future, you might be able to do this a little bit quicker. You might be able to do many of the steps in your head."}, {"video_title": "Example 4 Factor a polynomial with two variables by taking a common factor   Khan Academy.mp3", "Sentence": "Now, it looks like we did a lot of steps. And the reason why I kind of went through great pains to show you exactly what we're doing is so you know exactly what we're doing. In the future, you might be able to do this a little bit quicker. You might be able to do many of the steps in your head. You might say, okay, let me look at each of these. Well, the biggest coefficient that divides all of these is a 2. So let me factor a 2 out."}, {"video_title": "Example 4 Factor a polynomial with two variables by taking a common factor   Khan Academy.mp3", "Sentence": "You might be able to do many of the steps in your head. You might say, okay, let me look at each of these. Well, the biggest coefficient that divides all of these is a 2. So let me factor a 2 out. Well, all of these are divisible by x squared. That's the largest degree of x. Let me factor an x squared out."}, {"video_title": "Example 4 Factor a polynomial with two variables by taking a common factor   Khan Academy.mp3", "Sentence": "So let me factor a 2 out. Well, all of these are divisible by x squared. That's the largest degree of x. Let me factor an x squared out. And this guy doesn't have a y, so I can't factor a y out. So let's say it's going to be 2x squared times. And what's this guy divided by 2x squared?"}, {"video_title": "Example 4 Factor a polynomial with two variables by taking a common factor   Khan Academy.mp3", "Sentence": "Let me factor an x squared out. And this guy doesn't have a y, so I can't factor a y out. So let's say it's going to be 2x squared times. And what's this guy divided by 2x squared? Well, 4 divided by 2 is 2. x to the fourth divided by x squared is x squared. y divided by 1. There is no other y degree that we factored out, so it's just going to be y."}, {"video_title": "Example 4 Factor a polynomial with two variables by taking a common factor   Khan Academy.mp3", "Sentence": "And what's this guy divided by 2x squared? Well, 4 divided by 2 is 2. x to the fourth divided by x squared is x squared. y divided by 1. There is no other y degree that we factored out, so it's just going to be y. And then you have minus 8 divided by 2 is 4. x to the third divided by x squared is x. And you have y divided by, say, 1 is just y. And then you have minus 2 divided by 2 is 1. x squared divided by x squared is 1."}, {"video_title": "Example 4 Factor a polynomial with two variables by taking a common factor   Khan Academy.mp3", "Sentence": "There is no other y degree that we factored out, so it's just going to be y. And then you have minus 8 divided by 2 is 4. x to the third divided by x squared is x. And you have y divided by, say, 1 is just y. And then you have minus 2 divided by 2 is 1. x squared divided by x squared is 1. So 2x squared divided by 2x squared is just 1. So in the future, you'll do it more like this, where you kind of just factor it out in your head. But I really want you to understand what we did here."}, {"video_title": "Example 4 Factor a polynomial with two variables by taking a common factor   Khan Academy.mp3", "Sentence": "And then you have minus 2 divided by 2 is 1. x squared divided by x squared is 1. So 2x squared divided by 2x squared is just 1. So in the future, you'll do it more like this, where you kind of just factor it out in your head. But I really want you to understand what we did here. There is no magic. And to realize that there's no magic, you could just use the distributive property to multiply this out again. To multiply it out again, you're going to see that you get exactly this."}, {"video_title": "Factoring difference of squares missing values   High School Math   Khan Academy.mp3", "Sentence": "Brandon wrote that g could be equal to 10. Which student is correct? Now when you look at this, it seems really daunting, all this m's and g's here, but we just need to realize that they're factoring out, first they factor out a 4y from the 36y to the third minus 100y, and it looks like whatever's left is a difference of squares, which they then factor even further. So I encourage you to pause the video and just factor this out as much as you can, first factoring out a 4y, and then we can think about what g is going to be equal to or whether Sally or Brandon is correct. So now let's work through this together. So if we look at this expression right over here, we want to factor out a 4y. So 36y to the third minus 100y, that's the same thing as, 36y to the third is the same thing as 4y times, let's see, 4y times 9y squared, right?"}, {"video_title": "Factoring difference of squares missing values   High School Math   Khan Academy.mp3", "Sentence": "So I encourage you to pause the video and just factor this out as much as you can, first factoring out a 4y, and then we can think about what g is going to be equal to or whether Sally or Brandon is correct. So now let's work through this together. So if we look at this expression right over here, we want to factor out a 4y. So 36y to the third minus 100y, that's the same thing as, 36y to the third is the same thing as 4y times, let's see, 4y times 9y squared, right? Because four times nine is 36, and y times y squared is y to the third. So all I did to get the 9y squared is I divided 36 by four to get the nine, and I divided y to the third by the y to get y squared. So if you factor out a 4y, you're left with 9y squared for that first term, and then for this second term, let's see, if we, we're gonna subtract, if we factor out a 4y again, we factor out a 4y, what's left over?"}, {"video_title": "Factoring difference of squares missing values   High School Math   Khan Academy.mp3", "Sentence": "So 36y to the third minus 100y, that's the same thing as, 36y to the third is the same thing as 4y times, let's see, 4y times 9y squared, right? Because four times nine is 36, and y times y squared is y to the third. So all I did to get the 9y squared is I divided 36 by four to get the nine, and I divided y to the third by the y to get y squared. So if you factor out a 4y, you're left with 9y squared for that first term, and then for this second term, let's see, if we, we're gonna subtract, if we factor out a 4y again, we factor out a 4y, what's left over? 100 divided by four is 25, 25, and then y, and then y divided by y is just one, so we're just left with a 25 here. So just to be clear what's going on, this 36y to the third, I just rewrote it as 4y times 9y squared. One way to think about it is I wrote it with the 4y factored out, and then the 100y right over here, I wrote it with the 4y factored out."}, {"video_title": "Factoring difference of squares missing values   High School Math   Khan Academy.mp3", "Sentence": "So if you factor out a 4y, you're left with 9y squared for that first term, and then for this second term, let's see, if we, we're gonna subtract, if we factor out a 4y again, we factor out a 4y, what's left over? 100 divided by four is 25, 25, and then y, and then y divided by y is just one, so we're just left with a 25 here. So just to be clear what's going on, this 36y to the third, I just rewrote it as 4y times 9y squared. One way to think about it is I wrote it with the 4y factored out, and then the 100y right over here, I wrote it with the 4y factored out. So it's 4y times 25. And now it's very clear that we can factor out 4y from this entire thing. So we can factor out, you could think of it as undistributing the 4y, so this is going to be equal to, this is going to be equal to 4y, and what is left over?"}, {"video_title": "Factoring difference of squares missing values   High School Math   Khan Academy.mp3", "Sentence": "One way to think about it is I wrote it with the 4y factored out, and then the 100y right over here, I wrote it with the 4y factored out. So it's 4y times 25. And now it's very clear that we can factor out 4y from this entire thing. So we can factor out, you could think of it as undistributing the 4y, so this is going to be equal to, this is going to be equal to 4y, and what is left over? Well, if you factor out a 4y of this first term, we're gonna have a 9y squared, 9y squared, and then minus 25, and then we're gonna be left with minus 25. And when we write it like this, we see what we have in parentheses here, this is a difference of squares. And we could skip a step, but let me just rewrite it."}, {"video_title": "Factoring difference of squares missing values   High School Math   Khan Academy.mp3", "Sentence": "So we can factor out, you could think of it as undistributing the 4y, so this is going to be equal to, this is going to be equal to 4y, and what is left over? Well, if you factor out a 4y of this first term, we're gonna have a 9y squared, 9y squared, and then minus 25, and then we're gonna be left with minus 25. And when we write it like this, we see what we have in parentheses here, this is a difference of squares. And we could skip a step, but let me just rewrite it. So we could rewrite it as literally a difference of squares. 9y squared, that is the same thing as, that is the same thing as 3y, that whole thing to the second power, 3 squared is 9, y squared is y squared, and then we have minus 25, we can rewrite as 5 squared. So you see, we have a difference of squares, and we've seen this pattern multiple times."}, {"video_title": "Factoring difference of squares missing values   High School Math   Khan Academy.mp3", "Sentence": "And we could skip a step, but let me just rewrite it. So we could rewrite it as literally a difference of squares. 9y squared, that is the same thing as, that is the same thing as 3y, that whole thing to the second power, 3 squared is 9, y squared is y squared, and then we have minus 25, we can rewrite as 5 squared. So you see, we have a difference of squares, and we've seen this pattern multiple times. If this is the first time you're seeing it, I encourage you to watch the videos on Khan Academy on difference of squares, but we know anything of the form, anything of the form a squared minus b squared, minus b squared, let me do it in that color, minus b squared can be factored as being equal to, this is equal to, if I were to write it as a product of two binomials, this is going to be equal to a plus b, times a minus b, and you can verify that that works if you've never seen this before, or you can watch those videos for review. So this right over here can be rewritten as 4y, which we factored out at the beginning, is going to be times the product of two binomials for this part right over here, and so in this case, a is 3y, so it's going to be 3y plus 5 times 3y minus 5, so let me write that down. So 3y plus 5, plus 5 times 3y minus 5."}, {"video_title": "Factoring difference of squares missing values   High School Math   Khan Academy.mp3", "Sentence": "So you see, we have a difference of squares, and we've seen this pattern multiple times. If this is the first time you're seeing it, I encourage you to watch the videos on Khan Academy on difference of squares, but we know anything of the form, anything of the form a squared minus b squared, minus b squared, let me do it in that color, minus b squared can be factored as being equal to, this is equal to, if I were to write it as a product of two binomials, this is going to be equal to a plus b, times a minus b, and you can verify that that works if you've never seen this before, or you can watch those videos for review. So this right over here can be rewritten as 4y, which we factored out at the beginning, is going to be times the product of two binomials for this part right over here, and so in this case, a is 3y, so it's going to be 3y plus 5 times 3y minus 5, so let me write that down. So 3y plus 5, plus 5 times 3y minus 5. 3y minus 5. 3y minus 5. So now that we've factored this, let's go back to what they originally told us."}, {"video_title": "Factoring difference of squares missing values   High School Math   Khan Academy.mp3", "Sentence": "So 3y plus 5, plus 5 times 3y minus 5. 3y minus 5. 3y minus 5. So now that we've factored this, let's go back to what they originally told us. So that we have 4y, so this 4y corresponds to that 4y right over there, and then you have my plus g, and then you have my minus g. So you could view the my, the my is right over there, that's the 3y right over there, so we could say that m is equal to 3. m is equal to 3. And then we do plus 5 and minus 5, plus g and minus g. So g, if we're pattern matching right over here, g is going to be equal to 5. So g is equal to 5."}, {"video_title": "Factoring difference of squares missing values   High School Math   Khan Academy.mp3", "Sentence": "So now that we've factored this, let's go back to what they originally told us. So that we have 4y, so this 4y corresponds to that 4y right over there, and then you have my plus g, and then you have my minus g. So you could view the my, the my is right over there, that's the 3y right over there, so we could say that m is equal to 3. m is equal to 3. And then we do plus 5 and minus 5, plus g and minus g. So g, if we're pattern matching right over here, g is going to be equal to 5. So g is equal to 5. So what's interesting about this problem is that neither one of them, neither one of them are correct. So I could write neither is correct. g is equal to 5."}, {"video_title": "Squaring binomials of the form (x+a)\u00c3  \u00c2\u00b2   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "And I encourage you to pause the video and work through it on your own. All right, now let's work through this together. So we just have to remember, we're squaring the entire binomial. So this thing is going to be the same thing as x plus seven times x plus seven. I'm gonna write the second x plus seven in a different color, which is going to be helpful when we actually multiply things out. When we see it like this, then we can multiply these out the way we would multiply any binomials. And I'll first do it, I guess you could say the slower way, but the more intuitive way, applying the distributive property twice, and then we'll think about maybe some shortcuts or some patterns we might be able to recognize, especially when we are squaring binomials."}, {"video_title": "Squaring binomials of the form (x+a)\u00c3  \u00c2\u00b2   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "So this thing is going to be the same thing as x plus seven times x plus seven. I'm gonna write the second x plus seven in a different color, which is going to be helpful when we actually multiply things out. When we see it like this, then we can multiply these out the way we would multiply any binomials. And I'll first do it, I guess you could say the slower way, but the more intuitive way, applying the distributive property twice, and then we'll think about maybe some shortcuts or some patterns we might be able to recognize, especially when we are squaring binomials. So let's start with just applying the distributive property twice. So let's distribute this yellow x plus seven over this magenta x plus seven. So we can multiply it by the x, this magenta x."}, {"video_title": "Squaring binomials of the form (x+a)\u00c3  \u00c2\u00b2   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "And I'll first do it, I guess you could say the slower way, but the more intuitive way, applying the distributive property twice, and then we'll think about maybe some shortcuts or some patterns we might be able to recognize, especially when we are squaring binomials. So let's start with just applying the distributive property twice. So let's distribute this yellow x plus seven over this magenta x plus seven. So we can multiply it by the x, this magenta x. So it's going to be x, let me do it in that same color. So it's going to be magenta x times x plus seven, plus magenta seven times yellow x plus seven, x plus seven. And now we can apply the distributive property again."}, {"video_title": "Squaring binomials of the form (x+a)\u00c3  \u00c2\u00b2   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "So we can multiply it by the x, this magenta x. So it's going to be x, let me do it in that same color. So it's going to be magenta x times x plus seven, plus magenta seven times yellow x plus seven, x plus seven. And now we can apply the distributive property again. We can take this magenta x and distribute it over the x plus seven. So x times x is x squared. X times seven is seven x."}, {"video_title": "Squaring binomials of the form (x+a)\u00c3  \u00c2\u00b2   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "And now we can apply the distributive property again. We can take this magenta x and distribute it over the x plus seven. So x times x is x squared. X times seven is seven x. And then we can do it again over here. This seven, let me do it in a different color. So this seven times that x is going to be plus another seven x."}, {"video_title": "Squaring binomials of the form (x+a)\u00c3  \u00c2\u00b2   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "X times seven is seven x. And then we can do it again over here. This seven, let me do it in a different color. So this seven times that x is going to be plus another seven x. And then the seven times the seven is going to be 49. And we're in the home stretch. We can then simplify it."}, {"video_title": "Squaring binomials of the form (x+a)\u00c3  \u00c2\u00b2   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "So this seven times that x is going to be plus another seven x. And then the seven times the seven is going to be 49. And we're in the home stretch. We can then simplify it. This is going to be x squared. And then these two middle terms we can add together. Seven x, seven x, let me do this in orange."}, {"video_title": "Squaring binomials of the form (x+a)\u00c3  \u00c2\u00b2   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "We can then simplify it. This is going to be x squared. And then these two middle terms we can add together. Seven x, seven x, let me do this in orange. Seven x plus seven x, that's going to be 14 x, plus 14 x plus 49. So plus 49. And we're done."}, {"video_title": "Squaring binomials of the form (x+a)\u00c3  \u00c2\u00b2   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "Seven x, seven x, let me do this in orange. Seven x plus seven x, that's going to be 14 x, plus 14 x plus 49. So plus 49. And we're done. Now the key question is do we see some patterns here? Do we see some patterns that we can generalize and that might help us square binomials a little bit faster in the future? Well when we first looked at just multiplying binomials, we saw a pattern like x plus a times x plus b is going to be equal to x squared."}, {"video_title": "Squaring binomials of the form (x+a)\u00c3  \u00c2\u00b2   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "And we're done. Now the key question is do we see some patterns here? Do we see some patterns that we can generalize and that might help us square binomials a little bit faster in the future? Well when we first looked at just multiplying binomials, we saw a pattern like x plus a times x plus b is going to be equal to x squared. Let me write it this way. It's going to be equal to x squared plus a plus b, x plus b squared. And so if both a and b are the same thing, we could say that x plus a times x plus a is going to be equal to x squared."}, {"video_title": "Squaring binomials of the form (x+a)\u00c3  \u00c2\u00b2   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "Well when we first looked at just multiplying binomials, we saw a pattern like x plus a times x plus b is going to be equal to x squared. Let me write it this way. It's going to be equal to x squared plus a plus b, x plus b squared. And so if both a and b are the same thing, we could say that x plus a times x plus a is going to be equal to x squared. And this is the case when we have a coefficient of one on both of these x's, x squared. Now in this case, a and b are both a. So this is going to be a plus a times x, or we could just say plus two a x."}, {"video_title": "Squaring binomials of the form (x+a)\u00c3  \u00c2\u00b2   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "And so if both a and b are the same thing, we could say that x plus a times x plus a is going to be equal to x squared. And this is the case when we have a coefficient of one on both of these x's, x squared. Now in this case, a and b are both a. So this is going to be a plus a times x, or we could just say plus two a x. Let me be clear what I just did. Instead of writing a plus b, I could just view this as a plus a times x, and then plus a squared. Or that's the same thing as x squared, plus two a x plus a squared."}, {"video_title": "Squaring binomials of the form (x+a)\u00c3  \u00c2\u00b2   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "So this is going to be a plus a times x, or we could just say plus two a x. Let me be clear what I just did. Instead of writing a plus b, I could just view this as a plus a times x, and then plus a squared. Or that's the same thing as x squared, plus two a x plus a squared. This is a general way of expressing a squared binomial like this. A squared binomial where the coefficients on both x's are one. We can see that's exactly what we saw over here."}, {"video_title": "Squaring binomials of the form (x+a)\u00c3  \u00c2\u00b2   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "Or that's the same thing as x squared, plus two a x plus a squared. This is a general way of expressing a squared binomial like this. A squared binomial where the coefficients on both x's are one. We can see that's exactly what we saw over here. In the example we did, seven is our a. So we got x squared right over there. Let me circle it."}, {"video_title": "Squaring binomials of the form (x+a)\u00c3  \u00c2\u00b2   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "We can see that's exactly what we saw over here. In the example we did, seven is our a. So we got x squared right over there. Let me circle it. So we have this blue x squared. That corresponds to that over there. And then seven is our a, so two a x, two times seven is 14 x."}, {"video_title": "Squaring binomials of the form (x+a)\u00c3  \u00c2\u00b2   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "Let me circle it. So we have this blue x squared. That corresponds to that over there. And then seven is our a, so two a x, two times seven is 14 x. Notice we have the 14 x right over there. So this 14 x corresponds to two a x. And then finally, if a is seven, a squared is 49."}, {"video_title": "Squaring binomials of the form (x+a)\u00c3  \u00c2\u00b2   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "And then seven is our a, so two a x, two times seven is 14 x. Notice we have the 14 x right over there. So this 14 x corresponds to two a x. And then finally, if a is seven, a squared is 49. So in general, if you're squaring a binomial, a fast way to do it is to do this pattern here. We could do another example real fast just to make sure that we've understood things. If I were to tell you what is x, I'll throw a negative in here, x minus three squared."}, {"video_title": "Squaring binomials of the form (x+a)\u00c3  \u00c2\u00b2   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "And then finally, if a is seven, a squared is 49. So in general, if you're squaring a binomial, a fast way to do it is to do this pattern here. We could do another example real fast just to make sure that we've understood things. If I were to tell you what is x, I'll throw a negative in here, x minus three squared. I encourage you to pause the video and think about expressing this using this pattern. Well, this is going to be, in this case, our a, we have to be careful, our a is going to be negative three. So that is our a right over there."}, {"video_title": "Squaring binomials of the form (x+a)\u00c3  \u00c2\u00b2   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "If I were to tell you what is x, I'll throw a negative in here, x minus three squared. I encourage you to pause the video and think about expressing this using this pattern. Well, this is going to be, in this case, our a, we have to be careful, our a is going to be negative three. So that is our a right over there. So this is going to be equal to x squared. Now, two a x, let me do it in the same colors actually, just for fun. So it's going to be x squared."}, {"video_title": "Squaring binomials of the form (x+a)\u00c3  \u00c2\u00b2   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "So that is our a right over there. So this is going to be equal to x squared. Now, two a x, let me do it in the same colors actually, just for fun. So it's going to be x squared. Now, what is two times a times x? A is negative three. So two times a is negative six."}, {"video_title": "Squaring binomials of the form (x+a)\u00c3  \u00c2\u00b2   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "So it's going to be x squared. Now, what is two times a times x? A is negative three. So two times a is negative six. So it's going to be negative six x. So minus six x, that's two times a is the coefficient, and then we have our x there. And then plus a squared."}, {"video_title": "Squaring binomials of the form (x+a)\u00c3  \u00c2\u00b2   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "So two times a is negative six. So it's going to be negative six x. So minus six x, that's two times a is the coefficient, and then we have our x there. And then plus a squared. Well, if a is negative three, what is negative three times negative three? It's going to be positive nine. And just like that, when we looked at this pattern, we were able to very quickly figure out what this binomial squared actually is."}, {"video_title": "Combining like terms introduction   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "So I'm gonna add another three Chuck Norris's, and this might seem a little bit obvious, but how many Chuck Norris's do I now have? Well, two Chuck Norris's, we can represent this as literally a Chuck Norris plus a Chuck Norris. So let me do that, a Chuck Norris plus another Chuck Norris. Two Chuck Norris's, you could also do this two times Chuck Norris, and this is just another way of representing it. And three Chuck Norris's, three Chuck Norris's, you could do that as a Chuck Norris plus a Chuck Norris plus another Chuck Norris. And so we would have a grand total, and this might be very simple for you, but you would have a grand total of one, two, three, four, five Chuck Norris's. So this would be equal to five Chuck Norris's."}, {"video_title": "Combining like terms introduction   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Two Chuck Norris's, you could also do this two times Chuck Norris, and this is just another way of representing it. And three Chuck Norris's, three Chuck Norris's, you could do that as a Chuck Norris plus a Chuck Norris plus another Chuck Norris. And so we would have a grand total, and this might be very simple for you, but you would have a grand total of one, two, three, four, five Chuck Norris's. So this would be equal to five Chuck Norris's. Now let's get a little bit more abstract here. Chuck Norris is a very tangible thing. So let's go to a little bit more of traditional algebraic notation."}, {"video_title": "Combining like terms introduction   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "So this would be equal to five Chuck Norris's. Now let's get a little bit more abstract here. Chuck Norris is a very tangible thing. So let's go to a little bit more of traditional algebraic notation. If I have two X's, two X's, and remember, two X, you could do this as two X's or two times X, and to that I would add three X's. And to that I would add three X's. How many X's do I have?"}, {"video_title": "Combining like terms introduction   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "So let's go to a little bit more of traditional algebraic notation. If I have two X's, two X's, and remember, two X, you could do this as two X's or two times X, and to that I would add three X's. And to that I would add three X's. How many X's do I have? Well, once again, two X's, that's two times X. You could do that as an X plus an X. We don't know what the value of X is, but whatever that value is, we can add it to itself."}, {"video_title": "Combining like terms introduction   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "How many X's do I have? Well, once again, two X's, that's two times X. You could do that as an X plus an X. We don't know what the value of X is, but whatever that value is, we can add it to itself. And then three X's are going to be that value. Let me do that in that same green color. Three X's are going to be that value plus that value plus whatever that value is."}, {"video_title": "Combining like terms introduction   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "We don't know what the value of X is, but whatever that value is, we can add it to itself. And then three X's are going to be that value. Let me do that in that same green color. Three X's are going to be that value plus that value plus whatever that value is. And so how many X's do I now have? Well, I'm gonna have one, two, three, four, five X's. So two X plus three X is equal to five X."}, {"video_title": "Combining like terms introduction   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Three X's are going to be that value plus that value plus whatever that value is. And so how many X's do I now have? Well, I'm gonna have one, two, three, four, five X's. So two X plus three X is equal to five X. And if you think about it, all we really did, and hopefully you conceptually get it, is we just added the two numbers that we're multiplying the X. And these numbers, the two or the three, they're called coefficients. Very fancy word, but it's just this constant number, this regular number that's multiplied by the variable."}, {"video_title": "Combining like terms introduction   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "So two X plus three X is equal to five X. And if you think about it, all we really did, and hopefully you conceptually get it, is we just added the two numbers that we're multiplying the X. And these numbers, the two or the three, they're called coefficients. Very fancy word, but it's just this constant number, this regular number that's multiplied by the variable. You just added the two and the three to get your five X. Now let's think about this a little bit more. Let's go back to this original expression, the two Chuck Norris's plus three Chuck Norris's."}, {"video_title": "Combining like terms introduction   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Very fancy word, but it's just this constant number, this regular number that's multiplied by the variable. You just added the two and the three to get your five X. Now let's think about this a little bit more. Let's go back to this original expression, the two Chuck Norris's plus three Chuck Norris's. Let's say to that we were to add, let's say we were to add some type of a, let's say we were to add seven plums over here. So this is my drawing of a plum. So we have seven plums plus two Chuck Norris's plus three Chuck Norris's."}, {"video_title": "Combining like terms introduction   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Let's go back to this original expression, the two Chuck Norris's plus three Chuck Norris's. Let's say to that we were to add, let's say we were to add some type of a, let's say we were to add seven plums over here. So this is my drawing of a plum. So we have seven plums plus two Chuck Norris's plus three Chuck Norris's. And let's say that I add another two plums. I add another two plums here. So what would this whole thing be?"}, {"video_title": "Combining like terms introduction   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "So we have seven plums plus two Chuck Norris's plus three Chuck Norris's. And let's say that I add another two plums. I add another two plums here. So what would this whole thing be? Well, I wouldn't add the seven to the two to the three plus the two. We're adding different things here. You have two Chuck Norris's and three Chuck Norris's, so they're still going to simplify to five Chuck Norris's."}, {"video_title": "Combining like terms introduction   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "So what would this whole thing be? Well, I wouldn't add the seven to the two to the three plus the two. We're adding different things here. You have two Chuck Norris's and three Chuck Norris's, so they're still going to simplify to five Chuck Norris's. And then we would separately think about the plums. We have seven plums and we're adding another two plums. We're going to have nine plums plus nine plums."}, {"video_title": "Combining like terms introduction   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "You have two Chuck Norris's and three Chuck Norris's, so they're still going to simplify to five Chuck Norris's. And then we would separately think about the plums. We have seven plums and we're adding another two plums. We're going to have nine plums plus nine plums. So this simplifies to five Chuck Norris's and nine plums. Similarly over here, if I had, instead of just two X plus three X, if I had seven Y, seven Y plus two X plus three X plus two Y, what do I now have? Well, I can't add the X's and the Y's."}, {"video_title": "Combining like terms introduction   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "We're going to have nine plums plus nine plums. So this simplifies to five Chuck Norris's and nine plums. Similarly over here, if I had, instead of just two X plus three X, if I had seven Y, seven Y plus two X plus three X plus two Y, what do I now have? Well, I can't add the X's and the Y's. They could very well represent a different number. So all I can do is really add the X's and then I get the five X, and then I'd separately add the Y. If I have seven Y's and to that I add two Y's, I'm going to have nine Y's."}, {"video_title": "Combining like terms introduction   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Well, I can't add the X's and the Y's. They could very well represent a different number. So all I can do is really add the X's and then I get the five X, and then I'd separately add the Y. If I have seven Y's and to that I add two Y's, I'm going to have nine Y's. If I have seven of something and I add two of something, I now have nine of that something. So I'm going to have nine Y's. So you add that, we've done a different color."}, {"video_title": "Combining like terms introduction   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "If I have seven Y's and to that I add two Y's, I'm going to have nine Y's. If I have seven of something and I add two of something, I now have nine of that something. So I'm going to have nine Y's. So you add that, we've done a different color. You add this and this, you get that. You add the X's, you add the X's, you get that right over there. So hopefully that makes a little sense."}, {"video_title": "Combining like terms introduction   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "So you add that, we've done a different color. You add this and this, you get that. You add the X's, you add the X's, you get that right over there. So hopefully that makes a little sense. Actually, I'll throw out one more idea. So given this, what would happen if I were to have two X plus one plus seven X plus five? Well, once again, you might be tempted to add the two plus the one, but they're adding different things."}, {"video_title": "Combining like terms introduction   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "So hopefully that makes a little sense. Actually, I'll throw out one more idea. So given this, what would happen if I were to have two X plus one plus seven X plus five? Well, once again, you might be tempted to add the two plus the one, but they're adding different things. These are two X's. This is just the number one. So you really just have to add the X's together."}, {"video_title": "Combining like terms introduction   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Well, once again, you might be tempted to add the two plus the one, but they're adding different things. These are two X's. This is just the number one. So you really just have to add the X's together. So you're going to say, well, I got two X's and I'm going to add seven X's to that. Well, that means I now have nine X's. And then separately you would say, well, I've got just the abstract number one and then I got another five."}, {"video_title": "Slope-intercept equation from slope and point   Algebra I   Khan Academy.mp3", "Sentence": "What is the equation of this line in slope intercept form? So any line can be represented in slope intercept form as y is equal to mx plus b. Where this m right over here, that is the slope of the line. And this b over here, this is the y intercept of the line. Let me draw a quick line here just so that we can visualize that a little bit. So that is my y axis and then that is my x axis. Let me draw a line."}, {"video_title": "Slope-intercept equation from slope and point   Algebra I   Khan Academy.mp3", "Sentence": "And this b over here, this is the y intercept of the line. Let me draw a quick line here just so that we can visualize that a little bit. So that is my y axis and then that is my x axis. Let me draw a line. Since our line here has a negative slope, I'll draw a downward sloping line. Let's say our line looks something like that. Hopefully we're a little familiar with the slope already."}, {"video_title": "Slope-intercept equation from slope and point   Algebra I   Khan Academy.mp3", "Sentence": "Let me draw a line. Since our line here has a negative slope, I'll draw a downward sloping line. Let's say our line looks something like that. Hopefully we're a little familiar with the slope already. The slope essentially tells us start at some point on the line and go to some other point on the line. Measure how much you have to move in the x direction, that is your run. And then measure how much you have to move in the y direction, that is your rise."}, {"video_title": "Slope-intercept equation from slope and point   Algebra I   Khan Academy.mp3", "Sentence": "Hopefully we're a little familiar with the slope already. The slope essentially tells us start at some point on the line and go to some other point on the line. Measure how much you have to move in the x direction, that is your run. And then measure how much you have to move in the y direction, that is your rise. And our slope is equal to rise over run. What you're going to see over here would be downward sloping. Because if you move in the positive x direction, we have to go down."}, {"video_title": "Slope-intercept equation from slope and point   Algebra I   Khan Academy.mp3", "Sentence": "And then measure how much you have to move in the y direction, that is your rise. And our slope is equal to rise over run. What you're going to see over here would be downward sloping. Because if you move in the positive x direction, we have to go down. If our run is positive, our rise here is negative. So this would be a negative over positive, give you a negative number. That makes sense because we're downward sloping."}, {"video_title": "Slope-intercept equation from slope and point   Algebra I   Khan Academy.mp3", "Sentence": "Because if you move in the positive x direction, we have to go down. If our run is positive, our rise here is negative. So this would be a negative over positive, give you a negative number. That makes sense because we're downward sloping. The more we go down in this situation, for every step we move to the right, the more downward sloping we'll be, the more of a negative slope we'll have. That's slope. The y intercept just tells us where we intercept the y axis."}, {"video_title": "Slope-intercept equation from slope and point   Algebra I   Khan Academy.mp3", "Sentence": "That makes sense because we're downward sloping. The more we go down in this situation, for every step we move to the right, the more downward sloping we'll be, the more of a negative slope we'll have. That's slope. The y intercept just tells us where we intercept the y axis. The y intercept, this point right over here, this is where the line intersects with the y axis. This will be the point 0, b. This actually just falls straight out of this equation."}, {"video_title": "Slope-intercept equation from slope and point   Algebra I   Khan Academy.mp3", "Sentence": "The y intercept just tells us where we intercept the y axis. The y intercept, this point right over here, this is where the line intersects with the y axis. This will be the point 0, b. This actually just falls straight out of this equation. When x is equal to 0, let's evaluate this equation when x is equal to 0. y will be equal to m times 0 plus b. Anything times 0 is 0. y is equal to 0 plus b, or y will be equal to b when x is equal to 0. When x is equal to 0."}, {"video_title": "Slope-intercept equation from slope and point   Algebra I   Khan Academy.mp3", "Sentence": "This actually just falls straight out of this equation. When x is equal to 0, let's evaluate this equation when x is equal to 0. y will be equal to m times 0 plus b. Anything times 0 is 0. y is equal to 0 plus b, or y will be equal to b when x is equal to 0. When x is equal to 0. This is the point 0, b. They tell us what the slope of this line is. They tell us a line has a slope of negative 3 4ths."}, {"video_title": "Slope-intercept equation from slope and point   Algebra I   Khan Academy.mp3", "Sentence": "When x is equal to 0. This is the point 0, b. They tell us what the slope of this line is. They tell us a line has a slope of negative 3 4ths. We know that our slope is negative 3 4ths. They tell us that the line goes through the point 0, 8. They tell us we go through the point 0, 8."}, {"video_title": "Slope-intercept equation from slope and point   Algebra I   Khan Academy.mp3", "Sentence": "They tell us a line has a slope of negative 3 4ths. We know that our slope is negative 3 4ths. They tell us that the line goes through the point 0, 8. They tell us we go through the point 0, 8. Notice x is 0. We're on the y axis. When x is 0, we're on the y axis."}, {"video_title": "Slope-intercept equation from slope and point   Algebra I   Khan Academy.mp3", "Sentence": "They tell us we go through the point 0, 8. Notice x is 0. We're on the y axis. When x is 0, we're on the y axis. This is our y intercept. Our y intercept is the point 0, 8. We could say that b is equal to 8."}, {"video_title": "Slope-intercept equation from slope and point   Algebra I   Khan Academy.mp3", "Sentence": "When x is 0, we're on the y axis. This is our y intercept. Our y intercept is the point 0, 8. We could say that b is equal to 8. We know m is equal to negative 3 4ths. b is equal to 8. We can write the equation of this line in slope intercept form."}, {"video_title": "Systems of equations word problems example 4   Algebra I   Khan Academy.mp3", "Sentence": "You are at a Parisian cafe with a friend. A local in front of you buys a cup of coffee and a croissant for $5 or 5.30 euro. When you and your friend get two cups of coffee and two croissants, you are charged 14 euro. Can we solve for the price of a cup of coffee and croissant using the information in a system of linear equations and two variables? If yes, what is the solution? If no, what is the reason we cannot? So we're looking for two things, the price of a cup of coffee and the price of a croissant."}, {"video_title": "Systems of equations word problems example 4   Algebra I   Khan Academy.mp3", "Sentence": "Can we solve for the price of a cup of coffee and croissant using the information in a system of linear equations and two variables? If yes, what is the solution? If no, what is the reason we cannot? So we're looking for two things, the price of a cup of coffee and the price of a croissant. So let's define two variables here. Since we have all these c's here, let me just use x's and y's. So let's let x be equal to the price of the cup of coffee."}, {"video_title": "Systems of equations word problems example 4   Algebra I   Khan Academy.mp3", "Sentence": "So we're looking for two things, the price of a cup of coffee and the price of a croissant. So let's define two variables here. Since we have all these c's here, let me just use x's and y's. So let's let x be equal to the price of the cup of coffee. And let's let y be equal to the price of a croissant. So we first have this information of what the local in front of us did. The local in front of us buys one cup of coffee and one croissant for 5.30 euro."}, {"video_title": "Systems of equations word problems example 4   Algebra I   Khan Academy.mp3", "Sentence": "So let's let x be equal to the price of the cup of coffee. And let's let y be equal to the price of a croissant. So we first have this information of what the local in front of us did. The local in front of us buys one cup of coffee and one croissant for 5.30 euro. So how would we set that up as an equation? Well, we got one cup of coffee, so that's going to be 1x, or we could just write it x, plus 1y, because he got one croissant. And it cost 5.30."}, {"video_title": "Systems of equations word problems example 4   Algebra I   Khan Academy.mp3", "Sentence": "The local in front of us buys one cup of coffee and one croissant for 5.30 euro. So how would we set that up as an equation? Well, we got one cup of coffee, so that's going to be 1x, or we could just write it x, plus 1y, because he got one croissant. And it cost 5.30. So it cost, let me write this, so this is the amount that he paid. 5.30 euro. So this is what the local, this equation describes what happened to the local."}, {"video_title": "Systems of equations word problems example 4   Algebra I   Khan Academy.mp3", "Sentence": "And it cost 5.30. So it cost, let me write this, so this is the amount that he paid. 5.30 euro. So this is what the local, this equation describes what happened to the local. Bought one cup of coffee, one croissant, paid 5.30. Now, when you and your friend get two cups of coffee and two croissants, you're charged 14 euro. So what's an equation to describe this?"}, {"video_title": "Systems of equations word problems example 4   Algebra I   Khan Academy.mp3", "Sentence": "So this is what the local, this equation describes what happened to the local. Bought one cup of coffee, one croissant, paid 5.30. Now, when you and your friend get two cups of coffee and two croissants, you're charged 14 euro. So what's an equation to describe this? So we should be charged 2 times the price of a cup of coffee. So it should be 2x. And then we should be charged 2 times the price of a croissant, so plus 2y."}, {"video_title": "Systems of equations word problems example 4   Algebra I   Khan Academy.mp3", "Sentence": "So what's an equation to describe this? So we should be charged 2 times the price of a cup of coffee. So it should be 2x. And then we should be charged 2 times the price of a croissant, so plus 2y. And the sum of these should be the total amount that we're charged. We've been charged 14 euro. So let's see if we can solve this system of equations."}, {"video_title": "Systems of equations word problems example 4   Algebra I   Khan Academy.mp3", "Sentence": "And then we should be charged 2 times the price of a croissant, so plus 2y. And the sum of these should be the total amount that we're charged. We've been charged 14 euro. So let's see if we can solve this system of equations. So there's many, many, many ways to solve this. But the most obvious way, at least looking at this right over here, is we have x, we have 2x, we have y, we have 2y. Let's take this first equation that described the local and multiply it by 2."}, {"video_title": "Systems of equations word problems example 4   Algebra I   Khan Academy.mp3", "Sentence": "So let's see if we can solve this system of equations. So there's many, many, many ways to solve this. But the most obvious way, at least looking at this right over here, is we have x, we have 2x, we have y, we have 2y. Let's take this first equation that described the local and multiply it by 2. So let's just multiply it by 2. So we're going to multiply both sides, otherwise the equality won't hold anymore. So we would get 2x plus 2y is equal to 2 times 5.30 is 10 euro 60."}, {"video_title": "Systems of equations word problems example 4   Algebra I   Khan Academy.mp3", "Sentence": "Let's take this first equation that described the local and multiply it by 2. So let's just multiply it by 2. So we're going to multiply both sides, otherwise the equality won't hold anymore. So we would get 2x plus 2y is equal to 2 times 5.30 is 10 euro 60. 10 euro 60. Now something very interesting is going on here. If the local had bought just twice as many cups of coffee and twice as many croissants, he would have paid 10.60."}, {"video_title": "Systems of equations word problems example 4   Algebra I   Khan Academy.mp3", "Sentence": "So we would get 2x plus 2y is equal to 2 times 5.30 is 10 euro 60. 10 euro 60. Now something very interesting is going on here. If the local had bought just twice as many cups of coffee and twice as many croissants, he would have paid 10.60. And that would have been the exact amount of coffee and croissants you got, and you paid 14. So it looks pretty clear that you got charged a different amount. You got the tourist rate for the cup of coffee and the croissant, while he got the local rate."}, {"video_title": "Systems of equations word problems example 4   Algebra I   Khan Academy.mp3", "Sentence": "If the local had bought just twice as many cups of coffee and twice as many croissants, he would have paid 10.60. And that would have been the exact amount of coffee and croissants you got, and you paid 14. So it looks pretty clear that you got charged a different amount. You got the tourist rate for the cup of coffee and the croissant, while he got the local rate. And we can verify that there's no x and y that's going to satisfy this. And even logically it makes sense. Here, 2 times an x plus 2 times a y is 14."}, {"video_title": "Systems of equations word problems example 4   Algebra I   Khan Academy.mp3", "Sentence": "You got the tourist rate for the cup of coffee and the croissant, while he got the local rate. And we can verify that there's no x and y that's going to satisfy this. And even logically it makes sense. Here, 2 times an x plus 2 times a y is 14. Here, 2 times an x plus 2 times a y is 10 euro 60. And we can even show that mathematically, that this doesn't make sense. So if we were to subtract this bottom equation from this top, so essentially you could imagine multiplying the entire bottom equation times a negative 1."}, {"video_title": "Systems of equations word problems example 4   Algebra I   Khan Academy.mp3", "Sentence": "Here, 2 times an x plus 2 times a y is 14. Here, 2 times an x plus 2 times a y is 10 euro 60. And we can even show that mathematically, that this doesn't make sense. So if we were to subtract this bottom equation from this top, so essentially you could imagine multiplying the entire bottom equation times a negative 1. So let's multiply the entire bottom equation by a negative 1. And then we add these two equations. Remember, all we're doing is we're starting with, say, this equation, and we're adding the same thing to both sides."}, {"video_title": "Systems of equations word problems example 4   Algebra I   Khan Academy.mp3", "Sentence": "So if we were to subtract this bottom equation from this top, so essentially you could imagine multiplying the entire bottom equation times a negative 1. So let's multiply the entire bottom equation by a negative 1. And then we add these two equations. Remember, all we're doing is we're starting with, say, this equation, and we're adding the same thing to both sides. We're going to add this to this side. And we already know that negative 10.60 is the same thing as this. We're going to add it to that side."}, {"video_title": "Systems of equations word problems example 4   Algebra I   Khan Academy.mp3", "Sentence": "Remember, all we're doing is we're starting with, say, this equation, and we're adding the same thing to both sides. We're going to add this to this side. And we already know that negative 10.60 is the same thing as this. We're going to add it to that side. So on the left-hand side, this cancels with this. This cancels with this. We're left with 0."}, {"video_title": "Systems of equations word problems example 4   Algebra I   Khan Academy.mp3", "Sentence": "We're going to add it to that side. So on the left-hand side, this cancels with this. This cancels with this. We're left with 0. And on the right-hand side, 14 minus 10.60 will get you to 3.40. And there's no x and y that you can think of. There's no magical x and y that can all of a sudden make 0 equal 3.40."}, {"video_title": "How to determine the domain of a modeling function (example with a function)   Khan Academy.mp3", "Sentence": "Pooja had a beautiful plant. The plant began sprouting two days before Pooja bought it, and she had it for 98 days before it died. At its tallest, the plant was 30 centimeters tall. All right. Let h of t denote the height of Pooja's plant, h, measured in centimeters, t days from the time she bought it. Which number type is more appropriate for the domain of the function? And they tell us whether it's integers or real numbers."}, {"video_title": "How to determine the domain of a modeling function (example with a function)   Khan Academy.mp3", "Sentence": "All right. Let h of t denote the height of Pooja's plant, h, measured in centimeters, t days from the time she bought it. Which number type is more appropriate for the domain of the function? And they tell us whether it's integers or real numbers. So we just have to remind ourselves, the domain of a function, that's the set of all of the inputs for which the function is defined. And the inputs here, these are t, and it represents days. And if I input t, the number of days, into the function h, it'll output the height of the plant."}, {"video_title": "How to determine the domain of a modeling function (example with a function)   Khan Academy.mp3", "Sentence": "And they tell us whether it's integers or real numbers. So we just have to remind ourselves, the domain of a function, that's the set of all of the inputs for which the function is defined. And the inputs here, these are t, and it represents days. And if I input t, the number of days, into the function h, it'll output the height of the plant. So let's think about it. At first you might think in terms of, you know, zero days, one days, two days, and you might be tempted to say integers, but why not think about one and a half days, or 3.175 days? So I don't see any reason why t couldn't be a subset of real numbers, why you can think about, you know, the 97.99th day."}, {"video_title": "How to determine the domain of a modeling function (example with a function)   Khan Academy.mp3", "Sentence": "And if I input t, the number of days, into the function h, it'll output the height of the plant. So let's think about it. At first you might think in terms of, you know, zero days, one days, two days, and you might be tempted to say integers, but why not think about one and a half days, or 3.175 days? So I don't see any reason why t couldn't be a subset of real numbers, why you can think about, you know, the 97.99th day. So I would say real numbers. And then I say define the interval of the domain. So t, let's see, the t would be defined up to two days before she bought the plant, so I would say that t is equal to negative two, all the way to 98 days."}, {"video_title": "How to determine the domain of a modeling function (example with a function)   Khan Academy.mp3", "Sentence": "So I don't see any reason why t couldn't be a subset of real numbers, why you can think about, you know, the 97.99th day. So I would say real numbers. And then I say define the interval of the domain. So t, let's see, the t would be defined up to two days before she bought the plant, so I would say that t is equal to negative two, all the way to 98 days. And t could be equal to negative two, that would be two days before she bought it, or as high as 98. So let's think about it. So the interval, I would include negative two, I would include the low point, so that's why I'm using the brackets."}, {"video_title": "How to determine the domain of a modeling function (example with a function)   Khan Academy.mp3", "Sentence": "So t, let's see, the t would be defined up to two days before she bought the plant, so I would say that t is equal to negative two, all the way to 98 days. And t could be equal to negative two, that would be two days before she bought it, or as high as 98. So let's think about it. So the interval, I would include negative two, I would include the low point, so that's why I'm using the brackets. If I wanted t to be greater than, not greater than or equal to, I would do the parentheses. But since t could be equal to negative two, I'm gonna use the brackets. And at the high end, it's 98, and we're going to include 98."}, {"video_title": "How to determine the domain of a modeling function (example with a function)   Khan Academy.mp3", "Sentence": "So the interval, I would include negative two, I would include the low point, so that's why I'm using the brackets. If I wanted t to be greater than, not greater than or equal to, I would do the parentheses. But since t could be equal to negative two, I'm gonna use the brackets. And at the high end, it's 98, and we're going to include 98. So I'm gonna put the brackets there as well. So t would be a member of real numbers, such that it is a member of this interval right over here. So negative t could be, t could be negative two, but I guess you could say negative two is going to be less than or equal to t, which is going to be less than or equal to 98."}, {"video_title": "Solving equations with the distributive property   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "We have the equation negative 9 minus this whole expression, 9x minus 6, this whole thing is being subtracted from negative 9, is equal to 3 times this whole expression, 4x plus 6. Now a good place to start is to just get rid of these parentheses, and the best way to get rid of these parentheses is to kind of multiply them out. So this has a negative 1, you just see a minus here, but it's really the same thing as having a negative 1 times this quantity, and here you have a 3 times this quantity. So let's multiply it out using the distributive property. So the left hand side of our equation we have our negative 9, and then we want to multiply the negative 1 times each of these terms. So negative 1 times 9x is negative 9x, and then negative 1 times negative 6 is plus 6, or positive 6. And then that is going to be equal to, let's distribute the 3, 3 times 4x is 12x, and then 3 times 6 is 18."}, {"video_title": "Solving equations with the distributive property   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So let's multiply it out using the distributive property. So the left hand side of our equation we have our negative 9, and then we want to multiply the negative 1 times each of these terms. So negative 1 times 9x is negative 9x, and then negative 1 times negative 6 is plus 6, or positive 6. And then that is going to be equal to, let's distribute the 3, 3 times 4x is 12x, and then 3 times 6 is 18. Now what we want to do, let's combine our constant terms if we can, we have a negative 9 and a 6 here, on this side we've combined all of our like terms, we can't combine a 12x and an 18. So let's combine this, let's combine the negative 9 and the 6, our two constant terms on the left hand side of the equation. So we're going to have this negative 9x, so we're going to have negative 9x plus, let's see we have a negative 9 and then plus 6, so negative 9 plus 6 is negative 3."}, {"video_title": "Solving equations with the distributive property   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "And then that is going to be equal to, let's distribute the 3, 3 times 4x is 12x, and then 3 times 6 is 18. Now what we want to do, let's combine our constant terms if we can, we have a negative 9 and a 6 here, on this side we've combined all of our like terms, we can't combine a 12x and an 18. So let's combine this, let's combine the negative 9 and the 6, our two constant terms on the left hand side of the equation. So we're going to have this negative 9x, so we're going to have negative 9x plus, let's see we have a negative 9 and then plus 6, so negative 9 plus 6 is negative 3. So we're going to have a negative 9x and then we have a negative 3, so minus 3 right here, that's the negative 9 plus the 6, and that is equal to 12x plus 18. Now we want to group all of the x terms on one side of the equation and all of the constant terms, the negative 3 and the positive 18 on the other side. I like to always have my x terms on the left hand side if I can, you don't have to have them on the left, so let's do that."}, {"video_title": "Solving equations with the distributive property   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So we're going to have this negative 9x, so we're going to have negative 9x plus, let's see we have a negative 9 and then plus 6, so negative 9 plus 6 is negative 3. So we're going to have a negative 9x and then we have a negative 3, so minus 3 right here, that's the negative 9 plus the 6, and that is equal to 12x plus 18. Now we want to group all of the x terms on one side of the equation and all of the constant terms, the negative 3 and the positive 18 on the other side. I like to always have my x terms on the left hand side if I can, you don't have to have them on the left, so let's do that. So if I want all my x terms on the left, I have to get rid of this 12x from the right, and the best way to do that is to subtract 12x from both sides of the equation. So let me subtract 12x from the right, and subtract 12x from the left. Now on the left hand side I have negative 9x minus 12x, so negative 9 minus 12, that's negative 21x minus 3 is equal to 12x minus 12x, well that's just nothing, that's 0, so I could write a 0 here but I don't have to write anything, that was the whole point of subtracting the 12x from the left hand side."}, {"video_title": "Solving equations with the distributive property   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "I like to always have my x terms on the left hand side if I can, you don't have to have them on the left, so let's do that. So if I want all my x terms on the left, I have to get rid of this 12x from the right, and the best way to do that is to subtract 12x from both sides of the equation. So let me subtract 12x from the right, and subtract 12x from the left. Now on the left hand side I have negative 9x minus 12x, so negative 9 minus 12, that's negative 21x minus 3 is equal to 12x minus 12x, well that's just nothing, that's 0, so I could write a 0 here but I don't have to write anything, that was the whole point of subtracting the 12x from the left hand side. And that is going to be equal to, so on the right hand side we just are left with an 18, we are just left with that 18 here, these guys cancelled out. Now let's get rid of this negative 3 from the left hand side, so on the left hand side we only have x terms, on the right hand side we only have constant terms. So the best way to cancel out a negative 3 is to add 3, so it cancels out to 0, so we're going to add 3 to the left, let's add 3 to the right."}, {"video_title": "Solving equations with the distributive property   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Now on the left hand side I have negative 9x minus 12x, so negative 9 minus 12, that's negative 21x minus 3 is equal to 12x minus 12x, well that's just nothing, that's 0, so I could write a 0 here but I don't have to write anything, that was the whole point of subtracting the 12x from the left hand side. And that is going to be equal to, so on the right hand side we just are left with an 18, we are just left with that 18 here, these guys cancelled out. Now let's get rid of this negative 3 from the left hand side, so on the left hand side we only have x terms, on the right hand side we only have constant terms. So the best way to cancel out a negative 3 is to add 3, so it cancels out to 0, so we're going to add 3 to the left, let's add 3 to the right. And we get the left hand side of the equation, we have the negative 21x, no other x term to add or subtract here, so we have negative 21x, the negative 3 and the plus 3 or the positive 3 cancel out, that was the whole point, equals, what's 18 plus 3? 18 plus 3 is 21, so now we have negative 21x is equal to 21 and we want to solve for x. So if you have something times x and you just want it to be an x, let's divide by that something, and in this case that something is negative 21, so let's divide both sides of this equation by negative 21."}, {"video_title": "Solving equations with the distributive property   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So the best way to cancel out a negative 3 is to add 3, so it cancels out to 0, so we're going to add 3 to the left, let's add 3 to the right. And we get the left hand side of the equation, we have the negative 21x, no other x term to add or subtract here, so we have negative 21x, the negative 3 and the plus 3 or the positive 3 cancel out, that was the whole point, equals, what's 18 plus 3? 18 plus 3 is 21, so now we have negative 21x is equal to 21 and we want to solve for x. So if you have something times x and you just want it to be an x, let's divide by that something, and in this case that something is negative 21, so let's divide both sides of this equation by negative 21. Divide both sides by negative 21, the left hand side, negative 21 divided by negative 21, you're just left with an x, that was the whole point behind dividing by negative 21. And we get x is equal to, what's 21 divided by negative 21? Well that's just negative 1, right?"}, {"video_title": "Solving equations with the distributive property   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So if you have something times x and you just want it to be an x, let's divide by that something, and in this case that something is negative 21, so let's divide both sides of this equation by negative 21. Divide both sides by negative 21, the left hand side, negative 21 divided by negative 21, you're just left with an x, that was the whole point behind dividing by negative 21. And we get x is equal to, what's 21 divided by negative 21? Well that's just negative 1, right? You have the positive version divided by the negative version of itself, so it's just negative 1. So that is our answer. Now let's verify that this actually works for that original equation, so let's substitute negative 1 into that original equation."}, {"video_title": "Solving equations with the distributive property   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Well that's just negative 1, right? You have the positive version divided by the negative version of itself, so it's just negative 1. So that is our answer. Now let's verify that this actually works for that original equation, so let's substitute negative 1 into that original equation. So we have negative 9, I'll do it over here, I'll do it in a different color than we've been using, we have negative 9 minus, that 1 wasn't there originally, it's there implicitly, minus 9 times negative 1. 9 times, I'll put negative 1 in parentheses, minus 6 is equal to, well actually let me just solve for the left hand side when I substitute a negative 1 there. So the left hand side becomes negative 9 minus, 9 times negative 1 is negative 9 minus 6, and so this is negative 9 minus, in parentheses, negative 9 minus 6 is negative 15."}, {"video_title": "Solving equations with the distributive property   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Now let's verify that this actually works for that original equation, so let's substitute negative 1 into that original equation. So we have negative 9, I'll do it over here, I'll do it in a different color than we've been using, we have negative 9 minus, that 1 wasn't there originally, it's there implicitly, minus 9 times negative 1. 9 times, I'll put negative 1 in parentheses, minus 6 is equal to, well actually let me just solve for the left hand side when I substitute a negative 1 there. So the left hand side becomes negative 9 minus, 9 times negative 1 is negative 9 minus 6, and so this is negative 9 minus, in parentheses, negative 9 minus 6 is negative 15. So this is equal to negative 15, and so we get negative 9, let me make sure I did that, negative 9 minus 6, yep, negative 15. So I have negative 9 minus negative 15, that's the same thing as negative 9 plus 15, which is 6. So that's what we get on the left hand side of the equation when we substitute x is equal to negative 1, we get that it equals 6."}, {"video_title": "Solving equations with the distributive property   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So the left hand side becomes negative 9 minus, 9 times negative 1 is negative 9 minus 6, and so this is negative 9 minus, in parentheses, negative 9 minus 6 is negative 15. So this is equal to negative 15, and so we get negative 9, let me make sure I did that, negative 9 minus 6, yep, negative 15. So I have negative 9 minus negative 15, that's the same thing as negative 9 plus 15, which is 6. So that's what we get on the left hand side of the equation when we substitute x is equal to negative 1, we get that it equals 6. So let's see what happens when we substitute negative 1 to the right hand side of the equation. I'll do it in green. We get 3 times 4 times negative 1, plus 6, so that is 3 times negative 4 plus 6, negative 4 plus 6 is 2, so it's 3 times 2, which is also 6."}, {"video_title": "Example Solving simple quadratic   Quadratic equations   Algebra I   Khan Academy.mp3", "Sentence": "We're asked to solve the equation 2x squared plus 3 is equal to 75. So in this situation, it looks like we might be able to isolate the x squared pretty simply because there's only one term that involves an x here, it's only this x squared term. So let's try to do that. So let me just rewrite it. We have 2x squared plus 3 is equal to 75. I'm going to try to isolate this x squared over here. The best way to do that, or at least the first step, would be to subtract 3 from both sides of this equation."}, {"video_title": "Example Solving simple quadratic   Quadratic equations   Algebra I   Khan Academy.mp3", "Sentence": "So let me just rewrite it. We have 2x squared plus 3 is equal to 75. I'm going to try to isolate this x squared over here. The best way to do that, or at least the first step, would be to subtract 3 from both sides of this equation. So let's subtract 3 from both sides. The left-hand side, we're just left with 2x squared. That was the whole point of subtracting 3 from both sides."}, {"video_title": "Example Solving simple quadratic   Quadratic equations   Algebra I   Khan Academy.mp3", "Sentence": "The best way to do that, or at least the first step, would be to subtract 3 from both sides of this equation. So let's subtract 3 from both sides. The left-hand side, we're just left with 2x squared. That was the whole point of subtracting 3 from both sides. And on the right-hand side, 75 minus 3 is 72. Now I want to isolate this x squared. I have a 2x squared here, so I could have just an x squared here if I divide this side, or really both sides, by 2."}, {"video_title": "Example Solving simple quadratic   Quadratic equations   Algebra I   Khan Academy.mp3", "Sentence": "That was the whole point of subtracting 3 from both sides. And on the right-hand side, 75 minus 3 is 72. Now I want to isolate this x squared. I have a 2x squared here, so I could have just an x squared here if I divide this side, or really both sides, by 2. But if I do it to one side, I have to do it to the other side if I want to maintain the equality. So the left side just becomes x squared, and the right-hand side is 72 divided by 2, is 36. So we're left with x squared is equal to 36."}, {"video_title": "Example Solving simple quadratic   Quadratic equations   Algebra I   Khan Academy.mp3", "Sentence": "I have a 2x squared here, so I could have just an x squared here if I divide this side, or really both sides, by 2. But if I do it to one side, I have to do it to the other side if I want to maintain the equality. So the left side just becomes x squared, and the right-hand side is 72 divided by 2, is 36. So we're left with x squared is equal to 36. And then to solve for x, we can take the positive, the plus or minus square root of both sides. So we could say the plus or, let me write it this way. If we take the square root of both sides, we would get x is equal to the plus or minus square root of 36, which is equal to plus or minus 6."}, {"video_title": "Example Solving simple quadratic   Quadratic equations   Algebra I   Khan Academy.mp3", "Sentence": "So we're left with x squared is equal to 36. And then to solve for x, we can take the positive, the plus or minus square root of both sides. So we could say the plus or, let me write it this way. If we take the square root of both sides, we would get x is equal to the plus or minus square root of 36, which is equal to plus or minus 6. Let me just write that on another line. So x is equal to plus or minus 6. And remember here, if something squared is equal to 36, that something could be the negative version or the positive version."}, {"video_title": "Example Solving simple quadratic   Quadratic equations   Algebra I   Khan Academy.mp3", "Sentence": "If we take the square root of both sides, we would get x is equal to the plus or minus square root of 36, which is equal to plus or minus 6. Let me just write that on another line. So x is equal to plus or minus 6. And remember here, if something squared is equal to 36, that something could be the negative version or the positive version. It could be the principal root or it could be the negative root. Both negative 6 squared is 36 and positive 6 squared is 36. So both of these work."}, {"video_title": "Example Solving simple quadratic   Quadratic equations   Algebra I   Khan Academy.mp3", "Sentence": "And remember here, if something squared is equal to 36, that something could be the negative version or the positive version. It could be the principal root or it could be the negative root. Both negative 6 squared is 36 and positive 6 squared is 36. So both of these work. And you could put them back into the original equation to verify it. Let's do that. If you say 2 times 6 squared plus 3, that's 2 times 36, which is 72, plus 3 is 75."}, {"video_title": "Example Solving simple quadratic   Quadratic equations   Algebra I   Khan Academy.mp3", "Sentence": "So both of these work. And you could put them back into the original equation to verify it. Let's do that. If you say 2 times 6 squared plus 3, that's 2 times 36, which is 72, plus 3 is 75. So that works. If you put negative 6 in there, you're going to get the exact same result, because negative 6 squared is also 36. 2 times 36 is 72, plus 3 is 75."}, {"video_title": "Initial value & common ratio of exponential functions   High School Math   Khan Academy.mp3", "Sentence": "We have the variable, the input into our function, it's in the exponent. And a function like this is called an exponential function. So this is an exponential, exponential function. And that's because the variable, the input into our function is sitting in its definition of what is the output of that function going to be. The input is in the exponent. I could write another exponential function. I could write f of, let's say the input is the variable t, is equal to 5 times 3 to the t. Once again, this is an exponential function."}, {"video_title": "Initial value & common ratio of exponential functions   High School Math   Khan Academy.mp3", "Sentence": "And that's because the variable, the input into our function is sitting in its definition of what is the output of that function going to be. The input is in the exponent. I could write another exponential function. I could write f of, let's say the input is the variable t, is equal to 5 times 3 to the t. Once again, this is an exponential function. Now there's a couple of interesting things to think about an exponential function. In fact, we'll explore many, many of them, but I'll get a little used to the terminology. So one thing that you might see is the notion of an initial value."}, {"video_title": "Initial value & common ratio of exponential functions   High School Math   Khan Academy.mp3", "Sentence": "I could write f of, let's say the input is the variable t, is equal to 5 times 3 to the t. Once again, this is an exponential function. Now there's a couple of interesting things to think about an exponential function. In fact, we'll explore many, many of them, but I'll get a little used to the terminology. So one thing that you might see is the notion of an initial value. Initial value. Initial value. And this is essentially the value of the function when the input is zero."}, {"video_title": "Initial value & common ratio of exponential functions   High School Math   Khan Academy.mp3", "Sentence": "So one thing that you might see is the notion of an initial value. Initial value. Initial value. And this is essentially the value of the function when the input is zero. So for in these cases, the initial value for the function h is going to be h of zero. And when we evaluate that, that's going to be 1 4th times 2 to the zero. Well, 2 to the zero power is just one."}, {"video_title": "Initial value & common ratio of exponential functions   High School Math   Khan Academy.mp3", "Sentence": "And this is essentially the value of the function when the input is zero. So for in these cases, the initial value for the function h is going to be h of zero. And when we evaluate that, that's going to be 1 4th times 2 to the zero. Well, 2 to the zero power is just one. So it's equal to 1 4th. So the initial value, at least in this case, it seemed to just be that number that sits out here. We have the initial value times some number to this exponent."}, {"video_title": "Initial value & common ratio of exponential functions   High School Math   Khan Academy.mp3", "Sentence": "Well, 2 to the zero power is just one. So it's equal to 1 4th. So the initial value, at least in this case, it seemed to just be that number that sits out here. We have the initial value times some number to this exponent. And we'll come up with the name for this number as well. But let's see if this was true over here for f of t. So if we look at its initial value, f of zero is going to be 5 times 3 to the zero power. And the same thing again."}, {"video_title": "Initial value & common ratio of exponential functions   High School Math   Khan Academy.mp3", "Sentence": "We have the initial value times some number to this exponent. And we'll come up with the name for this number as well. But let's see if this was true over here for f of t. So if we look at its initial value, f of zero is going to be 5 times 3 to the zero power. And the same thing again. 3 to the zero is just one. 5 times one is just five. So the initial value is once again that."}, {"video_title": "Initial value & common ratio of exponential functions   High School Math   Khan Academy.mp3", "Sentence": "And the same thing again. 3 to the zero is just one. 5 times one is just five. So the initial value is once again that. So if you have exponential functions of this form, it makes sense. The initial value, well, if you put a zero in for the exponent, then the number raised to exponents is going to be one. And you're just going to be left with that thing that you're multiplying by that."}, {"video_title": "Initial value & common ratio of exponential functions   High School Math   Khan Academy.mp3", "Sentence": "So the initial value is once again that. So if you have exponential functions of this form, it makes sense. The initial value, well, if you put a zero in for the exponent, then the number raised to exponents is going to be one. And you're just going to be left with that thing that you're multiplying by that. Hopefully that makes sense. But since you're looking at it, hopefully it does make a little bit. Now, you might be saying, well, what do we call this number?"}, {"video_title": "Initial value & common ratio of exponential functions   High School Math   Khan Academy.mp3", "Sentence": "And you're just going to be left with that thing that you're multiplying by that. Hopefully that makes sense. But since you're looking at it, hopefully it does make a little bit. Now, you might be saying, well, what do we call this number? What do we call that number there? Or that number there? And that's called the common ratio."}, {"video_title": "Initial value & common ratio of exponential functions   High School Math   Khan Academy.mp3", "Sentence": "Now, you might be saying, well, what do we call this number? What do we call that number there? Or that number there? And that's called the common ratio. The common ratio. And in my brain, we say, well, why is it called a common ratio? Well, if you thought about integer inputs into this, especially sequential integer inputs into it, you would see a pattern."}, {"video_title": "Initial value & common ratio of exponential functions   High School Math   Khan Academy.mp3", "Sentence": "And that's called the common ratio. The common ratio. And in my brain, we say, well, why is it called a common ratio? Well, if you thought about integer inputs into this, especially sequential integer inputs into it, you would see a pattern. For example, h of, let me do this in that green color. H of zero is equal to, we already established 1 4th. Now, what is h of one going to be equal to?"}, {"video_title": "Initial value & common ratio of exponential functions   High School Math   Khan Academy.mp3", "Sentence": "Well, if you thought about integer inputs into this, especially sequential integer inputs into it, you would see a pattern. For example, h of, let me do this in that green color. H of zero is equal to, we already established 1 4th. Now, what is h of one going to be equal to? It's going to be 1 4th times two to the first power. So it's going to be 1 4th times two. What is h of two going to be equal to?"}, {"video_title": "Initial value & common ratio of exponential functions   High School Math   Khan Academy.mp3", "Sentence": "Now, what is h of one going to be equal to? It's going to be 1 4th times two to the first power. So it's going to be 1 4th times two. What is h of two going to be equal to? Well, it's going to be 1 4th times two squared. So it's going to be times two times two. Or we could just view this as, this is going to be two times h of one."}, {"video_title": "Initial value & common ratio of exponential functions   High School Math   Khan Academy.mp3", "Sentence": "What is h of two going to be equal to? Well, it's going to be 1 4th times two squared. So it's going to be times two times two. Or we could just view this as, this is going to be two times h of one. And actually, I should have done this when I wrote this one out. But this we could write as two times h of zero. So notice, if we were to take the ratio between h of two and h of one, it would be two."}, {"video_title": "Initial value & common ratio of exponential functions   High School Math   Khan Academy.mp3", "Sentence": "Or we could just view this as, this is going to be two times h of one. And actually, I should have done this when I wrote this one out. But this we could write as two times h of zero. So notice, if we were to take the ratio between h of two and h of one, it would be two. If we were to take the ratio between h of one and h of zero, it would be two. That is the common ratio between successive whole number inputs into our function. So h of, I could say, h of n plus one over h of n is going to be equal to, is going to be equal to, actually I can work it out mathematically, 1 4th times two to the n plus one over 1 4th times two to the n. That cancels, two to the n plus one divided by two to the n is just going to be equal to two."}, {"video_title": "Initial value & common ratio of exponential functions   High School Math   Khan Academy.mp3", "Sentence": "So notice, if we were to take the ratio between h of two and h of one, it would be two. If we were to take the ratio between h of one and h of zero, it would be two. That is the common ratio between successive whole number inputs into our function. So h of, I could say, h of n plus one over h of n is going to be equal to, is going to be equal to, actually I can work it out mathematically, 1 4th times two to the n plus one over 1 4th times two to the n. That cancels, two to the n plus one divided by two to the n is just going to be equal to two. That is your common ratio. So for the function h, for the function f, our common ratio is three. If we were to go the other way around, if someone said, hey, I have some function whose initial value, so let's say I have some function, I'll do this in a new color, I have some function g, and we know that its initial, initial value is five."}, {"video_title": "Initial value & common ratio of exponential functions   High School Math   Khan Academy.mp3", "Sentence": "So h of, I could say, h of n plus one over h of n is going to be equal to, is going to be equal to, actually I can work it out mathematically, 1 4th times two to the n plus one over 1 4th times two to the n. That cancels, two to the n plus one divided by two to the n is just going to be equal to two. That is your common ratio. So for the function h, for the function f, our common ratio is three. If we were to go the other way around, if someone said, hey, I have some function whose initial value, so let's say I have some function, I'll do this in a new color, I have some function g, and we know that its initial, initial value is five. And someone were to say its common ratio, its common ratio is six. What would this exponential function look like? And they're telling you this is an exponential function."}, {"video_title": "Initial value & common ratio of exponential functions   High School Math   Khan Academy.mp3", "Sentence": "If we were to go the other way around, if someone said, hey, I have some function whose initial value, so let's say I have some function, I'll do this in a new color, I have some function g, and we know that its initial, initial value is five. And someone were to say its common ratio, its common ratio is six. What would this exponential function look like? And they're telling you this is an exponential function. Well, g of, let's say x is the input, is going to be equal to our initial value, which is five. That's a, that's not a negative sign there. Our initial value is five, I'll write equals to make that clear, and then times our common ratio to the x power."}, {"video_title": "Solving quadratics by taking square roots examples   High School Math   Khan Academy.mp3", "Sentence": "So pause the video and see if you can solve for x here. Figure out which x values will satisfy this equation. All right, let's work through this. And the way I'm gonna do this is I'm gonna isolate the x plus three squared on one side. And the best way to do that is to add four to both sides. So adding four to both sides. We'll get rid of this four, this subtracting four, this negative four on the left-hand side."}, {"video_title": "Solving quadratics by taking square roots examples   High School Math   Khan Academy.mp3", "Sentence": "And the way I'm gonna do this is I'm gonna isolate the x plus three squared on one side. And the best way to do that is to add four to both sides. So adding four to both sides. We'll get rid of this four, this subtracting four, this negative four on the left-hand side. And so we're just left with x plus three squared. x plus three squared. And on the right-hand side, I'm just gonna have zero plus four."}, {"video_title": "Solving quadratics by taking square roots examples   High School Math   Khan Academy.mp3", "Sentence": "We'll get rid of this four, this subtracting four, this negative four on the left-hand side. And so we're just left with x plus three squared. x plus three squared. And on the right-hand side, I'm just gonna have zero plus four. So x plus three squared is equal to four. And so now I could take the square root of both sides. Or another way of thinking about it, if I have something squared equaling four, I could say that that something needs to either be positive or negative two."}, {"video_title": "Solving quadratics by taking square roots examples   High School Math   Khan Academy.mp3", "Sentence": "And on the right-hand side, I'm just gonna have zero plus four. So x plus three squared is equal to four. And so now I could take the square root of both sides. Or another way of thinking about it, if I have something squared equaling four, I could say that that something needs to either be positive or negative two. So one way of thinking about it is I'm saying that x plus three is going to be equal to the plus or minus square root of that four. And hopefully this makes intuitive sense for you. If something squared is equal to four, that means that the something, that means that this something right over here is going to be equal to the positive square root of four or the negative square root of four."}, {"video_title": "Solving quadratics by taking square roots examples   High School Math   Khan Academy.mp3", "Sentence": "Or another way of thinking about it, if I have something squared equaling four, I could say that that something needs to either be positive or negative two. So one way of thinking about it is I'm saying that x plus three is going to be equal to the plus or minus square root of that four. And hopefully this makes intuitive sense for you. If something squared is equal to four, that means that the something, that means that this something right over here is going to be equal to the positive square root of four or the negative square root of four. Or it's gonna be equal to positive or negative two. And so we could write that x plus three could be equal to positive two or x plus three could be equal to negative two. Notice, if x plus three was positive two, two squared is equal to four."}, {"video_title": "Solving quadratics by taking square roots examples   High School Math   Khan Academy.mp3", "Sentence": "If something squared is equal to four, that means that the something, that means that this something right over here is going to be equal to the positive square root of four or the negative square root of four. Or it's gonna be equal to positive or negative two. And so we could write that x plus three could be equal to positive two or x plus three could be equal to negative two. Notice, if x plus three was positive two, two squared is equal to four. If x plus three was negative two, negative two squared is equal to four. So either of these would satisfy our equation. So if x plus three is equal to two, we could just subtract three from both sides to solve for x."}, {"video_title": "Solving quadratics by taking square roots examples   High School Math   Khan Academy.mp3", "Sentence": "Notice, if x plus three was positive two, two squared is equal to four. If x plus three was negative two, negative two squared is equal to four. So either of these would satisfy our equation. So if x plus three is equal to two, we could just subtract three from both sides to solve for x. And we're left with x is equal to negative one. Or over here we could subtract three from both sides to solve for x. So or x is equal to negative two minus three is negative five."}, {"video_title": "Solving quadratics by taking square roots examples   High School Math   Khan Academy.mp3", "Sentence": "So if x plus three is equal to two, we could just subtract three from both sides to solve for x. And we're left with x is equal to negative one. Or over here we could subtract three from both sides to solve for x. So or x is equal to negative two minus three is negative five. So those are the two possible solutions. And you can verify that. Take these x values, substitute it back in, and then you can see when you substitute it back in, if you substitute x equals negative one, then x plus three is equal to two."}, {"video_title": "Solving quadratics by taking square roots examples   High School Math   Khan Academy.mp3", "Sentence": "So or x is equal to negative two minus three is negative five. So those are the two possible solutions. And you can verify that. Take these x values, substitute it back in, and then you can see when you substitute it back in, if you substitute x equals negative one, then x plus three is equal to two. Two squared is four, minus four is zero. And when x is equal to negative five, negative five plus three is negative two squared is positive four, minus four is also equal to zero. So these are the two possible x values that satisfy the equation."}, {"video_title": "Solving quadratics by taking square roots examples   High School Math   Khan Academy.mp3", "Sentence": "Take these x values, substitute it back in, and then you can see when you substitute it back in, if you substitute x equals negative one, then x plus three is equal to two. Two squared is four, minus four is zero. And when x is equal to negative five, negative five plus three is negative two squared is positive four, minus four is also equal to zero. So these are the two possible x values that satisfy the equation. Now let's do another one that's presented to us in a slightly different way. So we are told that f of x is equal to x minus two squared minus nine. And then we're asked at what x values does the graph of y equals f of x intersect the x-axis?"}, {"video_title": "Solving quadratics by taking square roots examples   High School Math   Khan Academy.mp3", "Sentence": "So these are the two possible x values that satisfy the equation. Now let's do another one that's presented to us in a slightly different way. So we are told that f of x is equal to x minus two squared minus nine. And then we're asked at what x values does the graph of y equals f of x intersect the x-axis? So if I'm just generally talking about some graph, so I'm not necessarily gonna draw that y equals f of x. So if I'm just, so that's our y-axis, this is our x-axis. And so if I just have the graph of some function, if I have the graph of some function that looks something like that, let's say that's the y is equal to some other function, not necessarily this f of x. Y is equal to g of x."}, {"video_title": "Solving quadratics by taking square roots examples   High School Math   Khan Academy.mp3", "Sentence": "And then we're asked at what x values does the graph of y equals f of x intersect the x-axis? So if I'm just generally talking about some graph, so I'm not necessarily gonna draw that y equals f of x. So if I'm just, so that's our y-axis, this is our x-axis. And so if I just have the graph of some function, if I have the graph of some function that looks something like that, let's say that's the y is equal to some other function, not necessarily this f of x. Y is equal to g of x. The x values where you intersect, where you intersect the x-axis, well in order to intersect the x-axis, y must be equal to zero. So y is equal to zero there. Notice our y-coordinate at either of those points are going to be equal to zero."}, {"video_title": "Solving quadratics by taking square roots examples   High School Math   Khan Academy.mp3", "Sentence": "And so if I just have the graph of some function, if I have the graph of some function that looks something like that, let's say that's the y is equal to some other function, not necessarily this f of x. Y is equal to g of x. The x values where you intersect, where you intersect the x-axis, well in order to intersect the x-axis, y must be equal to zero. So y is equal to zero there. Notice our y-coordinate at either of those points are going to be equal to zero. And that means that our function is equal to zero. So figuring out the x values where the graph of y equals f of x intersects the x-axis, this is equivalent to saying for what x values does f of x equal zero? So we could just say for what x values does this thing right over here equal zero?"}, {"video_title": "Solving quadratics by taking square roots examples   High School Math   Khan Academy.mp3", "Sentence": "Notice our y-coordinate at either of those points are going to be equal to zero. And that means that our function is equal to zero. So figuring out the x values where the graph of y equals f of x intersects the x-axis, this is equivalent to saying for what x values does f of x equal zero? So we could just say for what x values does this thing right over here equal zero? So let me just write that down. So we could rewrite this as x minus two squared minus nine equals zero. Well we could add nine to both sides and so we could get x minus two squared is equal to nine."}, {"video_title": "Solving quadratics by taking square roots examples   High School Math   Khan Academy.mp3", "Sentence": "So we could just say for what x values does this thing right over here equal zero? So let me just write that down. So we could rewrite this as x minus two squared minus nine equals zero. Well we could add nine to both sides and so we could get x minus two squared is equal to nine. And just like we saw before, that means that x minus two is equal to the positive or negative square root of nine. So we could say x minus two is equal to positive three or x minus two is equal to negative three. Well you add two to both sides of this, you get x is equal to five or if we add two to both sides of this equation, you'll get x is equal to negative one."}, {"video_title": "Solving quadratics by taking square roots examples   High School Math   Khan Academy.mp3", "Sentence": "Well we could add nine to both sides and so we could get x minus two squared is equal to nine. And just like we saw before, that means that x minus two is equal to the positive or negative square root of nine. So we could say x minus two is equal to positive three or x minus two is equal to negative three. Well you add two to both sides of this, you get x is equal to five or if we add two to both sides of this equation, you'll get x is equal to negative one. And you can verify that. If x is equal to five, five minus two is three squared is nine, minus nine is zero. So the point five comma zero is going to be on this graph."}, {"video_title": "Solving quadratics by taking square roots examples   High School Math   Khan Academy.mp3", "Sentence": "Well you add two to both sides of this, you get x is equal to five or if we add two to both sides of this equation, you'll get x is equal to negative one. And you can verify that. If x is equal to five, five minus two is three squared is nine, minus nine is zero. So the point five comma zero is going to be on this graph. And also if x is equal to negative one, negative one minus two, negative three squared is positive nine, minus nine is zero. So also the point negative one comma zero is on this graph. So those are the points where, those are the x values where the function intersects the x axis."}, {"video_title": "Recognize functions from verbal descriptions (example 1)   Algebra II   Khan Academy.mp3", "Sentence": "So three more than twice x. So it is three plus two x is another way of saying this first sentence. So is y a function of x? So whenever you're asked whether something is a function of something else, you're really just saying, look, for any input x, does it map to exactly one y? So if we say y is a function of x, in order for this to be a function, for any x that you input into this function, you must get exactly one y. So if you input an x, you must get exactly one y value. If you got two values, then it's no longer a function."}, {"video_title": "Recognize functions from verbal descriptions (example 1)   Algebra II   Khan Academy.mp3", "Sentence": "So whenever you're asked whether something is a function of something else, you're really just saying, look, for any input x, does it map to exactly one y? So if we say y is a function of x, in order for this to be a function, for any x that you input into this function, you must get exactly one y. So if you input an x, you must get exactly one y value. If you got two values, then it's no longer a function. For any input, you get exactly one y. You could have two inputs that get to the same y, but you can't have one input that results in two different outputs. You wouldn't know what the function is valued at at that input."}, {"video_title": "Recognize functions from verbal descriptions (example 1)   Algebra II   Khan Academy.mp3", "Sentence": "If you got two values, then it's no longer a function. For any input, you get exactly one y. You could have two inputs that get to the same y, but you can't have one input that results in two different outputs. You wouldn't know what the function is valued at at that input. Now here it looks pretty clear that for any input, you get exactly one output. Any input uniquely determines which y. It's not like if you put an x in here, you're not sure what y is going to be."}, {"video_title": "Recognize functions from verbal descriptions (example 1)   Algebra II   Khan Academy.mp3", "Sentence": "You wouldn't know what the function is valued at at that input. Now here it looks pretty clear that for any input, you get exactly one output. Any input uniquely determines which y. It's not like if you put an x in here, you're not sure what y is going to be. You know what y is going to be. If x is zero, y is three. If x is one, y is five."}, {"video_title": "Interpret a quadratic graph   Quadratic functions & equations   Algebra 1   Khan Academy.mp3", "Sentence": "The function f models the height of the ball, in meters, as a function of time, in seconds, after Katie threw it. And we could see that right over here. This is our function f. So at time t equals zero, the height looks like it's a couple of meters. And then as we go forward in time to a little under two and a half seconds, the ball's going up. And then after a little under two and a half seconds, the ball starts going down. And by the time you get to five seconds or close to five seconds, it looks like the ball is on the ground. Its height is zero meters."}, {"video_title": "Interpret a quadratic graph   Quadratic functions & equations   Algebra 1   Khan Academy.mp3", "Sentence": "And then as we go forward in time to a little under two and a half seconds, the ball's going up. And then after a little under two and a half seconds, the ball starts going down. And by the time you get to five seconds or close to five seconds, it looks like the ball is on the ground. Its height is zero meters. So then they ask us, which of these statements are true? Choose all that apply. So pause this video and see if you can work it out."}, {"video_title": "Interpret a quadratic graph   Quadratic functions & equations   Algebra 1   Khan Academy.mp3", "Sentence": "Its height is zero meters. So then they ask us, which of these statements are true? Choose all that apply. So pause this video and see if you can work it out. All right, now let's look through the choices. So the first one says, Katie threw the ball from a height of five meters. So let's see if that bears fruit or see if that's true."}, {"video_title": "Interpret a quadratic graph   Quadratic functions & equations   Algebra 1   Khan Academy.mp3", "Sentence": "So pause this video and see if you can work it out. All right, now let's look through the choices. So the first one says, Katie threw the ball from a height of five meters. So let's see if that bears fruit or see if that's true. So if she threw it from a height of five meters, that means that the y-intercept would have been at five meters. At time t equals zero, we would have been at a height of five meters. Clearly, that is not the y-intercept."}, {"video_title": "Interpret a quadratic graph   Quadratic functions & equations   Algebra 1   Khan Academy.mp3", "Sentence": "So let's see if that bears fruit or see if that's true. So if she threw it from a height of five meters, that means that the y-intercept would have been at five meters. At time t equals zero, we would have been at a height of five meters. Clearly, that is not the y-intercept. It looks like she threw it from a height of maybe one and a half or two meters. So I'm assuming that Katie is not five meters tall, so she wasn't on a ladder or anything. She just threw it from her regular height."}, {"video_title": "Interpret a quadratic graph   Quadratic functions & equations   Algebra 1   Khan Academy.mp3", "Sentence": "Clearly, that is not the y-intercept. It looks like she threw it from a height of maybe one and a half or two meters. So I'm assuming that Katie is not five meters tall, so she wasn't on a ladder or anything. She just threw it from her regular height. And so we can rule out this first choice. The second one says, at its highest point, the ball was about 31 meters above the ground. Let's see if that is true."}, {"video_title": "Interpret a quadratic graph   Quadratic functions & equations   Algebra 1   Khan Academy.mp3", "Sentence": "She just threw it from her regular height. And so we can rule out this first choice. The second one says, at its highest point, the ball was about 31 meters above the ground. Let's see if that is true. So the highest point is right over here. And yeah, that looks about 31 meters, so I like that choice. So I will select that one."}, {"video_title": "Interpret a quadratic graph   Quadratic functions & equations   Algebra 1   Khan Academy.mp3", "Sentence": "Let's see if that is true. So the highest point is right over here. And yeah, that looks about 31 meters, so I like that choice. So I will select that one. The ball was in the air for about two and a half seconds. So we can clearly see that that is not the case. The ball was going up for about two and a half seconds, but then it was going down for roughly another two and a half seconds."}, {"video_title": "Interpret a quadratic graph   Quadratic functions & equations   Algebra 1   Khan Academy.mp3", "Sentence": "So I will select that one. The ball was in the air for about two and a half seconds. So we can clearly see that that is not the case. The ball was going up for about two and a half seconds, but then it was going down for roughly another two and a half seconds. And so it was actually in the air for almost five seconds. So I would rule this one out. The ball reached its highest point in the air about two and a half seconds after Katie threw it."}, {"video_title": "Interpret a quadratic graph   Quadratic functions & equations   Algebra 1   Khan Academy.mp3", "Sentence": "The ball was going up for about two and a half seconds, but then it was going down for roughly another two and a half seconds. And so it was actually in the air for almost five seconds. So I would rule this one out. The ball reached its highest point in the air about two and a half seconds after Katie threw it. So let's see, two and a half seconds after Katie threw the ball, it's right over there. It looks like it reached its highest point a little bit before that, but they said about, so they're speaking in rough terms. So I think that statement can be true."}, {"video_title": "Interpret a quadratic graph   Quadratic functions & equations   Algebra 1   Khan Academy.mp3", "Sentence": "The ball reached its highest point in the air about two and a half seconds after Katie threw it. So let's see, two and a half seconds after Katie threw the ball, it's right over there. It looks like it reached its highest point a little bit before that, but they said about, so they're speaking in rough terms. So I think that statement can be true. It's about two and a half seconds. If they said exactly two and a half seconds, I wouldn't have selected it, because it seems like it happened at like 2.45 seconds or something like that. But there you go, those are the ones that seem true."}, {"video_title": "Introduction to geometric sequences   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "Now, a good starting point is just what is a sequence? And a sequence is, you can imagine, just a progression of numbers. So for example, and this isn't even a geometric series. If I just said 1, 2, 3, 4, 5, this is a sequence of numbers. It's not a geometric sequence, but it is a sequence. Now, a geometric sequence is a special progression or a special sequence of numbers where each successive number is a fixed multiple of the number before it. So let me explain what I'm saying."}, {"video_title": "Introduction to geometric sequences   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "If I just said 1, 2, 3, 4, 5, this is a sequence of numbers. It's not a geometric sequence, but it is a sequence. Now, a geometric sequence is a special progression or a special sequence of numbers where each successive number is a fixed multiple of the number before it. So let me explain what I'm saying. So let's say my first number is 2. And then I multiply 2 by the number 3. So I multiply it by 3, I get 6."}, {"video_title": "Introduction to geometric sequences   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "So let me explain what I'm saying. So let's say my first number is 2. And then I multiply 2 by the number 3. So I multiply it by 3, I get 6. And then I multiply 6 times the number 3, and I get 18. And then I multiply 18 times the number 3, and I get 54. And I just keep going that way."}, {"video_title": "Introduction to geometric sequences   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "So I multiply it by 3, I get 6. And then I multiply 6 times the number 3, and I get 18. And then I multiply 18 times the number 3, and I get 54. And I just keep going that way. So I just keep multiplying by the number 3. So I started, if we want to get some notation here, this is my first term. We'll call it a1 for my sequence."}, {"video_title": "Introduction to geometric sequences   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "And I just keep going that way. So I just keep multiplying by the number 3. So I started, if we want to get some notation here, this is my first term. We'll call it a1 for my sequence. And each time I'm multiplying it by a common number. And that number is often called the common ratio. So in this case, a1 is equal to 2."}, {"video_title": "Introduction to geometric sequences   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "We'll call it a1 for my sequence. And each time I'm multiplying it by a common number. And that number is often called the common ratio. So in this case, a1 is equal to 2. And my common ratio is equal to 3. So if someone were to tell you, hey, you've got a geometric sequence, a1 is equal to 90. And your common ratio is equal to negative 1 third."}, {"video_title": "Introduction to geometric sequences   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "So in this case, a1 is equal to 2. And my common ratio is equal to 3. So if someone were to tell you, hey, you've got a geometric sequence, a1 is equal to 90. And your common ratio is equal to negative 1 third. That means that the first term of your sequence is 90. The second term is negative 1 third times 90, which is what? That's negative 30."}, {"video_title": "Introduction to geometric sequences   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "And your common ratio is equal to negative 1 third. That means that the first term of your sequence is 90. The second term is negative 1 third times 90, which is what? That's negative 30. 1 third times 90 is 30. And then you put the negative number. Then the next number is going to be 1 third times this."}, {"video_title": "Introduction to geometric sequences   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "That's negative 30. 1 third times 90 is 30. And then you put the negative number. Then the next number is going to be 1 third times this. So 1 third, or negative 1 third times this. 1 third times 30 is 10. The negatives cancel out."}, {"video_title": "Introduction to geometric sequences   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "Then the next number is going to be 1 third times this. So 1 third, or negative 1 third times this. 1 third times 30 is 10. The negatives cancel out. So you get positive 10. Then the next number is going to be 10 times negative 1 third, or negative 10 thirds. And then the next number is going to be negative 10 thirds times negative 1 third."}, {"video_title": "Introduction to geometric sequences   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "The negatives cancel out. So you get positive 10. Then the next number is going to be 10 times negative 1 third, or negative 10 thirds. And then the next number is going to be negative 10 thirds times negative 1 third. So it's going to be positive 10 thirds. And you can just keep going on with this sequence. So that's what people talk about when they mean a geometric sequence."}, {"video_title": "Introduction to geometric sequences   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "And then the next number is going to be negative 10 thirds times negative 1 third. So it's going to be positive 10 thirds. And you can just keep going on with this sequence. So that's what people talk about when they mean a geometric sequence. And I want to make one little distinction here. This always used to confuse me because the terms are used very often in the same context. These are sequences."}, {"video_title": "Introduction to geometric sequences   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "So that's what people talk about when they mean a geometric sequence. And I want to make one little distinction here. This always used to confuse me because the terms are used very often in the same context. These are sequences. These are kind of a progression of numbers. 2, then 6, then 18, 90, then negative 30, then 10, then negative 10 over 3. Then, oh sorry, then this is positive 10 over 9."}, {"video_title": "Introduction to geometric sequences   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "These are sequences. These are kind of a progression of numbers. 2, then 6, then 18, 90, then negative 30, then 10, then negative 10 over 3. Then, oh sorry, then this is positive 10 over 9. Negative 1 third times negative 10 over 3. Negatives cancel out. 10 over 9."}, {"video_title": "Introduction to geometric sequences   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "Then, oh sorry, then this is positive 10 over 9. Negative 1 third times negative 10 over 3. Negatives cancel out. 10 over 9. I don't want to make a mistake here. These are sequences. You might also see the word a series."}, {"video_title": "Introduction to geometric sequences   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "10 over 9. I don't want to make a mistake here. These are sequences. You might also see the word a series. And you might even see a geometric series. A series, the most conventional use of the word series, means a sum of a sequence. So for example, this is a geometric sequence."}, {"video_title": "Introduction to geometric sequences   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "You might also see the word a series. And you might even see a geometric series. A series, the most conventional use of the word series, means a sum of a sequence. So for example, this is a geometric sequence. A geometric series would be 90 plus negative 30 plus 10 plus negative 10 over 3 plus 10 over 9. So a general way to view it is that a series is the sum of a sequence. I just want to make that clear because that used to confuse me a lot when I first learned about these things."}, {"video_title": "Introduction to geometric sequences   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "So for example, this is a geometric sequence. A geometric series would be 90 plus negative 30 plus 10 plus negative 10 over 3 plus 10 over 9. So a general way to view it is that a series is the sum of a sequence. I just want to make that clear because that used to confuse me a lot when I first learned about these things. But anyway, let's go back to the notion of a geometric sequence and actually do a word problem that deals with one of these. So they're telling us that Ann goes bungee jumping off of a bridge above water. So Ann bungee jumping."}, {"video_title": "Introduction to geometric sequences   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "I just want to make that clear because that used to confuse me a lot when I first learned about these things. But anyway, let's go back to the notion of a geometric sequence and actually do a word problem that deals with one of these. So they're telling us that Ann goes bungee jumping off of a bridge above water. So Ann bungee jumping. On the initial jump, the cord stretches by 120 feet. So on A1, our initial jump, the cord stretches by 120 feet. We could write it this way."}, {"video_title": "Introduction to geometric sequences   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "So Ann bungee jumping. On the initial jump, the cord stretches by 120 feet. So on A1, our initial jump, the cord stretches by 120 feet. We could write it this way. We could write jump and then how much the cord stretches. So on the initial jump, on jump 1, the cord stretches 120 feet. Then it says on the next bounce, the stretch is 60% of the original jump."}, {"video_title": "Introduction to geometric sequences   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "We could write it this way. We could write jump and then how much the cord stretches. So on the initial jump, on jump 1, the cord stretches 120 feet. Then it says on the next bounce, the stretch is 60% of the original jump. And then each additional bounce stretches the rope 60% of the previous stretch. So here, the common ratio where each successive term in our sequence is going to be 60% of the previous term. Or it's going to be 0.6 times the previous term."}, {"video_title": "Introduction to geometric sequences   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "Then it says on the next bounce, the stretch is 60% of the original jump. And then each additional bounce stretches the rope 60% of the previous stretch. So here, the common ratio where each successive term in our sequence is going to be 60% of the previous term. Or it's going to be 0.6 times the previous term. So on the second jump, we're going to stretch 60% of that, or 0.6 times 120, which is equal to what? That's equal to 72. Then on the third jump, we're going to stretch 0.6 of 72, or 0.6 times this."}, {"video_title": "Introduction to geometric sequences   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "Or it's going to be 0.6 times the previous term. So on the second jump, we're going to stretch 60% of that, or 0.6 times 120, which is equal to what? That's equal to 72. Then on the third jump, we're going to stretch 0.6 of 72, or 0.6 times this. So it would be 0.6 times 0.6 times 120. And then on the fourth jump, notice over here, so on the fourth jump, we're going to have 0.6 times 0.6 times 0.6 times 120. 60% of this jump."}, {"video_title": "Introduction to geometric sequences   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "Then on the third jump, we're going to stretch 0.6 of 72, or 0.6 times this. So it would be 0.6 times 0.6 times 120. And then on the fourth jump, notice over here, so on the fourth jump, we're going to have 0.6 times 0.6 times 0.6 times 120. 60% of this jump. So every time we're 60% of the previous jump. So if we wanted to make a general formula for this, just based on the way we've defined it right here, so the stretch on the nth jump, what would it be? So let's see, we start at 120 times 0.6 to the what?"}, {"video_title": "Introduction to geometric sequences   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "60% of this jump. So every time we're 60% of the previous jump. So if we wanted to make a general formula for this, just based on the way we've defined it right here, so the stretch on the nth jump, what would it be? So let's see, we start at 120 times 0.6 to the what? To the n minus 1. Now how did I get this? Let me write this a little bit neater."}, {"video_title": "Introduction to geometric sequences   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "So let's see, we start at 120 times 0.6 to the what? To the n minus 1. Now how did I get this? Let me write this a little bit neater. So this is equal to 0.6. Actually, let me write the 120 first. This is equal to 120 times 0.6 to the n minus 1."}, {"video_title": "Introduction to geometric sequences   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "Let me write this a little bit neater. So this is equal to 0.6. Actually, let me write the 120 first. This is equal to 120 times 0.6 to the n minus 1. How did I get that? Well, we're defining the first jump as stretching 120 feet. So when you put n is equal to 1 here, you get 1 minus 1, 0."}, {"video_title": "Introduction to geometric sequences   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "This is equal to 120 times 0.6 to the n minus 1. How did I get that? Well, we're defining the first jump as stretching 120 feet. So when you put n is equal to 1 here, you get 1 minus 1, 0. So you have 0.6 to the 0th power. So you just get a 1 here. And that's exactly what happened on the first jump."}, {"video_title": "Introduction to geometric sequences   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "So when you put n is equal to 1 here, you get 1 minus 1, 0. So you have 0.6 to the 0th power. So you just get a 1 here. And that's exactly what happened on the first jump. Then on the second jump, you put a 2 minus 1. And notice, 2 minus 1 is the first power. And we have exactly 1.6 here."}, {"video_title": "Introduction to geometric sequences   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "And that's exactly what happened on the first jump. Then on the second jump, you put a 2 minus 1. And notice, 2 minus 1 is the first power. And we have exactly 1.6 here. So I figured it was n minus 1 because when n is 2, we have 1.6. When n is 3, we have 2.6 is multiplied by themselves. When n is 4, we have 0.6 to the 3rd power."}, {"video_title": "Introduction to geometric sequences   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "And we have exactly 1.6 here. So I figured it was n minus 1 because when n is 2, we have 1.6. When n is 3, we have 2.6 is multiplied by themselves. When n is 4, we have 0.6 to the 3rd power. So whatever n is, we're taking 0.6 to the n minus 1th power. And of course, we're multiplying that times 120. Now, in the question, they also ask us, what will be the rope stretch on the 12th bounce?"}, {"video_title": "Introduction to geometric sequences   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "When n is 4, we have 0.6 to the 3rd power. So whatever n is, we're taking 0.6 to the n minus 1th power. And of course, we're multiplying that times 120. Now, in the question, they also ask us, what will be the rope stretch on the 12th bounce? And over here, I'm going to use the calculator. And actually, let me correct this a little bit. This is an incorrect, but they're talking about the bounce."}, {"video_title": "Introduction to geometric sequences   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "Now, in the question, they also ask us, what will be the rope stretch on the 12th bounce? And over here, I'm going to use the calculator. And actually, let me correct this a little bit. This is an incorrect, but they're talking about the bounce. And we could call the jump the 0th bounce. So let me change that. This isn't wrong, but I want to make it a little bit more, I think this is where they're going with the problem."}, {"video_title": "Introduction to geometric sequences   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "This is an incorrect, but they're talking about the bounce. And we could call the jump the 0th bounce. So let me change that. This isn't wrong, but I want to make it a little bit more, I think this is where they're going with the problem. So you can view the initial stretch as the 0th bounce. So instead of labeling it jump, let me label it bounce. So the initial stretch is the 0th bounce."}, {"video_title": "Introduction to geometric sequences   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "This isn't wrong, but I want to make it a little bit more, I think this is where they're going with the problem. So you can view the initial stretch as the 0th bounce. So instead of labeling it jump, let me label it bounce. So the initial stretch is the 0th bounce. Then this would be the first bounce, the second bounce, the third bounce. And then our formula becomes a lot simpler. Because if you said the stretch on nth bounce, then the formula just becomes 0.6 to the n times 120."}, {"video_title": "Introduction to geometric sequences   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "So the initial stretch is the 0th bounce. Then this would be the first bounce, the second bounce, the third bounce. And then our formula becomes a lot simpler. Because if you said the stretch on nth bounce, then the formula just becomes 0.6 to the n times 120. On the 0th bounce, that was our original stretch. You get 0.6 to the 0, that's 1 times 120. On the first bounce, 0.6 to the 1, you have 1.6 right here."}, {"video_title": "Introduction to geometric sequences   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "Because if you said the stretch on nth bounce, then the formula just becomes 0.6 to the n times 120. On the 0th bounce, that was our original stretch. You get 0.6 to the 0, that's 1 times 120. On the first bounce, 0.6 to the 1, you have 1.6 right here. 0.6 times the previous stretch or the previous bounce. So this has it in terms of bounces, which I think is what the questioner wants us to do. So what about the 12th bounce?"}, {"video_title": "Introduction to geometric sequences   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "On the first bounce, 0.6 to the 1, you have 1.6 right here. 0.6 times the previous stretch or the previous bounce. So this has it in terms of bounces, which I think is what the questioner wants us to do. So what about the 12th bounce? Using this convention right there. So if we do the 12th bounce, let's just get our calculator out. We're going to have 120 times 0.6 to the 12th power."}, {"video_title": "Introduction to geometric sequences   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "So what about the 12th bounce? Using this convention right there. So if we do the 12th bounce, let's just get our calculator out. We're going to have 120 times 0.6 to the 12th power. And hopefully we'll get order of operations right, because exponents take precedence over multiplication. So it'll just take the 0.6 to the 12th power only. And so this is equal to 0.26 feet."}, {"video_title": "Introduction to geometric sequences   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "We're going to have 120 times 0.6 to the 12th power. And hopefully we'll get order of operations right, because exponents take precedence over multiplication. So it'll just take the 0.6 to the 12th power only. And so this is equal to 0.26 feet. So after your 12th bounce, she's going to be barely moving. She's going to be moving about 3 inches on that 12th bounce. Well, anyway, hopefully you found this helpful."}, {"video_title": "Introduction to geometric sequences   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "And so this is equal to 0.26 feet. So after your 12th bounce, she's going to be barely moving. She's going to be moving about 3 inches on that 12th bounce. Well, anyway, hopefully you found this helpful. And I apologize for the slight divergence here. But I actually think on some level that's instructive, because you always have to make sure that your n matches well with what your results are. So when I talked about just your first jump, I said, OK, this is 1."}, {"video_title": "Introduction to geometric sequences   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "Well, anyway, hopefully you found this helpful. And I apologize for the slight divergence here. But I actually think on some level that's instructive, because you always have to make sure that your n matches well with what your results are. So when I talked about just your first jump, I said, OK, this is 1. And then I had 0.6 to the 0th power. So I did n minus 1. But then when I relabeled things in terms of bounces, this was the 0th bounce, then this makes sense that this is 0.6 to the 0th power."}, {"video_title": "Introduction to geometric sequences   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "So when I talked about just your first jump, I said, OK, this is 1. And then I had 0.6 to the 0th power. So I did n minus 1. But then when I relabeled things in terms of bounces, this was the 0th bounce, then this makes sense that this is 0.6 to the 0th power. This is the first bounce, then this would be 0.6 to the 1st power, second bounce, 0.6 to the 2nd power. And it made our equation a little bit simpler. Anyway, hopefully you found that interesting."}, {"video_title": "Rewriting square root of fraction.mp3", "Sentence": "When I say simplify it, I really mean, I want to, if there's any perfect squares here that I can factor out, to take it out from under the radical. And so I encourage you to pause the video and see if you can do that. All right, so there's a couple of ways that you could approach this. One way is to say, well this is going to be the same thing as the square root of 1 over the square root of 200. Square root of 1 is just 1 over the square root of 200. And there's a couple of ways to try to simplify the square root of 200. I will, I'll do it a couple of ways here."}, {"video_title": "Rewriting square root of fraction.mp3", "Sentence": "One way is to say, well this is going to be the same thing as the square root of 1 over the square root of 200. Square root of 1 is just 1 over the square root of 200. And there's a couple of ways to try to simplify the square root of 200. I will, I'll do it a couple of ways here. Square root of 200. You could realize that, okay, look, 100 is a perfect square, and it goes into 200. So this is the same thing as 2 times 100."}, {"video_title": "Rewriting square root of fraction.mp3", "Sentence": "I will, I'll do it a couple of ways here. Square root of 200. You could realize that, okay, look, 100 is a perfect square, and it goes into 200. So this is the same thing as 2 times 100. And so the square root of 200 is the square root of 2 times 100, which is the same thing as the square root of 2 times the square root of 100. And we know that the square root of 100 is 10. So it's the square root of 2 times 10, or we could write this as 10 square roots of 2."}, {"video_title": "Rewriting square root of fraction.mp3", "Sentence": "So this is the same thing as 2 times 100. And so the square root of 200 is the square root of 2 times 100, which is the same thing as the square root of 2 times the square root of 100. And we know that the square root of 100 is 10. So it's the square root of 2 times 10, or we could write this as 10 square roots of 2. That's one way to approach it. But if it didn't jump out at you immediately, that you have this large, perfect square that is a factor of 200, you could just start with small numbers. You could say, all right, let me do this in another, this alternate method in a different color."}, {"video_title": "Rewriting square root of fraction.mp3", "Sentence": "So it's the square root of 2 times 10, or we could write this as 10 square roots of 2. That's one way to approach it. But if it didn't jump out at you immediately, that you have this large, perfect square that is a factor of 200, you could just start with small numbers. You could say, all right, let me do this in another, this alternate method in a different color. You could say, that's the same color that I've been doing before. You could say that the square root of 200, say, well, it's divisible by 2, so it's 2 times 100. And if 100 didn't jump out at you as a perfect square, you could say, well, that's just gonna be 2 times 50."}, {"video_title": "Rewriting square root of fraction.mp3", "Sentence": "You could say, all right, let me do this in another, this alternate method in a different color. You could say, that's the same color that I've been doing before. You could say that the square root of 200, say, well, it's divisible by 2, so it's 2 times 100. And if 100 didn't jump out at you as a perfect square, you could say, well, that's just gonna be 2 times 50. Well, I could still divide 2 into that. That's 2 times 25. Let's see, and it's 25."}, {"video_title": "Rewriting square root of fraction.mp3", "Sentence": "And if 100 didn't jump out at you as a perfect square, you could say, well, that's just gonna be 2 times 50. Well, I could still divide 2 into that. That's 2 times 25. Let's see, and it's 25. If that doesn't jump out at you as a perfect square, you could say that, well, let's see, that's not divisible by 2, not divisible by 3, 4, but it is divisible by 5. That is 5 times 5. And to identify the perfect squares, you would say, all right, what, are there any factors where I have at least two of them?"}, {"video_title": "Rewriting square root of fraction.mp3", "Sentence": "Let's see, and it's 25. If that doesn't jump out at you as a perfect square, you could say that, well, let's see, that's not divisible by 2, not divisible by 3, 4, but it is divisible by 5. That is 5 times 5. And to identify the perfect squares, you would say, all right, what, are there any factors where I have at least two of them? Well, I have 2 times 2 here, and I also have, I also have 5 times 5 here. So I can rewrite the square root of 200 as being equal to the square root of 2 times 2, 2 times 2, let me just write it all out. So 2, actually, I think I'm gonna run out of space."}, {"video_title": "Rewriting square root of fraction.mp3", "Sentence": "And to identify the perfect squares, you would say, all right, what, are there any factors where I have at least two of them? Well, I have 2 times 2 here, and I also have, I also have 5 times 5 here. So I can rewrite the square root of 200 as being equal to the square root of 2 times 2, 2 times 2, let me just write it all out. So 2, actually, I think I'm gonna run out of space. So the square root, give myself more space under the radical, square root of 2 times 2 times 5 times 5, times 5 times 5, times 5 times 2, times 2. And I wrote it this order so you can see the perfect squares here. Well, this is going to be the same thing as the square root of 2 times 2."}, {"video_title": "Rewriting square root of fraction.mp3", "Sentence": "So 2, actually, I think I'm gonna run out of space. So the square root, give myself more space under the radical, square root of 2 times 2 times 5 times 5, times 5 times 5, times 5 times 2, times 2. And I wrote it this order so you can see the perfect squares here. Well, this is going to be the same thing as the square root of 2 times 2. This second method is a little bit more monotonous, but hopefully it's, you see that it works, I guess is what we would think. And they're really, they boil down to the same method. We're still going to get to the same answer."}, {"video_title": "Rewriting square root of fraction.mp3", "Sentence": "Well, this is going to be the same thing as the square root of 2 times 2. This second method is a little bit more monotonous, but hopefully it's, you see that it works, I guess is what we would think. And they're really, they boil down to the same method. We're still going to get to the same answer. So square root of 2 times 2 times the square root, times the square root of 5 times 5, times the square root of 5 times 5 times the squar roommates of 2. Times the square root of 2. Well the square root of 2 times 2 is just going to be, this is just 2, the square root of 5 times 5, well, that's just going to be 5."}, {"video_title": "Rewriting square root of fraction.mp3", "Sentence": "We're still going to get to the same answer. So square root of 2 times 2 times the square root, times the square root of 5 times 5, times the square root of 5 times 5 times the squar roommates of 2. Times the square root of 2. Well the square root of 2 times 2 is just going to be, this is just 2, the square root of 5 times 5, well, that's just going to be 5. So you have 2 times 5 times the square root of 2, which is 10 times the square root of 2. Equals 10 times the square root of 2. So this right over here, square root of 200, we can rewrite as 10 square roots of two."}, {"video_title": "Rewriting square root of fraction.mp3", "Sentence": "Well the square root of 2 times 2 is just going to be, this is just 2, the square root of 5 times 5, well, that's just going to be 5. So you have 2 times 5 times the square root of 2, which is 10 times the square root of 2. Equals 10 times the square root of 2. So this right over here, square root of 200, we can rewrite as 10 square roots of two. So this is going to be equal to one over 10 square roots of two. Now some people don't like having a radical in the denominator, and if you wanted to get rid of that, you could multiply both the numerator and the denominator by square root of two. Because notice, we're just multiplying by one."}, {"video_title": "Rewriting square root of fraction.mp3", "Sentence": "So this right over here, square root of 200, we can rewrite as 10 square roots of two. So this is going to be equal to one over 10 square roots of two. Now some people don't like having a radical in the denominator, and if you wanted to get rid of that, you could multiply both the numerator and the denominator by square root of two. Because notice, we're just multiplying by one. We're expressing one as square root of two over square root of two. And then what that does is we rewrite this as the square root of two over 10 times the square root of two times the square root of two. Well the square root of two times the square root of two is just going to be two."}, {"video_title": "Rewriting square root of fraction.mp3", "Sentence": "Because notice, we're just multiplying by one. We're expressing one as square root of two over square root of two. And then what that does is we rewrite this as the square root of two over 10 times the square root of two times the square root of two. Well the square root of two times the square root of two is just going to be two. So it's going to be 10 times two, which is 20. So it could also be written like that. So hopefully you found that helpful."}, {"video_title": "Factoring quadratics as (x+a)(x+b) (example 2)   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "To better understand how we can factor second degree expressions like this, I'm going to go through some examples. We'll factor this expression, and we'll factor this expression. And hopefully, it'll give you a background on how you could generally factor expressions like this. And to think about it, let's think about what happens if I were to multiply x plus something times x plus something else. Well, if I were to multiply this out, what do I get? Well, you're going to get x squared plus ax plus bx, which is the same thing as a plus bx, plus a times b, plus ab. So if you wanted to go from this form, which is what we have in these two examples, back to this, you really just have to think about, well, what's our coefficient on our x term?"}, {"video_title": "Factoring quadratics as (x+a)(x+b) (example 2)   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "And to think about it, let's think about what happens if I were to multiply x plus something times x plus something else. Well, if I were to multiply this out, what do I get? Well, you're going to get x squared plus ax plus bx, which is the same thing as a plus bx, plus a times b, plus ab. So if you wanted to go from this form, which is what we have in these two examples, back to this, you really just have to think about, well, what's our coefficient on our x term? And can I figure out two numbers that when I take their sum are equal to that coefficient? And what's my constant term? And can I think of two numbers, those same two numbers, that when I take the product equal that constant term?"}, {"video_title": "Factoring quadratics as (x+a)(x+b) (example 2)   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "So if you wanted to go from this form, which is what we have in these two examples, back to this, you really just have to think about, well, what's our coefficient on our x term? And can I figure out two numbers that when I take their sum are equal to that coefficient? And what's my constant term? And can I think of two numbers, those same two numbers, that when I take the product equal that constant term? So let's do that over here. If we look at our x coefficient, the coefficient on x, can we think of an a plus ab that is equal to that number, negative 14? And can we think of the same a and b that if we were to take its product, it would be equal to 40?"}, {"video_title": "Factoring quadratics as (x+a)(x+b) (example 2)   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "And can I think of two numbers, those same two numbers, that when I take the product equal that constant term? So let's do that over here. If we look at our x coefficient, the coefficient on x, can we think of an a plus ab that is equal to that number, negative 14? And can we think of the same a and b that if we were to take its product, it would be equal to 40? It would be equal to 40. So what's an a and a b that would work over here? Well, let's think about this a little bit."}, {"video_title": "Factoring quadratics as (x+a)(x+b) (example 2)   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "And can we think of the same a and b that if we were to take its product, it would be equal to 40? It would be equal to 40. So what's an a and a b that would work over here? Well, let's think about this a little bit. If I have 4 times 10 is 40, but 4 plus 10 is equal to positive 14. So that wouldn't quite work. But what happens if we make them both negative?"}, {"video_title": "Factoring quadratics as (x+a)(x+b) (example 2)   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "Well, let's think about this a little bit. If I have 4 times 10 is 40, but 4 plus 10 is equal to positive 14. So that wouldn't quite work. But what happens if we make them both negative? If we have negative 4 plus negative 10, well, that's going to be equal to negative 14. And negative 4 times negative 10 is equal to 40. The fact that this number right over here is positive tells you that these are going to be the same sign."}, {"video_title": "Factoring quadratics as (x+a)(x+b) (example 2)   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "But what happens if we make them both negative? If we have negative 4 plus negative 10, well, that's going to be equal to negative 14. And negative 4 times negative 10 is equal to 40. The fact that this number right over here is positive tells you that these are going to be the same sign. So these are going to be the same sign. These are going to have the exact same sign. If this number right over here was negative, then we would have different signs."}, {"video_title": "Factoring quadratics as (x+a)(x+b) (example 2)   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "The fact that this number right over here is positive tells you that these are going to be the same sign. So these are going to be the same sign. These are going to have the exact same sign. If this number right over here was negative, then we would have different signs. And so if you have two numbers that are going to be the same sign and they add up to a negative number, then that tells you that they're both going to be negative. So just going back to this, we know that a is going to be negative 4, b is equal to negative 10, and we are done factoring it. We can factor this expression as x plus negative 4 times x plus negative 10."}, {"video_title": "Factoring quadratics as (x+a)(x+b) (example 2)   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "If this number right over here was negative, then we would have different signs. And so if you have two numbers that are going to be the same sign and they add up to a negative number, then that tells you that they're both going to be negative. So just going back to this, we know that a is going to be negative 4, b is equal to negative 10, and we are done factoring it. We can factor this expression as x plus negative 4 times x plus negative 10. Or another way to write that, that's x minus 4 times x minus 10. Now let's do the same thing over here. Can we think of an a plus b that's equal to the coefficient on the x term?"}, {"video_title": "Factoring quadratics as (x+a)(x+b) (example 2)   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "We can factor this expression as x plus negative 4 times x plus negative 10. Or another way to write that, that's x minus 4 times x minus 10. Now let's do the same thing over here. Can we think of an a plus b that's equal to the coefficient on the x term? Well, the coefficient on the x term here, this is essentially negative 1 times x. So we could say the coefficient is negative 1. And can we think of an a times b where it's going to be equal to negative 12?"}, {"video_title": "Factoring quadratics as (x+a)(x+b) (example 2)   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "Can we think of an a plus b that's equal to the coefficient on the x term? Well, the coefficient on the x term here, this is essentially negative 1 times x. So we could say the coefficient is negative 1. And can we think of an a times b where it's going to be equal to negative 12? Well, let's think about this a little bit. The product of the two numbers is negative. So that means that they have different signs."}, {"video_title": "Factoring quadratics as (x+a)(x+b) (example 2)   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "And can we think of an a times b where it's going to be equal to negative 12? Well, let's think about this a little bit. The product of the two numbers is negative. So that means that they have different signs. So one will be positive and one will be negative. And so when I add them two together, I get to negative 1. Well, let's just think about the factors of negative 12."}, {"video_title": "Factoring quadratics as (x+a)(x+b) (example 2)   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "So that means that they have different signs. So one will be positive and one will be negative. And so when I add them two together, I get to negative 1. Well, let's just think about the factors of negative 12. Well, what about if 1 is 3 and maybe 1 is negative 4? Well, that seems to work. And you really just have to try these numbers out."}, {"video_title": "Factoring quadratics as (x+a)(x+b) (example 2)   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "Well, let's just think about the factors of negative 12. Well, what about if 1 is 3 and maybe 1 is negative 4? Well, that seems to work. And you really just have to try these numbers out. If a is 3, so if we're 3 plus negative 4, that indeed turns out to be negative 1. And if we have 3 times negative 4, that indeed is equal to negative 12. So that seems to work out."}, {"video_title": "Factoring quadratics as (x+a)(x+b) (example 2)   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "And you really just have to try these numbers out. If a is 3, so if we're 3 plus negative 4, that indeed turns out to be negative 1. And if we have 3 times negative 4, that indeed is equal to negative 12. So that seems to work out. And it's really a matter of trial and error. You could try negative 3 plus 4, but then that wouldn't have worked out over here. You could have tried 2 and 6, but that wouldn't have worked out on this number."}, {"video_title": "Factoring quadratics as (x+a)(x+b) (example 2)   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "So that seems to work out. And it's really a matter of trial and error. You could try negative 3 plus 4, but then that wouldn't have worked out over here. You could have tried 2 and 6, but that wouldn't have worked out on this number. Or 2 and negative 6, you wouldn't have gotten the sum to be equal to negative 1. But now that we've figured out what the a and b are, what is this expression factored? Well, it's going to be x plus 3 times x plus negative 4, or we could say x minus 4."}, {"video_title": "Solving systems of linear equations with substitution example   Algebra II   Khan Academy.mp3", "Sentence": "Negative 3x minus 4y is equal to negative 2. y is equal to 2x minus 5. So let me get out my little scratch pad, and let me rewrite the problem. So this is negative 3x minus 4y is equal to negative 2. And then they tell us y. y is equal to 2x minus 5. So what's neat about this is that they've already solved the second equation. They've already made it explicitly solved for y, which makes it very easy to substitute for. We could take this constraint, the constraint on y in terms of x, and substitute it for y in this first blue equation, and then solve for x."}, {"video_title": "Solving systems of linear equations with substitution example   Algebra II   Khan Academy.mp3", "Sentence": "And then they tell us y. y is equal to 2x minus 5. So what's neat about this is that they've already solved the second equation. They've already made it explicitly solved for y, which makes it very easy to substitute for. We could take this constraint, the constraint on y in terms of x, and substitute it for y in this first blue equation, and then solve for x. So let's try it out. So this first blue equation would then become negative 3x minus 4, but instead of putting a y there, the second constraint tells us that y needs to be equal to 2x minus 5. So it's 4 times 2x minus 5."}, {"video_title": "Solving systems of linear equations with substitution example   Algebra II   Khan Academy.mp3", "Sentence": "We could take this constraint, the constraint on y in terms of x, and substitute it for y in this first blue equation, and then solve for x. So let's try it out. So this first blue equation would then become negative 3x minus 4, but instead of putting a y there, the second constraint tells us that y needs to be equal to 2x minus 5. So it's 4 times 2x minus 5. And all of that is going to be equal to negative 2. So now we get just one equation with one unknown, and now we just have to solve for x. So let's see if we can do that."}, {"video_title": "Solving systems of linear equations with substitution example   Algebra II   Khan Academy.mp3", "Sentence": "So it's 4 times 2x minus 5. And all of that is going to be equal to negative 2. So now we get just one equation with one unknown, and now we just have to solve for x. So let's see if we can do that. So it is negative 3x, and then this part right over here, we have a negative 4. We'll be careful right over here. We have a negative 4 that we want to distribute."}, {"video_title": "Solving systems of linear equations with substitution example   Algebra II   Khan Academy.mp3", "Sentence": "So let's see if we can do that. So it is negative 3x, and then this part right over here, we have a negative 4. We'll be careful right over here. We have a negative 4 that we want to distribute. We're going to multiply negative 4 times 2x, which is negative 8x, and then negative 4 times negative 5 is positive 20, and that's going to equal negative 2. And now we can combine all of the x terms. So negative 3x minus 8x, that's going to be negative 11x."}, {"video_title": "Solving systems of linear equations with substitution example   Algebra II   Khan Academy.mp3", "Sentence": "We have a negative 4 that we want to distribute. We're going to multiply negative 4 times 2x, which is negative 8x, and then negative 4 times negative 5 is positive 20, and that's going to equal negative 2. And now we can combine all of the x terms. So negative 3x minus 8x, that's going to be negative 11x. And then we have plus 20 is equal to negative 2. Now to solve for x, we'll subtract 20 from both sides to get rid of the 20 on the left-hand side. Subtract 20 from both sides."}, {"video_title": "Solving systems of linear equations with substitution example   Algebra II   Khan Academy.mp3", "Sentence": "So negative 3x minus 8x, that's going to be negative 11x. And then we have plus 20 is equal to negative 2. Now to solve for x, we'll subtract 20 from both sides to get rid of the 20 on the left-hand side. Subtract 20 from both sides. On the left-hand side, we're just left with the negative 11x. And then on the right-hand side, we are left with negative 22. Now we can divide both sides by negative 11."}, {"video_title": "Solving systems of linear equations with substitution example   Algebra II   Khan Academy.mp3", "Sentence": "Subtract 20 from both sides. On the left-hand side, we're just left with the negative 11x. And then on the right-hand side, we are left with negative 22. Now we can divide both sides by negative 11. And we are left with x is equal to 22 divided by 11 is 2. And the negatives cancel out. x is equal to 2."}, {"video_title": "Solving systems of linear equations with substitution example   Algebra II   Khan Academy.mp3", "Sentence": "Now we can divide both sides by negative 11. And we are left with x is equal to 22 divided by 11 is 2. And the negatives cancel out. x is equal to 2. And so we're not quite done yet. We've done, I guess you could say, the hard part. We've solved for x, but now we have to solve for y."}, {"video_title": "Solving systems of linear equations with substitution example   Algebra II   Khan Academy.mp3", "Sentence": "x is equal to 2. And so we're not quite done yet. We've done, I guess you could say, the hard part. We've solved for x, but now we have to solve for y. And we could take this x value into either one of these equations and solve for y. But this second one has already explicitly solved for y, so let's use that one. So it says y is equal to 2 times, and instead of x, we now know that the x value where these two intersect, you could view it that way, is going to be equal to 2."}, {"video_title": "Solving systems of linear equations with substitution example   Algebra II   Khan Academy.mp3", "Sentence": "We've solved for x, but now we have to solve for y. And we could take this x value into either one of these equations and solve for y. But this second one has already explicitly solved for y, so let's use that one. So it says y is equal to 2 times, and instead of x, we now know that the x value where these two intersect, you could view it that way, is going to be equal to 2. So 2 times 2 minus 5. Let's figure out the corresponding y value. Minus 5."}, {"video_title": "Solving systems of linear equations with substitution example   Algebra II   Khan Academy.mp3", "Sentence": "So it says y is equal to 2 times, and instead of x, we now know that the x value where these two intersect, you could view it that way, is going to be equal to 2. So 2 times 2 minus 5. Let's figure out the corresponding y value. Minus 5. And so you get y is equal to, and that is the way the homes stretch, y is equal to 2 times 2 is 4 minus 5. So y is equal to negative 1. And you can verify that it will work in this top equation."}, {"video_title": "Solving systems of linear equations with substitution example   Algebra II   Khan Academy.mp3", "Sentence": "Minus 5. And so you get y is equal to, and that is the way the homes stretch, y is equal to 2 times 2 is 4 minus 5. So y is equal to negative 1. And you can verify that it will work in this top equation. If y is equal to negative 1 and x is equal to 2, this top equation becomes negative 3 times 2, which is negative 6, minus 4 times negative 1, which would be plus 4. And negative 6 plus 4 is indeed equal to negative 2. So it satisfies both of these equations."}, {"video_title": "Solving systems of linear equations with substitution example   Algebra II   Khan Academy.mp3", "Sentence": "And you can verify that it will work in this top equation. If y is equal to negative 1 and x is equal to 2, this top equation becomes negative 3 times 2, which is negative 6, minus 4 times negative 1, which would be plus 4. And negative 6 plus 4 is indeed equal to negative 2. So it satisfies both of these equations. And now we can type it in to verify that we got it right, although we know that we did. So x is equal to 2, y is equal to negative 1. So let's type it in."}, {"video_title": "Solving systems of linear equations with substitution example   Algebra II   Khan Academy.mp3", "Sentence": "So it satisfies both of these equations. And now we can type it in to verify that we got it right, although we know that we did. So x is equal to 2, y is equal to negative 1. So let's type it in. x is equal to 2, and y is equal to negative 1. Excellent. Now we're much less likely to be embarrassed by talking words."}, {"video_title": "Number of solutions to a system of equations algebraically   High School Math   Khan Academy.mp3", "Sentence": "There's a couple of ways to think about it. One way is to think about them graphically and think about, well, are they the same line? In which case they would have an infinite number of solutions, are they parallel? In which case they never intersect, you'd have no solutions. Or do they intersect exactly one place? In case you would have one solution. But instead we're gonna do this algebraically."}, {"video_title": "Number of solutions to a system of equations algebraically   High School Math   Khan Academy.mp3", "Sentence": "In which case they never intersect, you'd have no solutions. Or do they intersect exactly one place? In case you would have one solution. But instead we're gonna do this algebraically. So let's try to actually just solve the system and see what we get. So the first equation, I'm gonna leave that unchanged. Five x minus nine y is equal to 16."}, {"video_title": "Number of solutions to a system of equations algebraically   High School Math   Khan Academy.mp3", "Sentence": "But instead we're gonna do this algebraically. So let's try to actually just solve the system and see what we get. So the first equation, I'm gonna leave that unchanged. Five x minus nine y is equal to 16. Now this second equation right over here, let's say I wanna cancel out the x terms. So let me multiply the second equation by negative one. So I have a negative five x that I can cancel out with the five x."}, {"video_title": "Number of solutions to a system of equations algebraically   High School Math   Khan Academy.mp3", "Sentence": "Five x minus nine y is equal to 16. Now this second equation right over here, let's say I wanna cancel out the x terms. So let me multiply the second equation by negative one. So I have a negative five x that I can cancel out with the five x. So if I multiply the second equation by negative one, I'm gonna have negative five x plus nine y, plus nine y is equal to negative 36. Negative 36. Negative 36."}, {"video_title": "Number of solutions to a system of equations algebraically   High School Math   Khan Academy.mp3", "Sentence": "So I have a negative five x that I can cancel out with the five x. So if I multiply the second equation by negative one, I'm gonna have negative five x plus nine y, plus nine y is equal to negative 36. Negative 36. Negative 36. Now what I'm gonna do is I'm gonna add the left side of the equations and the right side of the equations to get a new equation. So five x minus five x, well that's gonna be zero. Negative nine y plus nine y, well that's gonna be zero again."}, {"video_title": "Number of solutions to a system of equations algebraically   High School Math   Khan Academy.mp3", "Sentence": "Negative 36. Now what I'm gonna do is I'm gonna add the left side of the equations and the right side of the equations to get a new equation. So five x minus five x, well that's gonna be zero. Negative nine y plus nine y, well that's gonna be zero again. I don't even have to write it. That's just gonna be zero on the left hand side. And on the right hand side, I'm gonna have 16 minus 36."}, {"video_title": "Number of solutions to a system of equations algebraically   High School Math   Khan Academy.mp3", "Sentence": "Negative nine y plus nine y, well that's gonna be zero again. I don't even have to write it. That's just gonna be zero on the left hand side. And on the right hand side, I'm gonna have 16 minus 36. So I have negative 20. So now I'm left with the somewhat bizarre looking equation that says that zero is equal to negative 20. Now one way you might say, well how does this make any sense?"}, {"video_title": "Number of solutions to a system of equations algebraically   High School Math   Khan Academy.mp3", "Sentence": "And on the right hand side, I'm gonna have 16 minus 36. So I have negative 20. So now I'm left with the somewhat bizarre looking equation that says that zero is equal to negative 20. Now one way you might say, well how does this make any sense? And the way to think about it is, well are there any xy values for which zero is going to be equal to negative 20? Well no, zero is never going to be equal to negative 20. And so it doesn't matter what xy values are."}, {"video_title": "Number of solutions to a system of equations algebraically   High School Math   Khan Academy.mp3", "Sentence": "Now one way you might say, well how does this make any sense? And the way to think about it is, well are there any xy values for which zero is going to be equal to negative 20? Well no, zero is never going to be equal to negative 20. And so it doesn't matter what xy values are. You can never find an x or xy pair that's going to make zero equal to negative 20. In fact, the x's and y's have disappeared from this equation. There's no way that this is going to be true."}, {"video_title": "Number of solutions to a system of equations algebraically   High School Math   Khan Academy.mp3", "Sentence": "And so it doesn't matter what xy values are. You can never find an x or xy pair that's going to make zero equal to negative 20. In fact, the x's and y's have disappeared from this equation. There's no way that this is going to be true. So we have no, we have no solutions. Now if you were to plot these, if you were to plot each of these lines, you would see that they are parallel lines. And that's why they have the same slope, different y intercepts, and that's why we have no solutions."}, {"video_title": "Number of solutions to a system of equations algebraically   High School Math   Khan Academy.mp3", "Sentence": "There's no way that this is going to be true. So we have no, we have no solutions. Now if you were to plot these, if you were to plot each of these lines, you would see that they are parallel lines. And that's why they have the same slope, different y intercepts, and that's why we have no solutions. They don't intersect. Let's do another one of these. This is fun."}, {"video_title": "Number of solutions to a system of equations algebraically   High School Math   Khan Academy.mp3", "Sentence": "And that's why they have the same slope, different y intercepts, and that's why we have no solutions. They don't intersect. Let's do another one of these. This is fun. All right. How many solutions does the following system of linear equations have? So let's do the same thing."}, {"video_title": "Number of solutions to a system of equations algebraically   High School Math   Khan Academy.mp3", "Sentence": "This is fun. All right. How many solutions does the following system of linear equations have? So let's do the same thing. I'm gonna keep the first equation the same. Negative 6x plus 4y is equal to 2. And the second equation, let me just see if I can cancel out the x term."}, {"video_title": "Number of solutions to a system of equations algebraically   High School Math   Khan Academy.mp3", "Sentence": "So let's do the same thing. I'm gonna keep the first equation the same. Negative 6x plus 4y is equal to 2. And the second equation, let me just see if I can cancel out the x term. So if I have a negative 6x, if I multiply this by 2, I'm gonna have a positive 6x. So I can, let's see, I'm gonna multiply this whole equation, both sides of it, by 2. So I'm gonna have 6x, 3x times 2 is 6x."}, {"video_title": "Number of solutions to a system of equations algebraically   High School Math   Khan Academy.mp3", "Sentence": "And the second equation, let me just see if I can cancel out the x term. So if I have a negative 6x, if I multiply this by 2, I'm gonna have a positive 6x. So I can, let's see, I'm gonna multiply this whole equation, both sides of it, by 2. So I'm gonna have 6x, 3x times 2 is 6x. Negative 2y times 2 is negative 4y. And that is going to be equal to negative 2. Now let's do the same thing."}, {"video_title": "Number of solutions to a system of equations algebraically   High School Math   Khan Academy.mp3", "Sentence": "So I'm gonna have 6x, 3x times 2 is 6x. Negative 2y times 2 is negative 4y. And that is going to be equal to negative 2. Now let's do the same thing. Let's add the left sides, and let's add the right sides. So negative 6x plus 6x, well that's gonna be 0. 4y minus 4y, that's 0."}, {"video_title": "Number of solutions to a system of equations algebraically   High School Math   Khan Academy.mp3", "Sentence": "Now let's do the same thing. Let's add the left sides, and let's add the right sides. So negative 6x plus 6x, well that's gonna be 0. 4y minus 4y, that's 0. We just have a 0 on the left-hand side. Now on the right-hand side, we have 2 plus negative 2. Well that's 0."}, {"video_title": "Number of solutions to a system of equations algebraically   High School Math   Khan Academy.mp3", "Sentence": "4y minus 4y, that's 0. We just have a 0 on the left-hand side. Now on the right-hand side, we have 2 plus negative 2. Well that's 0. So this is a little bit different. It still looks a little bit bizarre. 0 equals 0."}, {"video_title": "Number of solutions to a system of equations algebraically   High School Math   Khan Academy.mp3", "Sentence": "Well that's 0. So this is a little bit different. It still looks a little bit bizarre. 0 equals 0. Last time we had 0 is equal to, we had what, 0 is equal to negative 20. Now we have 0 equals 0. So one way to think about it is even though the x's and y's are no longer in this equation, we can say, well what xy pairs is it gonna be true for, that is gonna make it true that 0 is equal to 0?"}, {"video_title": "Number of solutions to a system of equations algebraically   High School Math   Khan Academy.mp3", "Sentence": "0 equals 0. Last time we had 0 is equal to, we had what, 0 is equal to negative 20. Now we have 0 equals 0. So one way to think about it is even though the x's and y's are no longer in this equation, we can say, well what xy pairs is it gonna be true for, that is gonna make it true that 0 is equal to 0? Well this is gonna be true no matter what x and y are. In fact, x and y are not involved in this equation anymore. 0 is always going to be equal to 0."}, {"video_title": "Number of solutions to a system of equations algebraically   High School Math   Khan Academy.mp3", "Sentence": "So one way to think about it is even though the x's and y's are no longer in this equation, we can say, well what xy pairs is it gonna be true for, that is gonna make it true that 0 is equal to 0? Well this is gonna be true no matter what x and y are. In fact, x and y are not involved in this equation anymore. 0 is always going to be equal to 0. So this is going to have an infinitely, this is going to have infinitely many solutions here. And that's because these are the same lines. I mean, they just look a little bit different algebraically, but if you scale one of them in the right way, in fact, if you just multiply both sides of this one, the second one, by negative 2, you're gonna get the top one."}, {"video_title": "Number of solutions to a system of equations algebraically   High School Math   Khan Academy.mp3", "Sentence": "0 is always going to be equal to 0. So this is going to have an infinitely, this is going to have infinitely many solutions here. And that's because these are the same lines. I mean, they just look a little bit different algebraically, but if you scale one of them in the right way, in fact, if you just multiply both sides of this one, the second one, by negative 2, you're gonna get the top one. And so they actually represent the exact same lines. You have an infinitely many solutions. All right."}, {"video_title": "Number of solutions to a system of equations algebraically   High School Math   Khan Academy.mp3", "Sentence": "I mean, they just look a little bit different algebraically, but if you scale one of them in the right way, in fact, if you just multiply both sides of this one, the second one, by negative 2, you're gonna get the top one. And so they actually represent the exact same lines. You have an infinitely many solutions. All right. When trying to find the solution to the following system of linear equations, Yvonne takes several correct steps that lead to the equation negative 5 is equal to 20. How many solutions does the system of linear equations have? I don't even have to look at the system right over here."}, {"video_title": "Number of solutions to a system of equations algebraically   High School Math   Khan Academy.mp3", "Sentence": "All right. When trying to find the solution to the following system of linear equations, Yvonne takes several correct steps that lead to the equation negative 5 is equal to 20. How many solutions does the system of linear equations have? I don't even have to look at the system right over here. The fact that she got the statement that can never be true, negative 5 is never going to be equal to 20, tells us that she has no solutions. And once again, if you were to plot these graphically, you would see that these are parallel lines. That's why they have no solutions."}, {"video_title": "Number of solutions to a system of equations algebraically   High School Math   Khan Academy.mp3", "Sentence": "I don't even have to look at the system right over here. The fact that she got the statement that can never be true, negative 5 is never going to be equal to 20, tells us that she has no solutions. And once again, if you were to plot these graphically, you would see that these are parallel lines. That's why they have no solutions. They never intersect. There's no xy pair that satisfies both of these constraints. Let's do a couple more of these."}, {"video_title": "Number of solutions to a system of equations algebraically   High School Math   Khan Academy.mp3", "Sentence": "That's why they have no solutions. They never intersect. There's no xy pair that satisfies both of these constraints. Let's do a couple more of these. When trying to find the solution to the following system of linear equations, Albus takes several correct steps that lead to the equation 5y is equal to negative 5. And I say, how many solutions does this system of linear equations have? Well, if 5y equals negative 5, we could divide both sides by 5, and we get y is equal to negative 1."}, {"video_title": "Number of solutions to a system of equations algebraically   High School Math   Khan Academy.mp3", "Sentence": "Let's do a couple more of these. When trying to find the solution to the following system of linear equations, Albus takes several correct steps that lead to the equation 5y is equal to negative 5. And I say, how many solutions does this system of linear equations have? Well, if 5y equals negative 5, we could divide both sides by 5, and we get y is equal to negative 1. And then if you substitute back in, y is equal to negative 1, if you did it in this first equation, if y is equal to negative 1, all of this becomes positive 2. You can subtract 2 from both sides, and you get 5x is equal to 4. Or you'd get what, x is equal to 4 fifths."}, {"video_title": "Number of solutions to a system of equations algebraically   High School Math   Khan Academy.mp3", "Sentence": "Well, if 5y equals negative 5, we could divide both sides by 5, and we get y is equal to negative 1. And then if you substitute back in, y is equal to negative 1, if you did it in this first equation, if y is equal to negative 1, all of this becomes positive 2. You can subtract 2 from both sides, and you get 5x is equal to 4. Or you'd get what, x is equal to 4 fifths. Or if you put negative 1 over here, you would get 5x minus 3 is equal to 1. You could add 3 to both sides, and you get 5x equals 4 again. x is equal to 4 fifths."}, {"video_title": "Number of solutions to a system of equations algebraically   High School Math   Khan Academy.mp3", "Sentence": "Or you'd get what, x is equal to 4 fifths. Or if you put negative 1 over here, you would get 5x minus 3 is equal to 1. You could add 3 to both sides, and you get 5x equals 4 again. x is equal to 4 fifths. So you have exactly one solution. You would have x is equal to 4 fifths. y is equal to negative 1."}, {"video_title": "Number of solutions to a system of equations algebraically   High School Math   Khan Academy.mp3", "Sentence": "x is equal to 4 fifths. So you have exactly one solution. You would have x is equal to 4 fifths. y is equal to negative 1. Let's do one more. When trying to find the solution of the following system of linear equations, Levan takes several correct steps that lead to the equation zero equals zero. So once again, I don't even need to look at this over here."}, {"video_title": "Coordinate plane quadrants   Negative numbers   6th grade   Khan Academy.mp3", "Sentence": "In which quadrant is the point (-7, 7), located? So let's just review what a quadrant is. A quadrant are each of the four sections of the coordinate plane. And when we talk about the sections, we're talking about the sections as divided by the coordinate axes. So this right here is the x-axis, and this up-down axis is the y-axis. And you can see it divides the coordinate plane into four sections. We call each of these sections quadrants."}, {"video_title": "Coordinate plane quadrants   Negative numbers   6th grade   Khan Academy.mp3", "Sentence": "And when we talk about the sections, we're talking about the sections as divided by the coordinate axes. So this right here is the x-axis, and this up-down axis is the y-axis. And you can see it divides the coordinate plane into four sections. We call each of these sections quadrants. This one over here, where both the x values and the y values are positive, we call the first quadrant. And we use the Roman numeral 1. Then if we kind of move counterclockwise around the coordinate plane, this quadrant, where the x values are negative and the y values are positive, we call this the second quadrant."}, {"video_title": "Coordinate plane quadrants   Negative numbers   6th grade   Khan Academy.mp3", "Sentence": "We call each of these sections quadrants. This one over here, where both the x values and the y values are positive, we call the first quadrant. And we use the Roman numeral 1. Then if we kind of move counterclockwise around the coordinate plane, this quadrant, where the x values are negative and the y values are positive, we call this the second quadrant. Then we go down here, where both the x values are negative and the y values are negative, we call this the third quadrant, once again using Roman numerals. Then finally, the quadrant where the x values are positive but the y values are negative, we call this the fourth quadrant. So let's see which quadrant the point (-7, 7), is located."}, {"video_title": "Coordinate plane quadrants   Negative numbers   6th grade   Khan Academy.mp3", "Sentence": "Then if we kind of move counterclockwise around the coordinate plane, this quadrant, where the x values are negative and the y values are positive, we call this the second quadrant. Then we go down here, where both the x values are negative and the y values are negative, we call this the third quadrant, once again using Roman numerals. Then finally, the quadrant where the x values are positive but the y values are negative, we call this the fourth quadrant. So let's see which quadrant the point (-7, 7), is located. So there's two ways to think about it. You could just say, look, we have a negative x value. Our x value is negative, so we're going to move to the left."}, {"video_title": "Coordinate plane quadrants   Negative numbers   6th grade   Khan Academy.mp3", "Sentence": "So let's see which quadrant the point (-7, 7), is located. So there's two ways to think about it. You could just say, look, we have a negative x value. Our x value is negative, so we're going to move to the left. So we're going to be on this side right here of the coordinate plane. So just by the fact that the x value is negative, we're going to be either in the second or the third quadrant. Now, we know that the y value is positive."}, {"video_title": "Coordinate plane quadrants   Negative numbers   6th grade   Khan Academy.mp3", "Sentence": "Our x value is negative, so we're going to move to the left. So we're going to be on this side right here of the coordinate plane. So just by the fact that the x value is negative, we're going to be either in the second or the third quadrant. Now, we know that the y value is positive. So if the x value is negative and the y value is positive, we're going to land someplace right over here in the second quadrant. The other way to think about it is you could literally just plot this point and see that it falls in the second quadrant. So let's do that."}, {"video_title": "Coordinate plane quadrants   Negative numbers   6th grade   Khan Academy.mp3", "Sentence": "Now, we know that the y value is positive. So if the x value is negative and the y value is positive, we're going to land someplace right over here in the second quadrant. The other way to think about it is you could literally just plot this point and see that it falls in the second quadrant. So let's do that. If x is negative 7, so that's negative 1, negative 2, negative 3, negative 4, negative 5, negative 6, negative 7. Did I do that right? 1, 2, 3, 4, 5, 6, 7."}, {"video_title": "Coordinate plane quadrants   Negative numbers   6th grade   Khan Academy.mp3", "Sentence": "So let's do that. If x is negative 7, so that's negative 1, negative 2, negative 3, negative 4, negative 5, negative 6, negative 7. Did I do that right? 1, 2, 3, 4, 5, 6, 7. So this is x is negative 7, and then we have to go up 7 because y is equal to positive 7. So 1, 2, 3, 4, 5, 6, 7. So the point (-7, 7), is right over here, clearly lies in the second quadrant."}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "So this might look a little daunting at first. We have x's on both sides of the equations. We're adding and subtracting numbers. How do you solve it? We'll do it a couple of different ways. The important thing to remember is we just want to isolate an x. Once you've isolated an x, you have x equals something, or x equals something."}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "How do you solve it? We'll do it a couple of different ways. The important thing to remember is we just want to isolate an x. Once you've isolated an x, you have x equals something, or x equals something. You're done. You've solved the equation. You can actually go back and check whether that works."}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "Once you've isolated an x, you have x equals something, or x equals something. You're done. You've solved the equation. You can actually go back and check whether that works. So what we're going to do is just do a bunch of operations on both sides of the equation to eventually isolate the x. But while we do those, I actually want to visualize what's occurring. Because I don't want you to just say, oh, what are the rules or the steps of solving equations, and I forgot whether this is allowed or that isn't allowed."}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "You can actually go back and check whether that works. So what we're going to do is just do a bunch of operations on both sides of the equation to eventually isolate the x. But while we do those, I actually want to visualize what's occurring. Because I don't want you to just say, oh, what are the rules or the steps of solving equations, and I forgot whether this is allowed or that isn't allowed. If you visualize what's happening, it'll actually be common sense what's allowed. So let's visualize it. So we have 2x right here on the left-hand side."}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "Because I don't want you to just say, oh, what are the rules or the steps of solving equations, and I forgot whether this is allowed or that isn't allowed. If you visualize what's happening, it'll actually be common sense what's allowed. So let's visualize it. So we have 2x right here on the left-hand side. So that's literally x plus x. And then you have plus 3. I'll do it like this."}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "So we have 2x right here on the left-hand side. So that's literally x plus x. And then you have plus 3. I'll do it like this. So that's equal to plus 1 plus 1 plus 1. That's the same thing as 3. I could have drawn three circles here as well."}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "I'll do it like this. So that's equal to plus 1 plus 1 plus 1. That's the same thing as 3. I could have drawn three circles here as well. Let me do the same color. Plus 3. And then that is equal to 5x's."}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "I could have drawn three circles here as well. Let me do the same color. Plus 3. And then that is equal to 5x's. I'll do that in blue. That is equal to 5x's. So 1, 2, 3, 4, 5."}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "And then that is equal to 5x's. I'll do that in blue. That is equal to 5x's. So 1, 2, 3, 4, 5. And I want to make it clear. You would never actually have to do it this way when you're solving the problem. You would just have to do the algebraic steps."}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "So 1, 2, 3, 4, 5. And I want to make it clear. You would never actually have to do it this way when you're solving the problem. You would just have to do the algebraic steps. But I'm doing this for you so you can actually visualize what this equation is saying. The left-hand side is these two orange x's plus 3. The right-hand side is 5x minus 2."}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "You would just have to do the algebraic steps. But I'm doing this for you so you can actually visualize what this equation is saying. The left-hand side is these two orange x's plus 3. The right-hand side is 5x minus 2. So minus 2 we could write as minus 1. Let me do this in a different color. I'll do it in pink."}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "The right-hand side is 5x minus 2. So minus 2 we could write as minus 1. Let me do this in a different color. I'll do it in pink. So minus 2 I'll do as minus 1 and minus 1. Now we want to isolate the x's on the same side of the equation. So how could we do that?"}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "I'll do it in pink. So minus 2 I'll do as minus 1 and minus 1. Now we want to isolate the x's on the same side of the equation. So how could we do that? Well, there's two ways of doing it. We could subtract these two x's from both sides of the equation. And that would be pretty reasonable."}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "So how could we do that? Well, there's two ways of doing it. We could subtract these two x's from both sides of the equation. And that would be pretty reasonable. Because then you'd have 5x's minus the 2x's. You'd have a positive number of x's on the right-hand side. Or you could actually subtract 5x from both sides."}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "And that would be pretty reasonable. Because then you'd have 5x's minus the 2x's. You'd have a positive number of x's on the right-hand side. Or you could actually subtract 5x from both sides. And that's what's neat about algebra. As long as you do legitimate operations, you will eventually get the right answer. So let's just start off subtracting 2x from both sides of the equation."}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "Or you could actually subtract 5x from both sides. And that's what's neat about algebra. As long as you do legitimate operations, you will eventually get the right answer. So let's just start off subtracting 2x from both sides of the equation. And when I mean that, I mean we're going to remove 2x's from the left-hand side. And if we remove 2x's from the left-hand side, we have to remove 2x's from the right-hand side, just like that. So what does that give us?"}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "So let's just start off subtracting 2x from both sides of the equation. And when I mean that, I mean we're going to remove 2x's from the left-hand side. And if we remove 2x's from the left-hand side, we have to remove 2x's from the right-hand side, just like that. So what does that give us? We're subtracting 2x's from the left. And we're also going to subtract 2x's from the right. Now what does our left-hand side simplify to?"}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "So what does that give us? We're subtracting 2x's from the left. And we're also going to subtract 2x's from the right. Now what does our left-hand side simplify to? We have 2x plus 3 minus 2x. The 2x's cancel out. So you're just left with the 3."}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "Now what does our left-hand side simplify to? We have 2x plus 3 minus 2x. The 2x's cancel out. So you're just left with the 3. And you see that over here. We took two of these x's away. We're just left with the plus 1, plus 1, plus 1."}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "So you're just left with the 3. And you see that over here. We took two of these x's away. We're just left with the plus 1, plus 1, plus 1. And then on the right-hand side, 5x minus 2x. We have it over here. We have 5x's minus 2x's."}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "We're just left with the plus 1, plus 1, plus 1. And then on the right-hand side, 5x minus 2x. We have it over here. We have 5x's minus 2x's. You only have 1, 2, 3x's left over. 3 is equal to 3x. And then you have your minus 2 there."}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "We have 5x's minus 2x's. You only have 1, 2, 3x's left over. 3 is equal to 3x. And then you have your minus 2 there. You have your minus 2. So normally, if you were to do the problem, you would just have to write what we have here on the left-hand side. So what can we do next?"}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "And then you have your minus 2 there. You have your minus 2. So normally, if you were to do the problem, you would just have to write what we have here on the left-hand side. So what can we do next? Remember, we want to isolate the x's. Well, we have all of our x's on the right-hand side right here. If we could get rid of this negative 2 off of the right-hand side, then the x's will be alone."}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "So what can we do next? Remember, we want to isolate the x's. Well, we have all of our x's on the right-hand side right here. If we could get rid of this negative 2 off of the right-hand side, then the x's will be alone. They'll be isolated. So how can we get rid of this negative 2? If we visualize it over here, this negative 1, this negative 1."}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "If we could get rid of this negative 2 off of the right-hand side, then the x's will be alone. They'll be isolated. So how can we get rid of this negative 2? If we visualize it over here, this negative 1, this negative 1. Well, we could add 2 to both sides of this equation. Think about what happens there. So if we add 2, so I'm going to do it like this, plus 1, plus 1."}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "If we visualize it over here, this negative 1, this negative 1. Well, we could add 2 to both sides of this equation. Think about what happens there. So if we add 2, so I'm going to do it like this, plus 1, plus 1. So you can literally see we're adding 2. And then we're going to add 2 to the left-hand side. 1 plus 1 plus."}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "So if we add 2, so I'm going to do it like this, plus 1, plus 1. So you can literally see we're adding 2. And then we're going to add 2 to the left-hand side. 1 plus 1 plus. What happens? Let me do it over here as well. So we're going to add 2."}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "1 plus 1 plus. What happens? Let me do it over here as well. So we're going to add 2. We're going to add 2. So what happens to the left-hand side? 3 plus 2 is going to be equal to 5."}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "So we're going to add 2. We're going to add 2. So what happens to the left-hand side? 3 plus 2 is going to be equal to 5. And that is going to be equal to 3x minus 2 plus 2. These guys cancel out, and you're just left with 3x. And we see it over here."}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "3 plus 2 is going to be equal to 5. And that is going to be equal to 3x minus 2 plus 2. These guys cancel out, and you're just left with 3x. And we see it over here. The left-hand side is 1 plus 1 plus 1 plus 1 plus 1. We have 5 1's or 5. And the right-hand side, we have the 3x's right over there."}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "And we see it over here. The left-hand side is 1 plus 1 plus 1 plus 1 plus 1. We have 5 1's or 5. And the right-hand side, we have the 3x's right over there. And then we have the negative 1, negative 1, plus 1, plus 1. Negative 1, these cancel out. They get us to 0."}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "And the right-hand side, we have the 3x's right over there. And then we have the negative 1, negative 1, plus 1, plus 1. Negative 1, these cancel out. They get us to 0. They cancel out. So we're just left with 5 is equal to 3x. So we have 1, 2, 3, 4, 5 is equal to 3x."}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "They get us to 0. They cancel out. So we're just left with 5 is equal to 3x. So we have 1, 2, 3, 4, 5 is equal to 3x. Let me clear everything that we've removed so it looks a little bit cleaner. Let me clear that out. And then let me clear."}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "So we have 1, 2, 3, 4, 5 is equal to 3x. Let me clear everything that we've removed so it looks a little bit cleaner. Let me clear that out. And then let me clear. These are all of the things that we've removed. Let me clear that out. And then let me clear that out like that."}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "And then let me clear. These are all of the things that we've removed. Let me clear that out. And then let me clear that out like that. Edit, clear. So now we are just left with 1, 2, 3, 4, 5. Actually, let me move this over."}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "And then let me clear that out like that. Edit, clear. So now we are just left with 1, 2, 3, 4, 5. Actually, let me move this over. So edit, cut, edit, paste. So I could just move this over right over here. We now have 1, 2, 3, 4, 5."}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "Actually, let me move this over. So edit, cut, edit, paste. So I could just move this over right over here. We now have 1, 2, 3, 4, 5. These are the 2 that we added here is equal to 3x's. These guys canceled out. That's why we have nothing there."}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "We now have 1, 2, 3, 4, 5. These are the 2 that we added here is equal to 3x's. These guys canceled out. That's why we have nothing there. Now, to solve this, we just divide both sides of this equation by 3. And this is going to be a little hard to visualize over here. But if we divide over here both sides by 3, what do we get?"}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "That's why we have nothing there. Now, to solve this, we just divide both sides of this equation by 3. And this is going to be a little hard to visualize over here. But if we divide over here both sides by 3, what do we get? We divide the left by 3. We divide the right by 3. The whole reason why we divided by 3 is because the x was being multiplied by 3."}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "But if we divide over here both sides by 3, what do we get? We divide the left by 3. We divide the right by 3. The whole reason why we divided by 3 is because the x was being multiplied by 3. 3 is the coefficient on the x. Fancy word. It literally just means the number multiplying the variable, the number we're solving, the variable we're solving for."}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "The whole reason why we divided by 3 is because the x was being multiplied by 3. 3 is the coefficient on the x. Fancy word. It literally just means the number multiplying the variable, the number we're solving, the variable we're solving for. So these 3's cancel out. The right-hand side of the equation is just x. The left-hand side is 5 thirds."}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "It literally just means the number multiplying the variable, the number we're solving, the variable we're solving for. So these 3's cancel out. The right-hand side of the equation is just x. The left-hand side is 5 thirds. So 5 thirds, we could say x is equal to 5 thirds. And this is different than everything we've seen so far. I now have the x on the right-hand side, the value on the left-hand side."}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "The left-hand side is 5 thirds. So 5 thirds, we could say x is equal to 5 thirds. And this is different than everything we've seen so far. I now have the x on the right-hand side, the value on the left-hand side. That's completely fine. This is the exact same thing as saying 5 thirds is equal to x is the same thing as saying x is equal to 5 thirds. Completely equivalent."}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "I now have the x on the right-hand side, the value on the left-hand side. That's completely fine. This is the exact same thing as saying 5 thirds is equal to x is the same thing as saying x is equal to 5 thirds. Completely equivalent. We sometimes get more used to this one, but this is completely the same thing. Now, if we wanted to write this as a mixed number, 3 goes into 5 one time with remainder 2. So it's going to be 1 and 2 thirds."}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "Completely equivalent. We sometimes get more used to this one, but this is completely the same thing. Now, if we wanted to write this as a mixed number, 3 goes into 5 one time with remainder 2. So it's going to be 1 and 2 thirds. So we could also write that x is equal to 1 and 2 thirds. And I'll leave it up for you to actually substitute back into this original equation and see that it works out. Now, to visualize it over here, how did you get 1 and 2 thirds?"}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "So it's going to be 1 and 2 thirds. So we could also write that x is equal to 1 and 2 thirds. And I'll leave it up for you to actually substitute back into this original equation and see that it works out. Now, to visualize it over here, how did you get 1 and 2 thirds? Let's think about it. Instead of doing 1, I'm going to do circles. I am going to do circles."}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "Now, to visualize it over here, how did you get 1 and 2 thirds? Let's think about it. Instead of doing 1, I'm going to do circles. I am going to do circles. Actually, even better, I'm going to do squares. So I'm going to have 5 squares on the left-hand side. I'll do it in this same yellow color right here."}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "I am going to do circles. Actually, even better, I'm going to do squares. So I'm going to have 5 squares on the left-hand side. I'll do it in this same yellow color right here. So I have 1, 2, 3, 4, 5. And that is going to be equal to the 3 x's. x plus x plus x."}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "I'll do it in this same yellow color right here. So I have 1, 2, 3, 4, 5. And that is going to be equal to the 3 x's. x plus x plus x. Now, we're dividing both sides of the equation by 3. Actually, that's what we did up here. We divided both sides by 3."}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "x plus x plus x. Now, we're dividing both sides of the equation by 3. Actually, that's what we did up here. We divided both sides by 3. So how do you do that? Right-hand side's pretty straightforward. You want to divide these 3 x's into 3 groups."}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "We divided both sides by 3. So how do you do that? Right-hand side's pretty straightforward. You want to divide these 3 x's into 3 groups. That's 1, 2, 3 groups. 1, 2, 3. Now, how do you divide 5 into 3?"}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "You want to divide these 3 x's into 3 groups. That's 1, 2, 3 groups. 1, 2, 3. Now, how do you divide 5 into 3? And they have to be even groups. And here, the answer tells us each group is going to be 1 and 2 thirds. So 1 and 2 thirds."}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "Now, how do you divide 5 into 3? And they have to be even groups. And here, the answer tells us each group is going to be 1 and 2 thirds. So 1 and 2 thirds. So it's going to be 2 thirds of this next one. And then we're going to have 1 and 2 thirds. So this is 1 third."}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "So 1 and 2 thirds. So it's going to be 2 thirds of this next one. And then we're going to have 1 and 2 thirds. So this is 1 third. We're going to need another 1. So this is 1 and 1 third. We're going to need 1 more third."}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "So this is 1 third. We're going to need another 1. So this is 1 and 1 third. We're going to need 1 more third. So it's going to be right here. And then we're left with 2 thirds and 1. So we've broken it up into 3 groups."}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "We're going to need 1 more third. So it's going to be right here. And then we're left with 2 thirds and 1. So we've broken it up into 3 groups. This right here, let me make it clear. This right here is 1 and 2 thirds. And then this right here, this is 1 third."}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "So we've broken it up into 3 groups. This right here, let me make it clear. This right here is 1 and 2 thirds. And then this right here, this is 1 third. That's another 1 third. So that's 2 thirds. And then that's 1 right there."}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "And then this right here, this is 1 third. That's another 1 third. So that's 2 thirds. And then that's 1 right there. So that's 1 and 2 thirds. And then finally, this is 2 thirds. And this is 1."}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "And then that's 1 right there. So that's 1 and 2 thirds. And then finally, this is 2 thirds. And this is 1. So this is 1 and 2 thirds. So when you divide both sides by 3, you get 1 and 2 thirds. Each section, each bucket is 1 and 2 thirds on the left-hand side."}, {"video_title": "Introduction to solving an equation with variables on both sides   Algebra I   Khan Academy.mp3", "Sentence": "And this is 1. So this is 1 and 2 thirds. So when you divide both sides by 3, you get 1 and 2 thirds. Each section, each bucket is 1 and 2 thirds on the left-hand side. On the left-hand side are 5 thirds. And on the right-hand side, we just have an x. So it still works."}, {"video_title": "How to simplify an expression by combining like terms and the distributive property   Khan Academy.mp3", "Sentence": "So for this video I've included something a little bit more soothing. So let's try to simplify some more expressions. And we'll see we're just applying ideas that we already knew about. So let's say I want to simplify the expression 2 times 3x plus 5. Well this literally means 2 3x plus 5s. So this is the exact same thing as, so this is 1 3x plus 5. And then to that I'm going to add another 3x plus 5."}, {"video_title": "How to simplify an expression by combining like terms and the distributive property   Khan Academy.mp3", "Sentence": "So let's say I want to simplify the expression 2 times 3x plus 5. Well this literally means 2 3x plus 5s. So this is the exact same thing as, so this is 1 3x plus 5. And then to that I'm going to add another 3x plus 5. This is literally what 2 times 3x plus 5 means. Well this is the same thing as, I mean if we just look at it right over here, we have now 2 3x's, so we could write it as 2 times 3x, plus we have 2 5's. Plus we have 2 5's."}, {"video_title": "How to simplify an expression by combining like terms and the distributive property   Khan Academy.mp3", "Sentence": "And then to that I'm going to add another 3x plus 5. This is literally what 2 times 3x plus 5 means. Well this is the same thing as, I mean if we just look at it right over here, we have now 2 3x's, so we could write it as 2 times 3x, plus we have 2 5's. Plus we have 2 5's. So plus 2 times 5. But you might say, hey Sal, isn't this just the distributive property that I know from arithmetic? I've essentially just distributed the 2."}, {"video_title": "How to simplify an expression by combining like terms and the distributive property   Khan Academy.mp3", "Sentence": "Plus we have 2 5's. So plus 2 times 5. But you might say, hey Sal, isn't this just the distributive property that I know from arithmetic? I've essentially just distributed the 2. 2 times 3x plus 2 times 5. And I would tell you, yes, it is. And the whole reason why I'm doing this is just to show you that it is exactly what you already know."}, {"video_title": "How to simplify an expression by combining like terms and the distributive property   Khan Academy.mp3", "Sentence": "I've essentially just distributed the 2. 2 times 3x plus 2 times 5. And I would tell you, yes, it is. And the whole reason why I'm doing this is just to show you that it is exactly what you already know. But with that out of the way, let's continue to simplify it. So when you multiply the 2 times the 3x, you get 6x. You multiply the 2 times the 5, you get 10."}, {"video_title": "How to simplify an expression by combining like terms and the distributive property   Khan Academy.mp3", "Sentence": "And the whole reason why I'm doing this is just to show you that it is exactly what you already know. But with that out of the way, let's continue to simplify it. So when you multiply the 2 times the 3x, you get 6x. You multiply the 2 times the 5, you get 10. So this simplified to 6x plus 10. Now let's try something that's a little bit more involved, but once again, really just things that you already know. So let's say I had 7 times 3y minus 5, minus 2 times 10 plus 4y."}, {"video_title": "How to simplify an expression by combining like terms and the distributive property   Khan Academy.mp3", "Sentence": "You multiply the 2 times the 5, you get 10. So this simplified to 6x plus 10. Now let's try something that's a little bit more involved, but once again, really just things that you already know. So let's say I had 7 times 3y minus 5, minus 2 times 10 plus 4y. Let's see if we can simplify this. Well, let's work on the left-hand side of the expression. The 7 times 3y minus 5."}, {"video_title": "How to simplify an expression by combining like terms and the distributive property   Khan Academy.mp3", "Sentence": "So let's say I had 7 times 3y minus 5, minus 2 times 10 plus 4y. Let's see if we can simplify this. Well, let's work on the left-hand side of the expression. The 7 times 3y minus 5. We just have to distribute the 7. So this is going to be 7 times 3y, which is going to give us 21y. Or if I had 3y's 7 times, it's going to be 21y's."}, {"video_title": "How to simplify an expression by combining like terms and the distributive property   Khan Academy.mp3", "Sentence": "The 7 times 3y minus 5. We just have to distribute the 7. So this is going to be 7 times 3y, which is going to give us 21y. Or if I had 3y's 7 times, it's going to be 21y's. Either way you want to think about it. And then I have 7 times, we've got to be careful with the sign. This is 7 times negative 5."}, {"video_title": "How to simplify an expression by combining like terms and the distributive property   Khan Academy.mp3", "Sentence": "Or if I had 3y's 7 times, it's going to be 21y's. Either way you want to think about it. And then I have 7 times, we've got to be careful with the sign. This is 7 times negative 5. 7 times negative 5 is negative 35. So we've simplified this part of it. Let's simplify the right-hand side."}, {"video_title": "How to simplify an expression by combining like terms and the distributive property   Khan Academy.mp3", "Sentence": "This is 7 times negative 5. 7 times negative 5 is negative 35. So we've simplified this part of it. Let's simplify the right-hand side. So you might be tempted to say, oh, 2 times 10 and 2 times 4y, and then subtract them. And if you do that right and you distribute the subtraction, it would work out. But I like to think of this as negative 2."}, {"video_title": "How to simplify an expression by combining like terms and the distributive property   Khan Academy.mp3", "Sentence": "Let's simplify the right-hand side. So you might be tempted to say, oh, 2 times 10 and 2 times 4y, and then subtract them. And if you do that right and you distribute the subtraction, it would work out. But I like to think of this as negative 2. And we're going to distribute the negative 2 times 10, and then we're going to distribute the times 10 and the negative 2 times 4y. So negative 2 times 10 is negative 20. So it's minus 20 right over here."}, {"video_title": "How to simplify an expression by combining like terms and the distributive property   Khan Academy.mp3", "Sentence": "But I like to think of this as negative 2. And we're going to distribute the negative 2 times 10, and then we're going to distribute the times 10 and the negative 2 times 4y. So negative 2 times 10 is negative 20. So it's minus 20 right over here. And then negative 2 times 4y is negative 8. So it's going to be negative 8y. So let's write a minus 8y right over here."}, {"video_title": "How to simplify an expression by combining like terms and the distributive property   Khan Academy.mp3", "Sentence": "So it's minus 20 right over here. And then negative 2 times 4y is negative 8. So it's going to be negative 8y. So let's write a minus 8y right over here. And are we done simplifying? Well, no, there's a little bit more that we can do. We can't add the 21y to the negative 35 or the negative 20 because these are adding different things or subtracting different things."}, {"video_title": "How to simplify an expression by combining like terms and the distributive property   Khan Academy.mp3", "Sentence": "So let's write a minus 8y right over here. And are we done simplifying? Well, no, there's a little bit more that we can do. We can't add the 21y to the negative 35 or the negative 20 because these are adding different things or subtracting different things. But we do have two things that are multiplying y. We have the 21y right over here. And then from that we are subtracting, or we can view it as from that we are subtracting 8y."}, {"video_title": "How to simplify an expression by combining like terms and the distributive property   Khan Academy.mp3", "Sentence": "We can't add the 21y to the negative 35 or the negative 20 because these are adding different things or subtracting different things. But we do have two things that are multiplying y. We have the 21y right over here. And then from that we are subtracting, or we can view it as from that we are subtracting 8y. So if I have 21 of something and I take 8 of them away, I'm left with 13 of that something. So those are going to simplify to 13y. And then I have negative 35."}, {"video_title": "How to simplify an expression by combining like terms and the distributive property   Khan Academy.mp3", "Sentence": "And then from that we are subtracting, or we can view it as from that we are subtracting 8y. So if I have 21 of something and I take 8 of them away, I'm left with 13 of that something. So those are going to simplify to 13y. And then I have negative 35. Let me use a new color. And then I have negative 35 minus 20. And so that's just going to simplify to negative 55."}, {"video_title": "Exponent example 2   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "Write 6 times 6 times 6 times 6 times 6 times 6 times 6 times 6 in exponential notation. So what's going on over here? We have 6 multiplied by itself how many times? Let's see, that's 6 times 1, then 6 times 2 is that, that's 6 times, well it's not 6 times 2, it's 6 times itself 2 times. 6 times itself 2 times would be 36, 6 times 2 would only be 12. So we have 1 6, 2 6, 3 6, 4 6, 5 6's, 6 6's, 7 6's, 8 6's. So we're multiplying 6 by itself 8 times."}, {"video_title": "Exponent example 2   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "Let's see, that's 6 times 1, then 6 times 2 is that, that's 6 times, well it's not 6 times 2, it's 6 times itself 2 times. 6 times itself 2 times would be 36, 6 times 2 would only be 12. So we have 1 6, 2 6, 3 6, 4 6, 5 6's, 6 6's, 7 6's, 8 6's. So we're multiplying 6 by itself 8 times. To write this in exponential notation we would say that this is equal to 6 to the 8th power, which is literally equal to 6 times 6 times 6 times 6 times 6 times 6 times 6. Now, I want to make it very clear, this is not, this is not, this is not equal to 6 times 8. 6 times 8, 6 times 8 will only be 48."}, {"video_title": "Exponent example 2   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "So we're multiplying 6 by itself 8 times. To write this in exponential notation we would say that this is equal to 6 to the 8th power, which is literally equal to 6 times 6 times 6 times 6 times 6 times 6 times 6. Now, I want to make it very clear, this is not, this is not, this is not equal to 6 times 8. 6 times 8, 6 times 8 will only be 48. 6 to the 8th power is a super huge number. 6 times 6 is 36, and you're going to multiply that times 6, which is what? That's like 36 times, that's what, 216, and then you keep multiplying it by 6, you get some huge number here."}, {"video_title": "Exponent example 2   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "6 times 8, 6 times 8 will only be 48. 6 to the 8th power is a super huge number. 6 times 6 is 36, and you're going to multiply that times 6, which is what? That's like 36 times, that's what, 216, and then you keep multiplying it by 6, you get some huge number here. This number right here, it's worthwhile to point out. This number is huge. This number is huge."}, {"video_title": "Exponent example 2   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "That's like 36 times, that's what, 216, and then you keep multiplying it by 6, you get some huge number here. This number right here, it's worthwhile to point out. This number is huge. This number is huge. This number right here, not so huge. It is not huge. So don't get confused."}, {"video_title": "Multiplying and dividing in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "So let's try to simplify this a little bit. And I'll start off by trying to simplify this denominator here. So the numerator is just 7 times 10 to the 5th. And the denominator, I just have a bunch of numbers that are being multiplied times each other. So I can do it in any order. So let me swap the order. So I'm going to do it 2 over 2 times 2.5 times 10 to the negative 2 times 10 to the 9th."}, {"video_title": "Multiplying and dividing in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "And the denominator, I just have a bunch of numbers that are being multiplied times each other. So I can do it in any order. So let me swap the order. So I'm going to do it 2 over 2 times 2.5 times 10 to the negative 2 times 10 to the 9th. And this is going to be equal to, so the numerator I haven't changed yet, 7 times 10 to the 5th over, and here in the denominator, 2 times, let me do this in a new color now, 2 times 2.5 is 5. And then 10 to the negative 2 times 10 to the 9th. When you multiply two numbers that are being raised to exponents and they have the exact same base, so it's 10 to the negative 2 times 10 to the negative 9, we can add the exponents."}, {"video_title": "Multiplying and dividing in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "So I'm going to do it 2 over 2 times 2.5 times 10 to the negative 2 times 10 to the 9th. And this is going to be equal to, so the numerator I haven't changed yet, 7 times 10 to the 5th over, and here in the denominator, 2 times, let me do this in a new color now, 2 times 2.5 is 5. And then 10 to the negative 2 times 10 to the 9th. When you multiply two numbers that are being raised to exponents and they have the exact same base, so it's 10 to the negative 2 times 10 to the negative 9, we can add the exponents. So this is going to be 10 to the 9 minus 2 or 10 to the 7th. So times 10 to the 7th. And now we can view this as being equal to 7 over 5 times 10 to the 5th over 10 to the 7th."}, {"video_title": "Multiplying and dividing in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "When you multiply two numbers that are being raised to exponents and they have the exact same base, so it's 10 to the negative 2 times 10 to the negative 9, we can add the exponents. So this is going to be 10 to the 9 minus 2 or 10 to the 7th. So times 10 to the 7th. And now we can view this as being equal to 7 over 5 times 10 to the 5th over 10 to the 7th. I'll do that in that orange color to keep track of the colors. 10 to the 7th. Now what is 7 divided by 5?"}, {"video_title": "Multiplying and dividing in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "And now we can view this as being equal to 7 over 5 times 10 to the 5th over 10 to the 7th. I'll do that in that orange color to keep track of the colors. 10 to the 7th. Now what is 7 divided by 5? 10 divided by 5 is equal to, let's see, it's 1 and 2 fifths or 1.4. So I'll just write it as 1.4. And then 10 to the 5th divided by 10 to the 7th."}, {"video_title": "Multiplying and dividing in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "Now what is 7 divided by 5? 10 divided by 5 is equal to, let's see, it's 1 and 2 fifths or 1.4. So I'll just write it as 1.4. And then 10 to the 5th divided by 10 to the 7th. So that's going to be the same thing as, and there's two ways to view this. You could view this as 10 to the 5th times 10 to the negative 7. You add the exponents, you get 10 to the negative 2."}, {"video_title": "Multiplying and dividing in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "And then 10 to the 5th divided by 10 to the 7th. So that's going to be the same thing as, and there's two ways to view this. You could view this as 10 to the 5th times 10 to the negative 7. You add the exponents, you get 10 to the negative 2. Or you say, hey, look, I'm dividing this by this. We have the same base. We can subtract exponents."}, {"video_title": "Multiplying and dividing in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "You add the exponents, you get 10 to the negative 2. Or you say, hey, look, I'm dividing this by this. We have the same base. We can subtract exponents. So it's going to be 10 to the 5 minus 7, which is 10 to the negative 2. So this part right over here is going to simplify to times 10 to the negative 2. Now are we done?"}, {"video_title": "Multiplying and dividing in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "We can subtract exponents. So it's going to be 10 to the 5 minus 7, which is 10 to the negative 2. So this part right over here is going to simplify to times 10 to the negative 2. Now are we done? Have we written what we have here in scientific notation? It looks like we have. This value right over here is greater than or equal to 1, but it is less than or equal to 9."}, {"video_title": "Multiplying and dividing in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "Now are we done? Have we written what we have here in scientific notation? It looks like we have. This value right over here is greater than or equal to 1, but it is less than or equal to 9. It's a digit between 1 and 9, including 1 and 9. And it's being multiplied by 10 to some power. So it looks like we're done."}, {"video_title": "Converting decimals to fractions 2 (ex 1)   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "Now let's just think about what places these are in. This is in the tenths place, tenths. This is in the hundredths place, hundredths, hundredths place. This too is in the thousandths place, thousandths. And this seven right here, this last seven, is in the ten thousandths place, ten thousandths. So there's a couple of ways we can do this. The way I like to think of this, this term right over here is in the ten thousandths place."}, {"video_title": "Converting decimals to fractions 2 (ex 1)   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "This too is in the thousandths place, thousandths. And this seven right here, this last seven, is in the ten thousandths place, ten thousandths. So there's a couple of ways we can do this. The way I like to think of this, this term right over here is in the ten thousandths place. We can view this whole thing right over here as 727 ten thousandths, because this is the smallest term, the smallest place right over here. So let's just rewrite it. This is equal to 727 over 10,000, over 10,000."}, {"video_title": "Converting decimals to fractions 2 (ex 1)   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "The way I like to think of this, this term right over here is in the ten thousandths place. We can view this whole thing right over here as 727 ten thousandths, because this is the smallest term, the smallest place right over here. So let's just rewrite it. This is equal to 727 over 10,000, over 10,000. And we've already written it as a fraction, and I think that's about as simplified as we can get. This number up here is not divisible by two, it's not divisible by five. In fact, it's not divisible by three, which means it wouldn't be divisible by six or nine."}, {"video_title": "Converting decimals to fractions 2 (ex 1)   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "This is equal to 727 over 10,000, over 10,000. And we've already written it as a fraction, and I think that's about as simplified as we can get. This number up here is not divisible by two, it's not divisible by five. In fact, it's not divisible by three, which means it wouldn't be divisible by six or nine. Doesn't even seem to be divisible by seven. It might be a prime number. But I think we are done."}, {"video_title": "Evaluating functions given their formula   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "So whenever you're dealing with a function, you take your input, in this case our input is going to be our 5, we input it into our little function box, and we need to get our output. And they define the function box here. Whatever your input is, take that, square it, and then subtract it from 49. So f of 5, every time I see an x here, since f of x is equal to this, every time I see an x, I would replace it with the input. So f of 5 is going to be equal to 49 minus, instead of writing x squared, I would write 5 squared. So this is equal to 49 minus 25, and 49 minus 25 is equal to 24. And we are done."}, {"video_title": "Graphs of systems of inequalities word problem   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "Her goal is to chop at least 20 vegetables with a time limit of 540 seconds. All right. The graph below represents the set of all combinations of carrots and broccoli. Inequality A, let's see, inequality A represents the range of all combinations Ksenia wants to chop, because she wants to chop at least 20 vegetables. So that's what inequality A is representing, that she wants to chop at least 20 vegetables. So all of this blue shaded area, and even the line is a solid line, so it includes point of the line. These are all of the scenarios where she's chopping at least 20 vegetables, all this blue area, including the blue line."}, {"video_title": "Graphs of systems of inequalities word problem   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "Inequality A, let's see, inequality A represents the range of all combinations Ksenia wants to chop, because she wants to chop at least 20 vegetables. So that's what inequality A is representing, that she wants to chop at least 20 vegetables. So all of this blue shaded area, and even the line is a solid line, so it includes point of the line. These are all of the scenarios where she's chopping at least 20 vegetables, all this blue area, including the blue line. And then it says inequality B represents the range of all combinations. She can chop with her time limit. So inequality B, this is all of the combinations where she is within her time limit, where she's not spending any more than 540 seconds."}, {"video_title": "Graphs of systems of inequalities word problem   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "These are all of the scenarios where she's chopping at least 20 vegetables, all this blue area, including the blue line. And then it says inequality B represents the range of all combinations. She can chop with her time limit. So inequality B, this is all of the combinations where she is within her time limit, where she's not spending any more than 540 seconds. What is the least number of carrots Ksenia can chop while achieving her goal? Well, her goal, remember, she wants to chop at least 20 vegetables, so you want to be in the blue area. You want to be in the solution set for inequality A, which would be the blue area, or on the blue line."}, {"video_title": "Graphs of systems of inequalities word problem   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "So inequality B, this is all of the combinations where she is within her time limit, where she's not spending any more than 540 seconds. What is the least number of carrots Ksenia can chop while achieving her goal? Well, her goal, remember, she wants to chop at least 20 vegetables, so you want to be in the blue area. You want to be in the solution set for inequality A, which would be the blue area, or on the blue line. And she wants to achieve her goal of meeting the time limit, so she needs to also be in the solution set for inequality B, so she also has to be in the green area, or on the green line. And so the overlap of the two, if she's meeting both constraints, it's going to be all of this area, all of this area right over here. This is the overlap, this is the overlap of the two solution sets."}, {"video_title": "Graphs of systems of inequalities word problem   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "You want to be in the solution set for inequality A, which would be the blue area, or on the blue line. And she wants to achieve her goal of meeting the time limit, so she needs to also be in the solution set for inequality B, so she also has to be in the green area, or on the green line. And so the overlap of the two, if she's meeting both constraints, it's going to be all of this area, all of this area right over here. This is the overlap, this is the overlap of the two solution sets. So in this overlap, where is the least number of carrots? What is the least number of carrots Ksenia can chop while achieving her goal? So if we see here the least number of carrots, you might be tempted to say, okay, 20 carrots, and that is in the solution set."}, {"video_title": "Graphs of systems of inequalities word problem   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "This is the overlap, this is the overlap of the two solution sets. So in this overlap, where is the least number of carrots? What is the least number of carrots Ksenia can chop while achieving her goal? So if we see here the least number of carrots, you might be tempted to say, okay, 20 carrots, and that is in the solution set. That would be 20 carrots and zero broccoli heads. But you can actually find a combination that has even fewer carrots. You could go all the way to this point, because remember, the points on the lines are also included in the solution sets because they are solid lines, not dashed lines."}, {"video_title": "Graphs of systems of inequalities word problem   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "So if we see here the least number of carrots, you might be tempted to say, okay, 20 carrots, and that is in the solution set. That would be 20 carrots and zero broccoli heads. But you can actually find a combination that has even fewer carrots. You could go all the way to this point, because remember, the points on the lines are also included in the solution sets because they are solid lines, not dashed lines. So this point right over here, 10 carrots and 10 broccoli heads actually meets her goal. So let me write that down. 10 carrots and 10 broccoli."}, {"video_title": "Graphs of systems of inequalities word problem   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "You could go all the way to this point, because remember, the points on the lines are also included in the solution sets because they are solid lines, not dashed lines. So this point right over here, 10 carrots and 10 broccoli heads actually meets her goal. So let me write that down. 10 carrots and 10 broccoli. 10 broccoli heads, broccoli. 10 broccoli heads, or let me just write that, 10 broccoli heads. So that's the least amount."}, {"video_title": "Graphs of systems of inequalities word problem   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "10 carrots and 10 broccoli. 10 broccoli heads, broccoli. 10 broccoli heads, or let me just write that, 10 broccoli heads. So that's the least amount. If you wanted to somehow figure out less than 10 carrots, in any of those scenarios, there's no overlap. If you said, oh, is there any way to do nine carrots? If you look over here, there's no overlap at C equals nine between the two solution sets."}, {"video_title": "Graphs of systems of inequalities word problem   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "So that's the least amount. If you wanted to somehow figure out less than 10 carrots, in any of those scenarios, there's no overlap. If you said, oh, is there any way to do nine carrots? If you look over here, there's no overlap at C equals nine between the two solution sets. So the minimum right over here is actually the point of intersection of these two lines. 10 carrots, 10 broccoli heads. That's the combination that has her chopping the minimum number of carrots while achieving, frankly, her goals, both of her goals, being under time and chopping at least 20 vegetables."}, {"video_title": "Recognizing rational and irrational expressions example   Algebra I   Khan Academy.mp3", "Sentence": "So if you have a rational number x, it can be expressed as the ratio of two integers, m and n. And if you have an irrational number, this cannot happen. So let's think about each of these. So nine is clearly a rational number. You can express nine as nine over one or 18 over two or 27 over three. So it can clearly be expressed as the ratio of two integers. But what about the square root of 45? So let's think about that a little bit."}, {"video_title": "Recognizing rational and irrational expressions example   Algebra I   Khan Academy.mp3", "Sentence": "You can express nine as nine over one or 18 over two or 27 over three. So it can clearly be expressed as the ratio of two integers. But what about the square root of 45? So let's think about that a little bit. Square root of 45, that's the same thing as the square root of nine times five, which is the same thing as the square root of nine times the square root of five. The principal root of nine is three. So it's three times the square root of five."}, {"video_title": "Recognizing rational and irrational expressions example   Algebra I   Khan Academy.mp3", "Sentence": "So let's think about that a little bit. Square root of 45, that's the same thing as the square root of nine times five, which is the same thing as the square root of nine times the square root of five. The principal root of nine is three. So it's three times the square root of five. So this is going to be nine plus three times the square root of five. So square root of five is irrational. You're taking the square root of a non-perfect square right over here, irrational."}, {"video_title": "Recognizing rational and irrational expressions example   Algebra I   Khan Academy.mp3", "Sentence": "So it's three times the square root of five. So this is going to be nine plus three times the square root of five. So square root of five is irrational. You're taking the square root of a non-perfect square right over here, irrational. Three is rational, but the product of a rational and an irrational is still going to be irrational. So that's going to be irrational. And then you're taking an irrational number and you're adding nine to it."}, {"video_title": "Recognizing rational and irrational expressions example   Algebra I   Khan Academy.mp3", "Sentence": "You're taking the square root of a non-perfect square right over here, irrational. Three is rational, but the product of a rational and an irrational is still going to be irrational. So that's going to be irrational. And then you're taking an irrational number and you're adding nine to it. You're adding a rational number to it. But you add a rational to an irrational and you're still going to have an irrational. So this whole thing is irrational."}, {"video_title": "Recognizing rational and irrational expressions example   Algebra I   Khan Academy.mp3", "Sentence": "And then you're taking an irrational number and you're adding nine to it. You're adding a rational number to it. But you add a rational to an irrational and you're still going to have an irrational. So this whole thing is irrational. Now let's think about this expression right over here. Well, the numerator can be rewritten as the square root of nine times five over three times the square root of five. Well, that's the same thing as the square root of nine times the square root of five over three times the square root of five."}, {"video_title": "Recognizing rational and irrational expressions example   Algebra I   Khan Academy.mp3", "Sentence": "So this whole thing is irrational. Now let's think about this expression right over here. Well, the numerator can be rewritten as the square root of nine times five over three times the square root of five. Well, that's the same thing as the square root of nine times the square root of five over three times the square root of five. Well, that's the same thing as three times the square root of five over three times the square root of 5, well, that's just going to be equal to 1. Or you could view it as 1 over 1. And 1 is clearly a rational number."}, {"video_title": "Recognizing rational and irrational expressions example   Algebra I   Khan Academy.mp3", "Sentence": "Well, that's the same thing as the square root of nine times the square root of five over three times the square root of five. Well, that's the same thing as three times the square root of five over three times the square root of 5, well, that's just going to be equal to 1. Or you could view it as 1 over 1. And 1 is clearly a rational number. You could write it as 1 over 1, 2 over 2, 3 over 3, really any integer over itself. So this is going to be rational. Now let's do this last expression right over here."}, {"video_title": "Recognizing rational and irrational expressions example   Algebra I   Khan Academy.mp3", "Sentence": "And 1 is clearly a rational number. You could write it as 1 over 1, 2 over 2, 3 over 3, really any integer over itself. So this is going to be rational. Now let's do this last expression right over here. 3 times the principal root of 9. Well, what's the principal root of 9? Well, it's 3."}, {"video_title": "Recognizing rational and irrational expressions example   Algebra I   Khan Academy.mp3", "Sentence": "Now let's do this last expression right over here. 3 times the principal root of 9. Well, what's the principal root of 9? Well, it's 3. So this is going to be 3 times 3, which is equal to 9. And we've already talked about the fact that 9 can clearly be expressed as the ratio of two integers. 9 over 1, 27 over 3, 45 over 5, all different ways of expressing 9."}, {"video_title": "Slope from two ordered pairs example 2   Algebra I   Khan Academy.mp3", "Sentence": "So let me draw a quick graph over here. So our first point is 7, negative 1. So 1, 2, 3, 4, 5, 6, 7. This is the x-axis. 7, negative 1. So 7, negative 1 is right over there. 7, negative 1."}, {"video_title": "Slope from two ordered pairs example 2   Algebra I   Khan Academy.mp3", "Sentence": "This is the x-axis. 7, negative 1. So 7, negative 1 is right over there. 7, negative 1. This of course is the y-axis. And then the next point is negative 3, negative 1. So we go back 3 in the horizontal direction."}, {"video_title": "Slope from two ordered pairs example 2   Algebra I   Khan Academy.mp3", "Sentence": "7, negative 1. This of course is the y-axis. And then the next point is negative 3, negative 1. So we go back 3 in the horizontal direction. Negative 3. But the y-coordinate is still negative 1. It's still negative 1."}, {"video_title": "Slope from two ordered pairs example 2   Algebra I   Khan Academy.mp3", "Sentence": "So we go back 3 in the horizontal direction. Negative 3. But the y-coordinate is still negative 1. It's still negative 1. So the line that connects these two points will look like this. It will look like that. Now, they're asking us to find the slope of the line that goes through the ordered pairs."}, {"video_title": "Slope from two ordered pairs example 2   Algebra I   Khan Academy.mp3", "Sentence": "It's still negative 1. So the line that connects these two points will look like this. It will look like that. Now, they're asking us to find the slope of the line that goes through the ordered pairs. Find the slope of this line. And just to give a little bit of intuition here, slope is a measure of a line's inclination. And the way that it's defined, slope is defined as rise over run, or change in y over change in x, or sometimes you'll see it defined as the variable m, and then they'll define change in y as just being the second y-coordinate minus the first y-coordinate, and then the change in x as the second x-coordinate minus the first x-coordinate."}, {"video_title": "Slope from two ordered pairs example 2   Algebra I   Khan Academy.mp3", "Sentence": "Now, they're asking us to find the slope of the line that goes through the ordered pairs. Find the slope of this line. And just to give a little bit of intuition here, slope is a measure of a line's inclination. And the way that it's defined, slope is defined as rise over run, or change in y over change in x, or sometimes you'll see it defined as the variable m, and then they'll define change in y as just being the second y-coordinate minus the first y-coordinate, and then the change in x as the second x-coordinate minus the first x-coordinate. These are all different variations in slope, but hopefully you appreciate that these are measuring inclination. If I rise a ton when I run a little bit, if I move a little bit in the x-direction and I rise a bunch, then I have a very steep line. I have a very steep upward sloping line."}, {"video_title": "Slope from two ordered pairs example 2   Algebra I   Khan Academy.mp3", "Sentence": "And the way that it's defined, slope is defined as rise over run, or change in y over change in x, or sometimes you'll see it defined as the variable m, and then they'll define change in y as just being the second y-coordinate minus the first y-coordinate, and then the change in x as the second x-coordinate minus the first x-coordinate. These are all different variations in slope, but hopefully you appreciate that these are measuring inclination. If I rise a ton when I run a little bit, if I move a little bit in the x-direction and I rise a bunch, then I have a very steep line. I have a very steep upward sloping line. If I don't change at all when I run a bit, then I have a very low slope, and that's actually what's happening here. I'm going from, you could either view this as the starting point or view this as the starting point, but let's view this as the starting point. So this negative 3 comma 1."}, {"video_title": "Slope from two ordered pairs example 2   Algebra I   Khan Academy.mp3", "Sentence": "I have a very steep upward sloping line. If I don't change at all when I run a bit, then I have a very low slope, and that's actually what's happening here. I'm going from, you could either view this as the starting point or view this as the starting point, but let's view this as the starting point. So this negative 3 comma 1. If I go from negative 3 comma negative 1 to 7 comma negative 1, I'm running a good bit. I'm going from negative 3, my x value is negative 3 here, and it goes all the way to 7. So my change in x here is 10."}, {"video_title": "Slope from two ordered pairs example 2   Algebra I   Khan Academy.mp3", "Sentence": "So this negative 3 comma 1. If I go from negative 3 comma negative 1 to 7 comma negative 1, I'm running a good bit. I'm going from negative 3, my x value is negative 3 here, and it goes all the way to 7. So my change in x here is 10. To go from negative 3 to 7, I change my x value by 10. But what's my change in y? Well, my y value here is negative 1, and my y value over here is still negative 1."}, {"video_title": "Slope from two ordered pairs example 2   Algebra I   Khan Academy.mp3", "Sentence": "So my change in x here is 10. To go from negative 3 to 7, I change my x value by 10. But what's my change in y? Well, my y value here is negative 1, and my y value over here is still negative 1. So my change in y is 0. My change in y is going to be 0. My y value does not change no matter how much I change my x value."}, {"video_title": "Slope from two ordered pairs example 2   Algebra I   Khan Academy.mp3", "Sentence": "Well, my y value here is negative 1, and my y value over here is still negative 1. So my change in y is 0. My change in y is going to be 0. My y value does not change no matter how much I change my x value. So the slope here is going to be, when we run 10, what was our rise? How much did we change in y? Well, we didn't rise at all."}, {"video_title": "Slope from two ordered pairs example 2   Algebra I   Khan Academy.mp3", "Sentence": "My y value does not change no matter how much I change my x value. So the slope here is going to be, when we run 10, what was our rise? How much did we change in y? Well, we didn't rise at all. We didn't go up or down. So the slope here is 0. Or another way to think about it is this line has no inclination."}, {"video_title": "Slope from two ordered pairs example 2   Algebra I   Khan Academy.mp3", "Sentence": "Well, we didn't rise at all. We didn't go up or down. So the slope here is 0. Or another way to think about it is this line has no inclination. It's a completely flat, it's a completely horizontal line. So this should make sense. This is a 0."}, {"video_title": "Slope from two ordered pairs example 2   Algebra I   Khan Academy.mp3", "Sentence": "Or another way to think about it is this line has no inclination. It's a completely flat, it's a completely horizontal line. So this should make sense. This is a 0. The slope here is 0. And just to make sure that this gels with all of these other formulas that you might know, but I want to make it very clear, these are all just telling you rise over run, or change in y over change in x, a way to measure inclination. But let's just apply them just so hopefully it all makes sense to you."}, {"video_title": "Slope from two ordered pairs example 2   Algebra I   Khan Academy.mp3", "Sentence": "This is a 0. The slope here is 0. And just to make sure that this gels with all of these other formulas that you might know, but I want to make it very clear, these are all just telling you rise over run, or change in y over change in x, a way to measure inclination. But let's just apply them just so hopefully it all makes sense to you. So we could also say slope is change in y over change in x. If we take this to be our start, and if we take this to be our end point, then we would call this over here x1, and then this is over here, this is y1, and then we would call this x2, and we would call this y2. If this is our start point and that is our end point."}, {"video_title": "Slope from two ordered pairs example 2   Algebra I   Khan Academy.mp3", "Sentence": "But let's just apply them just so hopefully it all makes sense to you. So we could also say slope is change in y over change in x. If we take this to be our start, and if we take this to be our end point, then we would call this over here x1, and then this is over here, this is y1, and then we would call this x2, and we would call this y2. If this is our start point and that is our end point. And so the slope here, the change in y, y2 minus y1, so it's negative 1 minus negative 1. All of that over x2, negative 3 minus x1, minus 7. So the numerator, negative 1 minus negative 1, that's the same thing as negative 1 plus 1."}, {"video_title": "Slope from two ordered pairs example 2   Algebra I   Khan Academy.mp3", "Sentence": "If this is our start point and that is our end point. And so the slope here, the change in y, y2 minus y1, so it's negative 1 minus negative 1. All of that over x2, negative 3 minus x1, minus 7. So the numerator, negative 1 minus negative 1, that's the same thing as negative 1 plus 1. And our denominator is negative 3 minus 7, which is negative 10. So once again, negative 1 plus 1 is 0 over negative 10. And this is still going to be 0."}, {"video_title": "Slope from two ordered pairs example 2   Algebra I   Khan Academy.mp3", "Sentence": "So the numerator, negative 1 minus negative 1, that's the same thing as negative 1 plus 1. And our denominator is negative 3 minus 7, which is negative 10. So once again, negative 1 plus 1 is 0 over negative 10. And this is still going to be 0. And the only reason why we have a negative 10 here and a positive 10 there is because we swapped the starting and the ending points. In this example right over here, we took this as the start point and made this coordinate over here as the end point. Over here we swapped them around."}, {"video_title": "Slope from two ordered pairs example 2   Algebra I   Khan Academy.mp3", "Sentence": "And this is still going to be 0. And the only reason why we have a negative 10 here and a positive 10 there is because we swapped the starting and the ending points. In this example right over here, we took this as the start point and made this coordinate over here as the end point. Over here we swapped them around. 7, negative 1 was our start point, and negative 3, negative 1 is our end point. So if we start over here, our change in x is going to be negative 10, but our change in y is still going to be 0. So regardless of how you do it, the slope of this line is 0."}, {"video_title": "Linear vs. exponential growth from data (example 2)   High School Math   Khan Academy.mp3", "Sentence": "So we have time in minutes, and then we have the corresponding temperature at different times in minutes. Which model for C of T, the temperature of the glass of water T minutes after it's served, best fits the data? So pause the video and see which of these models best fit the data. Alright, now let's work through this together. So in order for it, we see these choices, some of these are exponential models, some of these are linear models. In order for it to be a linear, in order for a linear model to be a good description, when you have a fixed change in time, you should have a fixed change in temperature. If you're dealing with an exponential model, then as you have a fixed change in time, you should be changing by the same factor."}, {"video_title": "Linear vs. exponential growth from data (example 2)   High School Math   Khan Academy.mp3", "Sentence": "Alright, now let's work through this together. So in order for it, we see these choices, some of these are exponential models, some of these are linear models. In order for it to be a linear, in order for a linear model to be a good description, when you have a fixed change in time, you should have a fixed change in temperature. If you're dealing with an exponential model, then as you have a fixed change in time, you should be changing by the same factor. So the amount you change from, say, minute one to minute two, or from minute two to minute three, it's not going to be the exact same amount, but it should be the same factor of where you started. So let's think about this. So here, our change in time is two minutes."}, {"video_title": "Linear vs. exponential growth from data (example 2)   High School Math   Khan Academy.mp3", "Sentence": "If you're dealing with an exponential model, then as you have a fixed change in time, you should be changing by the same factor. So the amount you change from, say, minute one to minute two, or from minute two to minute three, it's not going to be the exact same amount, but it should be the same factor of where you started. So let's think about this. So here, our change in time is two minutes. What is the absolute change in temperature? So our absolute change in temperature is negative, what, 15.7? Negative 15.7."}, {"video_title": "Linear vs. exponential growth from data (example 2)   High School Math   Khan Academy.mp3", "Sentence": "So here, our change in time is two minutes. What is the absolute change in temperature? So our absolute change in temperature is negative, what, 15.7? Negative 15.7. And what if we viewed it as a multiplication? So what do we multiply 80 by to get 64.3? Well, I can get a calculator out for that."}, {"video_title": "Linear vs. exponential growth from data (example 2)   High School Math   Khan Academy.mp3", "Sentence": "Negative 15.7. And what if we viewed it as a multiplication? So what do we multiply 80 by to get 64.3? Well, I can get a calculator out for that. So 64.3 divided by 80 is equal to 0.8. I'll just say approximately 0.8. So we could multiply by 0.8."}, {"video_title": "Linear vs. exponential growth from data (example 2)   High School Math   Khan Academy.mp3", "Sentence": "Well, I can get a calculator out for that. So 64.3 divided by 80 is equal to 0.8. I'll just say approximately 0.8. So we could multiply by 0.8. This is going to be approximate. So to get from 80 to 64.3, I could either subtract by 15.7 if I'm dealing with a linear model, or I can multiply by 0.8. Now, if I increase my time again by two, I'm going from minute two to minute four, so delta T is equal to two, the absolute change here is what?"}, {"video_title": "Linear vs. exponential growth from data (example 2)   High School Math   Khan Academy.mp3", "Sentence": "So we could multiply by 0.8. This is going to be approximate. So to get from 80 to 64.3, I could either subtract by 15.7 if I'm dealing with a linear model, or I can multiply by 0.8. Now, if I increase my time again by two, I'm going from minute two to minute four, so delta T is equal to two, the absolute change here is what? This is going to be not 12, this is going to be, my brain isn't functioning optimally. If this was 64.7, then this would be 12, but it's four less than that, so it's 11.6, negative 11.6. But if you looked at it as multiplying it by a factor, what would you have to multiply it by approximately?"}, {"video_title": "Linear vs. exponential growth from data (example 2)   High School Math   Khan Academy.mp3", "Sentence": "Now, if I increase my time again by two, I'm going from minute two to minute four, so delta T is equal to two, the absolute change here is what? This is going to be not 12, this is going to be, my brain isn't functioning optimally. If this was 64.7, then this would be 12, but it's four less than that, so it's 11.6, negative 11.6. But if you looked at it as multiplying it by a factor, what would you have to multiply it by approximately? Let's get the calculator back out. So if I said 52.7 divided by 64.3, divided by 64.3 is equal to, it's about 0.82. So times 0.82."}, {"video_title": "Linear vs. exponential growth from data (example 2)   High School Math   Khan Academy.mp3", "Sentence": "But if you looked at it as multiplying it by a factor, what would you have to multiply it by approximately? Let's get the calculator back out. So if I said 52.7 divided by 64.3, divided by 64.3 is equal to, it's about 0.82. So times 0.82. So just by looking at this, I could keep going, but it looks like for a given change in time, my absolute change in the number is not even close to being the same. If this was like 15.6, I'd be like, okay, there's a little bit of error here, data that you're collecting in the real world is never going to be perfect, these are models that try to get as close to describing the data. But over here, we keep multiplying it by a factor of roughly 0.8, roughly 0.8."}, {"video_title": "Linear vs. exponential growth from data (example 2)   High School Math   Khan Academy.mp3", "Sentence": "So times 0.82. So just by looking at this, I could keep going, but it looks like for a given change in time, my absolute change in the number is not even close to being the same. If this was like 15.6, I'd be like, okay, there's a little bit of error here, data that you're collecting in the real world is never going to be perfect, these are models that try to get as close to describing the data. But over here, we keep multiplying it by a factor of roughly 0.8, roughly 0.8. Now, you might be tempted to immediately say, okay, well, that means that C of T is going to be equal to our initial temperature, 80 times a common ratio of 0.8 to the number of minutes that pass by. Now, this is very tempting, and it wouldn't be the case if this was one minute and if this was two minutes, but our change in temperature each time is, our change in temperature each time is two minutes. So what we really should say is, this is, one way to think about it is that it takes, it takes two minutes to have a 0.8 change, so, or to be multiplied by 0.8."}, {"video_title": "Linear vs. exponential growth from data (example 2)   High School Math   Khan Academy.mp3", "Sentence": "But over here, we keep multiplying it by a factor of roughly 0.8, roughly 0.8. Now, you might be tempted to immediately say, okay, well, that means that C of T is going to be equal to our initial temperature, 80 times a common ratio of 0.8 to the number of minutes that pass by. Now, this is very tempting, and it wouldn't be the case if this was one minute and if this was two minutes, but our change in temperature each time is, our change in temperature each time is two minutes. So what we really should say is, this is, one way to think about it is that it takes, it takes two minutes to have a 0.8 change, so, or to be multiplied by 0.8. So the real way to describe this would be T over two. Every two minutes, when T is zero, we'd be at 80. After two minutes, we would take 80 times 0.8, which is what we got over here."}, {"video_title": "Linear vs. exponential growth from data (example 2)   High School Math   Khan Academy.mp3", "Sentence": "So what we really should say is, this is, one way to think about it is that it takes, it takes two minutes to have a 0.8 change, so, or to be multiplied by 0.8. So the real way to describe this would be T over two. Every two minutes, when T is zero, we'd be at 80. After two minutes, we would take 80 times 0.8, which is what we got over here. After four minutes, it would be 80 times 0.8 squared. In fact, let's just verify that we feel pretty good about this. So, if we had something like this, so, T and C of T, so when T is zero, C of T is 80."}, {"video_title": "Linear vs. exponential growth from data (example 2)   High School Math   Khan Academy.mp3", "Sentence": "After two minutes, we would take 80 times 0.8, which is what we got over here. After four minutes, it would be 80 times 0.8 squared. In fact, let's just verify that we feel pretty good about this. So, if we had something like this, so, T and C of T, so when T is zero, C of T is 80. When T is, well, let me just do the same data that we have here. When T is two, we have 80 times, two over two is one, so it's 80 times 0.8, which is pretty close to what we have over here. When T is four, it would be 80.0.8 squared, which is pretty close to what we have right over here."}, {"video_title": "Linear vs. exponential growth from data (example 2)   High School Math   Khan Academy.mp3", "Sentence": "So, if we had something like this, so, T and C of T, so when T is zero, C of T is 80. When T is, well, let me just do the same data that we have here. When T is two, we have 80 times, two over two is one, so it's 80 times 0.8, which is pretty close to what we have over here. When T is four, it would be 80.0.8 squared, which is pretty close to what we have right over here. I can just calculate it for you. If I have 0.8 squared times 80, 51.2, getting pretty close. This is a pretty good approximation, pretty good model."}, {"video_title": "Linear vs. exponential growth from data (example 2)   High School Math   Khan Academy.mp3", "Sentence": "When T is four, it would be 80.0.8 squared, which is pretty close to what we have right over here. I can just calculate it for you. If I have 0.8 squared times 80, 51.2, getting pretty close. This is a pretty good approximation, pretty good model. So, I'm liking this model. This isn't exactly one of the choices. So, how do we manipulate this a little bit?"}, {"video_title": "Linear vs. exponential growth from data (example 2)   High School Math   Khan Academy.mp3", "Sentence": "This is a pretty good approximation, pretty good model. So, I'm liking this model. This isn't exactly one of the choices. So, how do we manipulate this a little bit? Well, we can remind ourselves that this is the same thing as 80 times 0.8 to the 1.5 and then that to the T power. And what's 0.8 to the 1.5? So, 0.8, that's the same thing as the square root of 0.8."}, {"video_title": "Linear vs. exponential growth from data (example 2)   High School Math   Khan Academy.mp3", "Sentence": "So, how do we manipulate this a little bit? Well, we can remind ourselves that this is the same thing as 80 times 0.8 to the 1.5 and then that to the T power. And what's 0.8 to the 1.5? So, 0.8, that's the same thing as the square root of 0.8. It's roughly 0.89. So, this is approximately 80 times 0.89 to the T power. And if you look at all of these choices, this one is pretty close to this."}, {"video_title": "Linear vs. exponential growth from data (example 2)   High School Math   Khan Academy.mp3", "Sentence": "So, 0.8, that's the same thing as the square root of 0.8. It's roughly 0.89. So, this is approximately 80 times 0.89 to the T power. And if you look at all of these choices, this one is pretty close to this. This model best fits the data, especially of the choices going. This is pretty close to the model that I just thought about. Now, another way of doing it that might have been a little bit simpler."}, {"video_title": "Linear vs. exponential growth from data (example 2)   High School Math   Khan Academy.mp3", "Sentence": "And if you look at all of these choices, this one is pretty close to this. This model best fits the data, especially of the choices going. This is pretty close to the model that I just thought about. Now, another way of doing it that might have been a little bit simpler. I like to do it this way because even if I didn't have choices, we would have gotten to something reasonable. Another way to do it is say, okay, 80 is our initial state. All of these, whether you're talking about an exponential or a linear model, start with 80 when T is equal to zero."}, {"video_title": "Linear vs. exponential growth from data (example 2)   High School Math   Khan Academy.mp3", "Sentence": "Now, another way of doing it that might have been a little bit simpler. I like to do it this way because even if I didn't have choices, we would have gotten to something reasonable. Another way to do it is say, okay, 80 is our initial state. All of these, whether you're talking about an exponential or a linear model, start with 80 when T is equal to zero. But it's clearly not a linear model because we're not changing by even roughly the same amount every time. But it looks like every two minutes we're changing by a factor of 0.8. So, we're going to have an exponential model."}, {"video_title": "Linear vs. exponential growth from data (example 2)   High School Math   Khan Academy.mp3", "Sentence": "All of these, whether you're talking about an exponential or a linear model, start with 80 when T is equal to zero. But it's clearly not a linear model because we're not changing by even roughly the same amount every time. But it looks like every two minutes we're changing by a factor of 0.8. So, we're going to have an exponential model. So, you say, okay, it'd be one of these two choices. Now, this one down here you could rule out because we're not changing by a factor of 0.8 or 0.81 every minute. We're changing by a factor of 0.81 every two minutes."}, {"video_title": "Linear vs. exponential growth from data (example 2)   High School Math   Khan Academy.mp3", "Sentence": "So, we're going to have an exponential model. So, you say, okay, it'd be one of these two choices. Now, this one down here you could rule out because we're not changing by a factor of 0.8 or 0.81 every minute. We're changing by a factor of 0.81 every two minutes. So, you could have ruled that one out and then you could have deduced this right over here. And you could have said, look, if I'm changing by a factor of 0.9 every minute, then that would be 0.81 every two minutes, which is pretty close to what we're seeing here, changing by a factor of about 0.8 or 0.81 every two minutes. So, once again, that's why we like that first choice."}, {"video_title": "Order of operations examples exponents   Arithmetic operations   6th grade   Khan Academy.mp3", "Sentence": "I'm assuming you've given a go at it. Now let's work through them. So when we see something like this, we have to remember our order of operations. We have 2 times 3 squared. And we have to remember that the first thing we need to think about are the parentheses. I'll just write paren for short. Then we worry about exponents."}, {"video_title": "Order of operations examples exponents   Arithmetic operations   6th grade   Khan Academy.mp3", "Sentence": "We have 2 times 3 squared. And we have to remember that the first thing we need to think about are the parentheses. I'll just write paren for short. Then we worry about exponents. Then we worry about multiplication and division. And actually, let me write it this way. We worry about multiplication and division."}, {"video_title": "Order of operations examples exponents   Arithmetic operations   6th grade   Khan Academy.mp3", "Sentence": "Then we worry about exponents. Then we worry about multiplication and division. And actually, let me write it this way. We worry about multiplication and division. And then we worry about addition and subtraction. So in this expression right over here, there are no parentheses. So we do the exponents first."}, {"video_title": "Order of operations examples exponents   Arithmetic operations   6th grade   Khan Academy.mp3", "Sentence": "We worry about multiplication and division. And then we worry about addition and subtraction. So in this expression right over here, there are no parentheses. So we do the exponents first. So we calculate what 3 squared is. 3 times 3 is 9. So this becomes 2 times 9, which is equal to 18."}, {"video_title": "Order of operations examples exponents   Arithmetic operations   6th grade   Khan Academy.mp3", "Sentence": "So we do the exponents first. So we calculate what 3 squared is. 3 times 3 is 9. So this becomes 2 times 9, which is equal to 18. Now let's look at this one. And this one is interesting because it looks like the same expression. But now there are parentheses."}, {"video_title": "Order of operations examples exponents   Arithmetic operations   6th grade   Khan Academy.mp3", "Sentence": "So this becomes 2 times 9, which is equal to 18. Now let's look at this one. And this one is interesting because it looks like the same expression. But now there are parentheses. And because of these parentheses, we were going to do the multiplication before we take the exponent. So 2 times 3 is going to be 6. And we're going to take that to the second power."}, {"video_title": "Order of operations examples exponents   Arithmetic operations   6th grade   Khan Academy.mp3", "Sentence": "But now there are parentheses. And because of these parentheses, we were going to do the multiplication before we take the exponent. So 2 times 3 is going to be 6. And we're going to take that to the second power. So that's 6 times 6, which is equal to 36. Now let's think about this one right over here. Once again, we want to do our multiplication and our division first."}, {"video_title": "Order of operations examples exponents   Arithmetic operations   6th grade   Khan Academy.mp3", "Sentence": "And we're going to take that to the second power. So that's 6 times 6, which is equal to 36. Now let's think about this one right over here. Once again, we want to do our multiplication and our division first. So we have a division right over here. 81 over 9 is the same thing as 81 divided by 9. That's going to be 9."}, {"video_title": "Order of operations examples exponents   Arithmetic operations   6th grade   Khan Academy.mp3", "Sentence": "Once again, we want to do our multiplication and our division first. So we have a division right over here. 81 over 9 is the same thing as 81 divided by 9. That's going to be 9. And then we have, so it becomes 1 plus 5 times 9. Now we want to do the multiplication before we do the addition. So we're going to do our 5 times 9, which is 45."}, {"video_title": "Order of operations examples exponents   Arithmetic operations   6th grade   Khan Academy.mp3", "Sentence": "That's going to be 9. And then we have, so it becomes 1 plus 5 times 9. Now we want to do the multiplication before we do the addition. So we're going to do our 5 times 9, which is 45. So this becomes 1 plus 45, which of course is equal to 46. Now let's tackle this one right over here. So we would want to do the exponents first."}, {"video_title": "Order of operations examples exponents   Arithmetic operations   6th grade   Khan Academy.mp3", "Sentence": "So we're going to do our 5 times 9, which is 45. So this becomes 1 plus 45, which of course is equal to 46. Now let's tackle this one right over here. So we would want to do the exponents first. So 1 squared, well, that's just going to be equal to 1. So that's just going to be equal to 1. And so you have 2 times 4 plus 1."}, {"video_title": "Order of operations examples exponents   Arithmetic operations   6th grade   Khan Academy.mp3", "Sentence": "So we would want to do the exponents first. So 1 squared, well, that's just going to be equal to 1. So that's just going to be equal to 1. And so you have 2 times 4 plus 1. What should you do? Should you add first or do the multiplication first? Well, multiplication takes precedence over addition."}, {"video_title": "Order of operations examples exponents   Arithmetic operations   6th grade   Khan Academy.mp3", "Sentence": "And so you have 2 times 4 plus 1. What should you do? Should you add first or do the multiplication first? Well, multiplication takes precedence over addition. So you're going to do the 2 times 4 first. 2 times 4 is 8. So you're going to have 8 plus 1, which of course is equal to 9."}, {"video_title": "Order of operations examples exponents   Arithmetic operations   6th grade   Khan Academy.mp3", "Sentence": "Well, multiplication takes precedence over addition. So you're going to do the 2 times 4 first. 2 times 4 is 8. So you're going to have 8 plus 1, which of course is equal to 9. Now you have a very similar expression, but you have parentheses. So that's going to force you to do what's in the parentheses before you take the exponent. But within the parentheses, we have multiplication and addition."}, {"video_title": "Order of operations examples exponents   Arithmetic operations   6th grade   Khan Academy.mp3", "Sentence": "So you're going to have 8 plus 1, which of course is equal to 9. Now you have a very similar expression, but you have parentheses. So that's going to force you to do what's in the parentheses before you take the exponent. But within the parentheses, we have multiplication and addition. And we have to remember that we do the multiplication first. So we're going to do the 2 times 4 first. So that's going to be 8 plus 1 to the second power."}, {"video_title": "Order of operations examples exponents   Arithmetic operations   6th grade   Khan Academy.mp3", "Sentence": "But within the parentheses, we have multiplication and addition. And we have to remember that we do the multiplication first. So we're going to do the 2 times 4 first. So that's going to be 8 plus 1 to the second power. 8 plus 1 is 9, so that's 9 to the second power. 9 squared is the same thing as 9 times 9, which is equal to 81. Now we have one more right over here that looked very similar to this one, except once again, we have parentheses."}, {"video_title": "Order of operations examples exponents   Arithmetic operations   6th grade   Khan Academy.mp3", "Sentence": "So that's going to be 8 plus 1 to the second power. 8 plus 1 is 9, so that's 9 to the second power. 9 squared is the same thing as 9 times 9, which is equal to 81. Now we have one more right over here that looked very similar to this one, except once again, we have parentheses. So it's making us do the addition first. Without parentheses, we would do the multiplication and the division first. But here we see that 1 plus 5 is 6."}, {"video_title": "Find measure of complementary angles   Angles and intersecting lines   Geometry   Khan Academy.mp3", "Sentence": "So we're told that ray OL is perpendicular to ray ON. So OL is perpendicular to ON. So we know that this right over here is a right angle, or it's a 90 degree angle. Then they tell us that the measure of angle LOM is equal to 2x plus 46. So LOM, so this angle right over here, is equal to 2x plus 46. And then this angle, then they tell us that the measure of angle MON is equal to 3x minus 6. So MON, so this angle right over here is 3x minus 6."}, {"video_title": "Find measure of complementary angles   Angles and intersecting lines   Geometry   Khan Academy.mp3", "Sentence": "Then they tell us that the measure of angle LOM is equal to 2x plus 46. So LOM, so this angle right over here, is equal to 2x plus 46. And then this angle, then they tell us that the measure of angle MON is equal to 3x minus 6. So MON, so this angle right over here is 3x minus 6. And so they ask us, find angle MON. So we need to find this angle right over here. And we'd be able to find it if we knew what x is."}, {"video_title": "Find measure of complementary angles   Angles and intersecting lines   Geometry   Khan Academy.mp3", "Sentence": "So MON, so this angle right over here is 3x minus 6. And so they ask us, find angle MON. So we need to find this angle right over here. And we'd be able to find it if we knew what x is. If we knew what x is, then it's going to be 3 times that minus 6. Now the one thing that we know is that if we add up these two angles, these are adjacent angles. And their outside rays form a right angle."}, {"video_title": "Find measure of complementary angles   Angles and intersecting lines   Geometry   Khan Academy.mp3", "Sentence": "And we'd be able to find it if we knew what x is. If we knew what x is, then it's going to be 3 times that minus 6. Now the one thing that we know is that if we add up these two angles, these are adjacent angles. And their outside rays form a right angle. So these two angles are going to add up to 90 degrees. And I guess we can assume here that we're dealing in degrees. So let's say that 2x plus 46 plus 3x minus 6 is going to add up to 90 degrees."}, {"video_title": "Find measure of complementary angles   Angles and intersecting lines   Geometry   Khan Academy.mp3", "Sentence": "And their outside rays form a right angle. So these two angles are going to add up to 90 degrees. And I guess we can assume here that we're dealing in degrees. So let's say that 2x plus 46 plus 3x minus 6 is going to add up to 90 degrees. These two angles are complementary. So it's going to add up to 90 degrees, or it's going to add up to 90. And now we just have to simplify."}, {"video_title": "Find measure of complementary angles   Angles and intersecting lines   Geometry   Khan Academy.mp3", "Sentence": "So let's say that 2x plus 46 plus 3x minus 6 is going to add up to 90 degrees. These two angles are complementary. So it's going to add up to 90 degrees, or it's going to add up to 90. And now we just have to simplify. We have two x's. We have another three x's. So you add those two, you're going to get five x's."}, {"video_title": "Find measure of complementary angles   Angles and intersecting lines   Geometry   Khan Academy.mp3", "Sentence": "And now we just have to simplify. We have two x's. We have another three x's. So you add those two, you're going to get five x's. And then you have 46. And you're going to subtract 6. So it's going to be plus 40 is equal to 90."}, {"video_title": "Find measure of complementary angles   Angles and intersecting lines   Geometry   Khan Academy.mp3", "Sentence": "So you add those two, you're going to get five x's. And then you have 46. And you're going to subtract 6. So it's going to be plus 40 is equal to 90. Subtract 40 from both sides. Let's do that. Subtract 40 from both sides."}, {"video_title": "Find measure of complementary angles   Angles and intersecting lines   Geometry   Khan Academy.mp3", "Sentence": "So it's going to be plus 40 is equal to 90. Subtract 40 from both sides. Let's do that. Subtract 40 from both sides. And on the left-hand side, you're just left with a 5x. And on the right-hand side, you're left with a 50. Now just divide both sides by 5."}, {"video_title": "Find measure of complementary angles   Angles and intersecting lines   Geometry   Khan Academy.mp3", "Sentence": "Subtract 40 from both sides. And on the left-hand side, you're just left with a 5x. And on the right-hand side, you're left with a 50. Now just divide both sides by 5. And we are left with x is equal to 10. Now our answer is not 10. We've just figured out what x is."}, {"video_title": "Find measure of complementary angles   Angles and intersecting lines   Geometry   Khan Academy.mp3", "Sentence": "Now just divide both sides by 5. And we are left with x is equal to 10. Now our answer is not 10. We've just figured out what x is. What they ask us for is what's the measure of angle MON? Well, we already know that the measure of angle MON is equal to 3x minus 6, which is equal to 3 times 10 minus 6. I don't want to do that minus 6 in that green color."}, {"video_title": "Exponent example 1   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "Let me rewrite that. We have 5 to the third power. Now it's important to remember this does not mean 5 times 3. This means 5 times itself 3 times. So this is equal to 5 times 5 times 5. 5 times 3, just as a bit of a refresher so you realize the difference, 5 times 3, let me write it over here. 5 times 3 is equal to 5 plus 5 plus 5."}, {"video_title": "Exponent example 1   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "This means 5 times itself 3 times. So this is equal to 5 times 5 times 5. 5 times 3, just as a bit of a refresher so you realize the difference, 5 times 3, let me write it over here. 5 times 3 is equal to 5 plus 5 plus 5. So when you multiply by 3, you're adding the number to itself 3 times. When you take it to the third power, you're multiplying the number by itself 3 times. So 5 times 3, you've seen that before, that's 15."}, {"video_title": "Exponent example 1   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "5 times 3 is equal to 5 plus 5 plus 5. So when you multiply by 3, you're adding the number to itself 3 times. When you take it to the third power, you're multiplying the number by itself 3 times. So 5 times 3, you've seen that before, that's 15. But 5 to the third power, 5 times itself 3 times is equal to, well 5 times 5 is 25, and then 25 times 5 is 125. So this is equal to 125. And we're done."}, {"video_title": "Example 2 Factoring quadratics by grouping   Algebra I   Khan Academy.mp3", "Sentence": "And to factor by grouping, we need to look for 2 numbers whose product is equal to 4 times negative 15. So we're looking for 2 numbers whose product, let's call those a and b, is going to be equal to 4 times negative 15. 4 times negative 15, or negative 60. And the sum of those 2 numbers, a plus b, needs to be equal to this 4 right there. Needs to be equal to 4. So let's think about all the factors of negative 60 or 60. And we're looking for ones that are essentially 4 apart, because the numbers are going to be of different signs, because their product is negative."}, {"video_title": "Example 2 Factoring quadratics by grouping   Algebra I   Khan Academy.mp3", "Sentence": "And the sum of those 2 numbers, a plus b, needs to be equal to this 4 right there. Needs to be equal to 4. So let's think about all the factors of negative 60 or 60. And we're looking for ones that are essentially 4 apart, because the numbers are going to be of different signs, because their product is negative. So when you take 2 numbers of different signs and you sum them, you kind of view it as the difference of their absolute values. If that confuses you, don't worry about it. But this tells you that the numbers, since they're going to be of different size, their absolute values are going to be roughly 4 apart."}, {"video_title": "Example 2 Factoring quadratics by grouping   Algebra I   Khan Academy.mp3", "Sentence": "And we're looking for ones that are essentially 4 apart, because the numbers are going to be of different signs, because their product is negative. So when you take 2 numbers of different signs and you sum them, you kind of view it as the difference of their absolute values. If that confuses you, don't worry about it. But this tells you that the numbers, since they're going to be of different size, their absolute values are going to be roughly 4 apart. So we could try out things like 5 and 12. 5 and negative 12, because 1 has to be negative. If you add these 2, you get negative 7."}, {"video_title": "Example 2 Factoring quadratics by grouping   Algebra I   Khan Academy.mp3", "Sentence": "But this tells you that the numbers, since they're going to be of different size, their absolute values are going to be roughly 4 apart. So we could try out things like 5 and 12. 5 and negative 12, because 1 has to be negative. If you add these 2, you get negative 7. If you did negative 5 and 12, you'd get positive 7. They're just still too far apart. What if we tried 6 and negative 10?"}, {"video_title": "Example 2 Factoring quadratics by grouping   Algebra I   Khan Academy.mp3", "Sentence": "If you add these 2, you get negative 7. If you did negative 5 and 12, you'd get positive 7. They're just still too far apart. What if we tried 6 and negative 10? Then you'd get a negative 4 if you added these 2. We want a positive 4, so let's do negative 6 and 10. Negative 6 plus 10 is positive 4."}, {"video_title": "Example 2 Factoring quadratics by grouping   Algebra I   Khan Academy.mp3", "Sentence": "What if we tried 6 and negative 10? Then you'd get a negative 4 if you added these 2. We want a positive 4, so let's do negative 6 and 10. Negative 6 plus 10 is positive 4. So those will be our 2 numbers, negative 6 and positive 10. Now, what we want to do is we want to break up this middle term here. The whole point of figuring out the negative 6 and the 10 is to break up the 4y into a negative 6y and a 10y."}, {"video_title": "Example 2 Factoring quadratics by grouping   Algebra I   Khan Academy.mp3", "Sentence": "Negative 6 plus 10 is positive 4. So those will be our 2 numbers, negative 6 and positive 10. Now, what we want to do is we want to break up this middle term here. The whole point of figuring out the negative 6 and the 10 is to break up the 4y into a negative 6y and a 10y. So let's do that. So this 4y can be rewritten as negative 6y plus 10y. Because if you add those, you get 4y."}, {"video_title": "Example 2 Factoring quadratics by grouping   Algebra I   Khan Academy.mp3", "Sentence": "The whole point of figuring out the negative 6 and the 10 is to break up the 4y into a negative 6y and a 10y. So let's do that. So this 4y can be rewritten as negative 6y plus 10y. Because if you add those, you get 4y. And on the other sides of it, you have your 4y squared, and then you have your minus 15. All I did is expand this into these 2 numbers as being the coefficients on the y. If you add these, you get the 4y again."}, {"video_title": "Example 2 Factoring quadratics by grouping   Algebra I   Khan Academy.mp3", "Sentence": "Because if you add those, you get 4y. And on the other sides of it, you have your 4y squared, and then you have your minus 15. All I did is expand this into these 2 numbers as being the coefficients on the y. If you add these, you get the 4y again. Now, this is where the grouping comes in. You group the terms. So let's see."}, {"video_title": "Example 2 Factoring quadratics by grouping   Algebra I   Khan Academy.mp3", "Sentence": "If you add these, you get the 4y again. Now, this is where the grouping comes in. You group the terms. So let's see. Let me do it in a different color. So if I take these 2 guys, what can I factor out of those 2 guys? Well, there's a common factor."}, {"video_title": "Example 2 Factoring quadratics by grouping   Algebra I   Khan Academy.mp3", "Sentence": "So let's see. Let me do it in a different color. So if I take these 2 guys, what can I factor out of those 2 guys? Well, there's a common factor. It looks like there's a common factor of 2y. So if we factor out 2y, we get 2y times 4y squared divided by 2y is 2y. And then negative 6y divided by 2y is negative 3."}, {"video_title": "Example 2 Factoring quadratics by grouping   Algebra I   Khan Academy.mp3", "Sentence": "Well, there's a common factor. It looks like there's a common factor of 2y. So if we factor out 2y, we get 2y times 4y squared divided by 2y is 2y. And then negative 6y divided by 2y is negative 3. Negative 3. So this group gets factored into 2y times 2y minus 3. Now, let's look at this other group right here."}, {"video_title": "Example 2 Factoring quadratics by grouping   Algebra I   Khan Academy.mp3", "Sentence": "And then negative 6y divided by 2y is negative 3. Negative 3. So this group gets factored into 2y times 2y minus 3. Now, let's look at this other group right here. This was the whole point about breaking it up like this. And in other videos, I've explained why this works. Now, here, the greatest common factor is a 5."}, {"video_title": "Example 2 Factoring quadratics by grouping   Algebra I   Khan Academy.mp3", "Sentence": "Now, let's look at this other group right here. This was the whole point about breaking it up like this. And in other videos, I've explained why this works. Now, here, the greatest common factor is a 5. So we can factor out a 5. So this is equal to plus 5 times 10y divided by 5 is 2y. Negative 15 divided by 5 is 3."}, {"video_title": "Example 2 Factoring quadratics by grouping   Algebra I   Khan Academy.mp3", "Sentence": "Now, here, the greatest common factor is a 5. So we can factor out a 5. So this is equal to plus 5 times 10y divided by 5 is 2y. Negative 15 divided by 5 is 3. And so we have 2y times 2y minus 3 plus 5 times 2y minus 3. So now you have 2 terms, and 2y minus 3 is a common factor to both. So let's factor out a 2y minus 3."}, {"video_title": "Example 2 Factoring quadratics by grouping   Algebra I   Khan Academy.mp3", "Sentence": "Negative 15 divided by 5 is 3. And so we have 2y times 2y minus 3 plus 5 times 2y minus 3. So now you have 2 terms, and 2y minus 3 is a common factor to both. So let's factor out a 2y minus 3. So this is equal to 2y minus 3 times 2y, times that 2y, plus that 5. There's no magic happening here. All I did is undistribute the 2y minus 3."}, {"video_title": "Example 2 Factoring quadratics by grouping   Algebra I   Khan Academy.mp3", "Sentence": "So let's factor out a 2y minus 3. So this is equal to 2y minus 3 times 2y, times that 2y, plus that 5. There's no magic happening here. All I did is undistribute the 2y minus 3. I factored it out of both of these guys and took it out of the parentheses. If I distributed it in, you'd get back to this expression. But we're done."}, {"video_title": "Example 2 Factoring quadratics by grouping   Algebra I   Khan Academy.mp3", "Sentence": "All I did is undistribute the 2y minus 3. I factored it out of both of these guys and took it out of the parentheses. If I distributed it in, you'd get back to this expression. But we're done. We factored it. We factored it into 2 binomial expressions. 4y squared plus 4y minus 15 is 2y minus 3 times 2y plus 5."}, {"video_title": "Coordinate plane graphing points and naming quadrants   Negative numbers   6th grade   Khan Academy.mp3", "Sentence": "That says how far to move in the, in the horizontal or the x direction. It's a positive 4, so I'm going to go 4 to the right. And then the second coordinate says, what do we do in the vertical direction or in the y direction? It's a negative 1. Since it's negative, we're going to go down. And it's a negative 1, so we're going to go down 1. So that right over there is the point 4, negative 1."}, {"video_title": "Coordinate plane graphing points and naming quadrants   Negative numbers   6th grade   Khan Academy.mp3", "Sentence": "It's a negative 1. Since it's negative, we're going to go down. And it's a negative 1, so we're going to go down 1. So that right over there is the point 4, negative 1. So I've plotted it, but now I have to select which quadrant the point lies in. And this is just a naming convention. This is the first quadrant, this is the second quadrant, this is the third quadrant, and this is the fourth quadrant."}, {"video_title": "Coordinate plane graphing points and naming quadrants   Negative numbers   6th grade   Khan Academy.mp3", "Sentence": "So that right over there is the point 4, negative 1. So I've plotted it, but now I have to select which quadrant the point lies in. And this is just a naming convention. This is the first quadrant, this is the second quadrant, this is the third quadrant, and this is the fourth quadrant. So the point lies in the fourth quadrant, quadrant 4. And I guess you have to know your Roman numerals a little bit to know that that's the fo, that's representing quadrant 4. Let's do a couple more of these."}, {"video_title": "Coordinate plane graphing points and naming quadrants   Negative numbers   6th grade   Khan Academy.mp3", "Sentence": "This is the first quadrant, this is the second quadrant, this is the third quadrant, and this is the fourth quadrant. So the point lies in the fourth quadrant, quadrant 4. And I guess you have to know your Roman numerals a little bit to know that that's the fo, that's representing quadrant 4. Let's do a couple more of these. Plot 8, negative 4 and select the quadrant in which the point lies. Well, my x coordinate is 8, so I move 8 or go 8 in the positive x direction. And then my y coordinate is negative 4, so I go 4 down."}, {"video_title": "Coordinate plane graphing points and naming quadrants   Negative numbers   6th grade   Khan Academy.mp3", "Sentence": "Let's do a couple more of these. Plot 8, negative 4 and select the quadrant in which the point lies. Well, my x coordinate is 8, so I move 8 or go 8 in the positive x direction. And then my y coordinate is negative 4, so I go 4 down. And this is sitting again in not the first, not the second, not the third, but the fourth quadrant in quadrant 4. Let's do one more of these. Hopefully we get a different quadrant."}, {"video_title": "Coordinate plane graphing points and naming quadrants   Negative numbers   6th grade   Khan Academy.mp3", "Sentence": "And then my y coordinate is negative 4, so I go 4 down. And this is sitting again in not the first, not the second, not the third, but the fourth quadrant in quadrant 4. Let's do one more of these. Hopefully we get a different quadrant. So we want to plot the point negative 5, 5. So now my x coordinate is negative, it's negative 5. So I'm going to move to the left in the x direction."}, {"video_title": "Coordinate plane graphing points and naming quadrants   Negative numbers   6th grade   Khan Academy.mp3", "Sentence": "Hopefully we get a different quadrant. So we want to plot the point negative 5, 5. So now my x coordinate is negative, it's negative 5. So I'm going to move to the left in the x direction. So I go to negative 5. And my y coordinate is positive, so I go up 5. So negative 5, 5."}, {"video_title": "Coordinate plane graphing points and naming quadrants   Negative numbers   6th grade   Khan Academy.mp3", "Sentence": "So I'm going to move to the left in the x direction. So I go to negative 5. And my y coordinate is positive, so I go up 5. So negative 5, 5. And this is sitting not in the first quadrant, but the second quadrant. And of course this is the third and the fourth. So this is sitting in the second quadrant."}, {"video_title": "Multiplying in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "So let's multiply first, and then let's try to get what we have in scientific notation. Actually, before we do that, let's just even remember what it means to be in scientific notation. So to be in scientific notation, and actually each of these numbers right here are in scientific notation, it's going to be the form a times 10 to some power, where a can be greater than or equal to 1, and it is going to be less than 10. So both of these numbers are greater than or equal to 1, and they are less than 10, and they're being multiplied by some power of 10. So let's see how we can multiply this. So this over here, this is just the exact same thing. So if I do this in magenta, this is the exact same thing as 9.1 times 10 to the 6th times 3.2 times, well, actually, I don't have to write it."}, {"video_title": "Multiplying in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "So both of these numbers are greater than or equal to 1, and they are less than 10, and they're being multiplied by some power of 10. So let's see how we can multiply this. So this over here, this is just the exact same thing. So if I do this in magenta, this is the exact same thing as 9.1 times 10 to the 6th times 3.2 times, well, actually, I don't have to write it. Let me write it all with a dot notation to make it a little bit more straightforward. So this is equal to 9.1 times 10 to the 6th times 3.2 times 10 to the negative 5th power. Now in multiplication, this comes from the associative property, essentially allows us to remove these parentheses."}, {"video_title": "Multiplying in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "So if I do this in magenta, this is the exact same thing as 9.1 times 10 to the 6th times 3.2 times, well, actually, I don't have to write it. Let me write it all with a dot notation to make it a little bit more straightforward. So this is equal to 9.1 times 10 to the 6th times 3.2 times 10 to the negative 5th power. Now in multiplication, this comes from the associative property, essentially allows us to remove these parentheses. It says, look, you can multiply like that first, or you can actually multiply these guys first. You can reassociate them. And the commutative property tells us that we can rearrange this thing right here."}, {"video_title": "Multiplying in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "Now in multiplication, this comes from the associative property, essentially allows us to remove these parentheses. It says, look, you can multiply like that first, or you can actually multiply these guys first. You can reassociate them. And the commutative property tells us that we can rearrange this thing right here. And what I want to rearrange is I want to multiply the 9.1 times the 3.2 first, and then multiply that times 10 to the 6th times 10 to the negative 5th. So I'm just going to rearrange this using the commutative property. So this is the same thing as 9.1 times 3.2, and I'm going to reassociate, so I'm going to do these first."}, {"video_title": "Multiplying in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "And the commutative property tells us that we can rearrange this thing right here. And what I want to rearrange is I want to multiply the 9.1 times the 3.2 first, and then multiply that times 10 to the 6th times 10 to the negative 5th. So I'm just going to rearrange this using the commutative property. So this is the same thing as 9.1 times 3.2, and I'm going to reassociate, so I'm going to do these first. And then that times 10 to the 6th times 10 to the negative 5, and the reason why this is useful is that this is really easy to multiply. We have the same base here, base 10, and we're taking the product, so we can add the exponents. So this part right over here, 10 to the 6th times 10 to the negative 5, that's going to be 10 to the 6 minus 5 power, or essentially just 10 to the first power, which is really just equal to 10."}, {"video_title": "Multiplying in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "So this is the same thing as 9.1 times 3.2, and I'm going to reassociate, so I'm going to do these first. And then that times 10 to the 6th times 10 to the negative 5, and the reason why this is useful is that this is really easy to multiply. We have the same base here, base 10, and we're taking the product, so we can add the exponents. So this part right over here, 10 to the 6th times 10 to the negative 5, that's going to be 10 to the 6 minus 5 power, or essentially just 10 to the first power, which is really just equal to 10. And that's going to be multiplied by 9.1 times 3.2. So let me do that over here. So if I have 9.1 times 3.2, so at first I'm going to ignore the decimals, I'm just going to treat it like 91 times 32."}, {"video_title": "Multiplying in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "So this part right over here, 10 to the 6th times 10 to the negative 5, that's going to be 10 to the 6 minus 5 power, or essentially just 10 to the first power, which is really just equal to 10. And that's going to be multiplied by 9.1 times 3.2. So let me do that over here. So if I have 9.1 times 3.2, so at first I'm going to ignore the decimals, I'm just going to treat it like 91 times 32. So if 2 times 1 is 2, 2 times 9 is 18. Take a 0 here, because I'm in the tenths place now, multiplying everything really by 30, not just by 3. That's why my 0 is there."}, {"video_title": "Multiplying in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "So if I have 9.1 times 3.2, so at first I'm going to ignore the decimals, I'm just going to treat it like 91 times 32. So if 2 times 1 is 2, 2 times 9 is 18. Take a 0 here, because I'm in the tenths place now, multiplying everything really by 30, not just by 3. That's why my 0 is there. And I multiply 3 times 1 to get 3, and then 3 times 9 is 27. And so it is 2, 2 plus 0, so I'm adding here, 2 plus 0 is 2, 8 plus 3 is 11, carry or regroup that 1. 1 plus 1 is 2, 2 plus 7 is 9, and then I have a 2 here."}, {"video_title": "Multiplying in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "That's why my 0 is there. And I multiply 3 times 1 to get 3, and then 3 times 9 is 27. And so it is 2, 2 plus 0, so I'm adding here, 2 plus 0 is 2, 8 plus 3 is 11, carry or regroup that 1. 1 plus 1 is 2, 2 plus 7 is 9, and then I have a 2 here. So 91 times 32 is 2,912, but I didn't multiply 91 times 32. I multiplied 9.1 times 3.2. So what I want to do is count the number of digits I have behind the decimal point."}, {"video_title": "Multiplying in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "1 plus 1 is 2, 2 plus 7 is 9, and then I have a 2 here. So 91 times 32 is 2,912, but I didn't multiply 91 times 32. I multiplied 9.1 times 3.2. So what I want to do is count the number of digits I have behind the decimal point. I have one, two digits behind the decimal point, and so I'll have to have two digits behind the decimal point in the answer. So one, two, I'll stick the decimal right over there. So this part right over here comes out to be 29.12."}, {"video_title": "Multiplying in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "So what I want to do is count the number of digits I have behind the decimal point. I have one, two digits behind the decimal point, and so I'll have to have two digits behind the decimal point in the answer. So one, two, I'll stick the decimal right over there. So this part right over here comes out to be 29.12. So you might say, you might feel like we're done. This kind of looks like scientific notation. I have a number times a power of 10."}, {"video_title": "Multiplying in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "So this part right over here comes out to be 29.12. So you might say, you might feel like we're done. This kind of looks like scientific notation. I have a number times a power of 10. But remember, this number has to be greater than or equal to 1, which it is, and less than 10. But this number is not less than 10. It's not in scientific notation."}, {"video_title": "Multiplying in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "I have a number times a power of 10. But remember, this number has to be greater than or equal to 1, which it is, and less than 10. But this number is not less than 10. It's not in scientific notation. So what we can do is let's just write this number in scientific notation, and then we could use the power of 10 part to multiply this power of 10. So 29.12, this is the same thing as 2.912. Notice, what did I do to go from there to there?"}, {"video_title": "Multiplying in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "It's not in scientific notation. So what we can do is let's just write this number in scientific notation, and then we could use the power of 10 part to multiply this power of 10. So 29.12, this is the same thing as 2.912. Notice, what did I do to go from there to there? I just moved the decimal to the left. Or another way to think about it, if I wanted to go from here to there, what could I do to this? Well, I would multiply it by 10."}, {"video_title": "Multiplying in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "Notice, what did I do to go from there to there? I just moved the decimal to the left. Or another way to think about it, if I wanted to go from here to there, what could I do to this? Well, I would multiply it by 10. If I multiplied by 10, I would move the decimal to the right. It would go from 2.9 to 29. So if I want to write this value, this is just this times 10, so 29.12 is the same thing as 2.912 times 10."}, {"video_title": "Multiplying in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "Well, I would multiply it by 10. If I multiplied by 10, I would move the decimal to the right. It would go from 2.9 to 29. So if I want to write this value, this is just this times 10, so 29.12 is the same thing as 2.912 times 10. And now this is in scientific notation, but that's just this part, and I still have to multiply it by another 10. So times another 10. And so to finish up this problem, we get 2.912 times 10 times 10, or 10 to the first times 10 to the first."}, {"video_title": "Multiplying in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "So if I want to write this value, this is just this times 10, so 29.12 is the same thing as 2.912 times 10. And now this is in scientific notation, but that's just this part, and I still have to multiply it by another 10. So times another 10. And so to finish up this problem, we get 2.912 times 10 times 10, or 10 to the first times 10 to the first. Well, what's that? Well, that's going to be this part right over here. That's just 10 squared."}, {"video_title": "Multiplying binomials area model   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "So I have this big rectangle here that's divided into four smaller rectangles. So what I want to do is I want to express the area of this larger rectangle, and I want to do it two ways. The first way I want to express it as the product of two binomials, and then I want to express it as a trinomial. So let's think about this a little bit. So one way to say, well look, the height of this larger rectangle from here to here, we see that that distance is x, and then from here to here it's two. So the entire height right over here, the entire height right over here is going to be x plus two. So the height is x plus two, and what's the width?"}, {"video_title": "Multiplying binomials area model   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "So let's think about this a little bit. So one way to say, well look, the height of this larger rectangle from here to here, we see that that distance is x, and then from here to here it's two. So the entire height right over here, the entire height right over here is going to be x plus two. So the height is x plus two, and what's the width? Well, the width is, we go from there to there is x, and then from there to there is three. So the entire width is x plus three, x plus three. So just like that, I've expressed the area of the entire rectangle, and it's the product of two binomials."}, {"video_title": "Multiplying binomials area model   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "So the height is x plus two, and what's the width? Well, the width is, we go from there to there is x, and then from there to there is three. So the entire width is x plus three, x plus three. So just like that, I've expressed the area of the entire rectangle, and it's the product of two binomials. But now let's express it as a trinomial. Well, to do that, we can break down the larger area into the areas of each of these smaller rectangles. So what's the area of this purple rectangle right over here?"}, {"video_title": "Multiplying binomials area model   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "So just like that, I've expressed the area of the entire rectangle, and it's the product of two binomials. But now let's express it as a trinomial. Well, to do that, we can break down the larger area into the areas of each of these smaller rectangles. So what's the area of this purple rectangle right over here? Well, the purple rectangle, its height is x, and its width is x, so its area is x squared. Let me write that, that's x squared. What's the area of this yellow rectangle?"}, {"video_title": "Multiplying binomials area model   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "So what's the area of this purple rectangle right over here? Well, the purple rectangle, its height is x, and its width is x, so its area is x squared. Let me write that, that's x squared. What's the area of this yellow rectangle? Well, its height is x, same height as right over here. Its height is x, and its width is three. So it's gonna be x times three, or three x."}, {"video_title": "Multiplying binomials area model   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "What's the area of this yellow rectangle? Well, its height is x, same height as right over here. Its height is x, and its width is three. So it's gonna be x times three, or three x. It'll have an area of three x. So that area is three x. If we're summing up the area of the entire thing, this would be plus three x."}, {"video_title": "Multiplying binomials area model   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "So it's gonna be x times three, or three x. It'll have an area of three x. So that area is three x. If we're summing up the area of the entire thing, this would be plus three x. So this expression right over here, that's the area of this purple region plus the area of this yellow region. And then we can move on to this green region. What's the area going to be here?"}, {"video_title": "Multiplying binomials area model   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "If we're summing up the area of the entire thing, this would be plus three x. So this expression right over here, that's the area of this purple region plus the area of this yellow region. And then we can move on to this green region. What's the area going to be here? Well, the height is two, and the width is x. So multiplying height times width is gonna be two times x. And we can just add that, plus two times x."}, {"video_title": "Multiplying binomials area model   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "What's the area going to be here? Well, the height is two, and the width is x. So multiplying height times width is gonna be two times x. And we can just add that, plus two times x. And then finally, this little gray box here, its height is two, we see that right over there. Its height is two, and its width is three. We see it right over there."}, {"video_title": "Multiplying binomials area model   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "And we can just add that, plus two times x. And then finally, this little gray box here, its height is two, we see that right over there. Its height is two, and its width is three. We see it right over there. So it has an area of six, two times three. So plus six, and you might say, well, this isn't a trinomial. This has four terms right over here."}, {"video_title": "Multiplying binomials area model   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "We see it right over there. So it has an area of six, two times three. So plus six, and you might say, well, this isn't a trinomial. This has four terms right over here. But you might notice that we can add, that we can add these middle two terms. Three x plus two x. I have three x's, and I add two x's to that, I'm gonna have five x's. So this entire thing simplifies to x squared, x squared, plus five x, plus six, plus six."}, {"video_title": "Multiplying binomials area model   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "This has four terms right over here. But you might notice that we can add, that we can add these middle two terms. Three x plus two x. I have three x's, and I add two x's to that, I'm gonna have five x's. So this entire thing simplifies to x squared, x squared, plus five x, plus six, plus six. So this and this are two ways of expressing the area, so they're going to be equal. And that makes sense, because if you multiply it out, these binomials, and simplify it, you would get this trinomial. We could do that really fast."}, {"video_title": "Multiplying binomials area model   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "So this entire thing simplifies to x squared, x squared, plus five x, plus six, plus six. So this and this are two ways of expressing the area, so they're going to be equal. And that makes sense, because if you multiply it out, these binomials, and simplify it, you would get this trinomial. We could do that really fast. You multiply the x times the x. Actually, let me do it in the same colors. You multiply, you multiply the x times the x, you get the x squared."}, {"video_title": "Multiplying binomials area model   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "We could do that really fast. You multiply the x times the x. Actually, let me do it in the same colors. You multiply, you multiply the x times the x, you get the x squared. You multiply this x times the three, you get your three x. You multiply, you multiply the two times the x, you get your two x. And then you multiply the two times the three, and you get your six."}, {"video_title": "Fraction decimal and percent from visual model.mp3", "Sentence": "So let's assume that this entire square represents a whole. And we can see that part of it is shaded in in blue. What we're going to do in this video is try to represent the part that is shaded in blue as a fraction, as a decimal, decimal, decimal, and as a percent. So pause the video and see if you can do that. Well let's first think about it as a fraction. So the whole is split into one, two, three, four, five, six, seven, eight, nine, 10 equal sections. And six of them are filled in."}, {"video_title": "Fraction decimal and percent from visual model.mp3", "Sentence": "So pause the video and see if you can do that. Well let's first think about it as a fraction. So the whole is split into one, two, three, four, five, six, seven, eight, nine, 10 equal sections. And six of them are filled in. So the blue represents 6 tenths of a whole, or represents, you could just say 6 tenths. And you could also rewrite that if you divide the numerator and the denominator by two, that's the same thing as three over five. Fair enough."}, {"video_title": "Fraction decimal and percent from visual model.mp3", "Sentence": "And six of them are filled in. So the blue represents 6 tenths of a whole, or represents, you could just say 6 tenths. And you could also rewrite that if you divide the numerator and the denominator by two, that's the same thing as three over five. Fair enough. Now let's represent it as a decimal. What decimal would it be? Pause the video again and see if you can do that."}, {"video_title": "Fraction decimal and percent from visual model.mp3", "Sentence": "Fair enough. Now let's represent it as a decimal. What decimal would it be? Pause the video again and see if you can do that. Well 6 tenths, we could literally just go to our place value. So that's the ones place, we have a decimal. And then you have your tenths place."}, {"video_title": "Fraction decimal and percent from visual model.mp3", "Sentence": "Pause the video again and see if you can do that. Well 6 tenths, we could literally just go to our place value. So that's the ones place, we have a decimal. And then you have your tenths place. And so we have 6 tenths, so you could just put it right over there. We are putting a six in the tenths place to represent 6 tenths. Now what about a percentage?"}, {"video_title": "Fraction decimal and percent from visual model.mp3", "Sentence": "And then you have your tenths place. And so we have 6 tenths, so you could just put it right over there. We are putting a six in the tenths place to represent 6 tenths. Now what about a percentage? Well percent means per hundred. So one way to think about it is, six over 10 is the same thing as what per hundred? That is equal to, if we multiply the numerator and the denominator by 10, that's the same thing as 60 per hundred."}, {"video_title": "Fraction decimal and percent from visual model.mp3", "Sentence": "Now what about a percentage? Well percent means per hundred. So one way to think about it is, six over 10 is the same thing as what per hundred? That is equal to, if we multiply the numerator and the denominator by 10, that's the same thing as 60 per hundred. Per hundred. Or another way of thinking about it, 60 per, instead of hundred, you could say cent. And so this would be equal to, this would be equal to 60%."}, {"video_title": "Fraction decimal and percent from visual model.mp3", "Sentence": "That is equal to, if we multiply the numerator and the denominator by 10, that's the same thing as 60 per hundred. Per hundred. Or another way of thinking about it, 60 per, instead of hundred, you could say cent. And so this would be equal to, this would be equal to 60%. Let's do another example. So here once again, our entire square represents a whole. So see if you can represent this as the part that's shaded in blue as a fraction."}, {"video_title": "Fraction decimal and percent from visual model.mp3", "Sentence": "And so this would be equal to, this would be equal to 60%. Let's do another example. So here once again, our entire square represents a whole. So see if you can represent this as the part that's shaded in blue as a fraction. Pause the video and do that. Well you can see that this is a 10 by 10 grid. There's 100 equal sections here."}, {"video_title": "Fraction decimal and percent from visual model.mp3", "Sentence": "So see if you can represent this as the part that's shaded in blue as a fraction. Pause the video and do that. Well you can see that this is a 10 by 10 grid. There's 100 equal sections here. 100 equal sections. Each of these squares represents 1 hundredth. And how many of them are there?"}, {"video_title": "Fraction decimal and percent from visual model.mp3", "Sentence": "There's 100 equal sections here. 100 equal sections. Each of these squares represents 1 hundredth. And how many of them are there? Well let's see, this row is 10, 20, 30, 40, and then one, two, three, four. So this is 44 over 100. 44 hundredths."}, {"video_title": "Fraction decimal and percent from visual model.mp3", "Sentence": "And how many of them are there? Well let's see, this row is 10, 20, 30, 40, and then one, two, three, four. So this is 44 over 100. 44 hundredths. And we could actually represent this in other ways. We could divide the numerator and the denominator by four, in which case you would get 11 over 25. That's another way to represent this same fraction."}, {"video_title": "Fraction decimal and percent from visual model.mp3", "Sentence": "44 hundredths. And we could actually represent this in other ways. We could divide the numerator and the denominator by four, in which case you would get 11 over 25. That's another way to represent this same fraction. Now what about as a decimal? Well 44 hundredths, you could say, well you have your ones place, and then this is the same thing. You literally just say this is 44 hundredths."}, {"video_title": "Fraction decimal and percent from visual model.mp3", "Sentence": "That's another way to represent this same fraction. Now what about as a decimal? Well 44 hundredths, you could say, well you have your ones place, and then this is the same thing. You literally just say this is 44 hundredths. This is another way of representing 44 hundredths. It's 4 tenths and 4 hundredths is 44 hundredths. And then if you were to do a percent, well this is 44 per hundred, or 44 hundredths, but even here I like looking at it as 44 per hundred, or 44 per cent."}, {"video_title": "Absolute value of integers   Negative numbers and absolute value   Pre-Algebra   Khan Academy.mp3", "Sentence": "Find the absolute value of x when x is equal to 5, x is equal to negative 10, and x is equal to negative 12. So the absolute value, the way of writing it is almost more complicated than what it really is. The absolute value is really just the distance of x from 0. Distance from 0. So let me just draw a fast number line over here. So if I were to first, let's just put 0 right over here since we're thinking about the distance from 0. So let's just think about the absolute value of x when x is equal to 5."}, {"video_title": "Absolute value of integers   Negative numbers and absolute value   Pre-Algebra   Khan Academy.mp3", "Sentence": "Distance from 0. So let me just draw a fast number line over here. So if I were to first, let's just put 0 right over here since we're thinking about the distance from 0. So let's just think about the absolute value of x when x is equal to 5. So that's equivalent to the absolute value of 5. We just substitute 5 for x. The absolute value of 5 is the distance of 5 from 0."}, {"video_title": "Absolute value of integers   Negative numbers and absolute value   Pre-Algebra   Khan Academy.mp3", "Sentence": "So let's just think about the absolute value of x when x is equal to 5. So that's equivalent to the absolute value of 5. We just substitute 5 for x. The absolute value of 5 is the distance of 5 from 0. So you go 1, 2, 3, 4, 5. 5 is exactly 5 to the right of 0. So the absolute value of 5 is just 5."}, {"video_title": "Absolute value of integers   Negative numbers and absolute value   Pre-Algebra   Khan Academy.mp3", "Sentence": "The absolute value of 5 is the distance of 5 from 0. So you go 1, 2, 3, 4, 5. 5 is exactly 5 to the right of 0. So the absolute value of 5 is just 5. Now, I think you already get to see this is a pretty straightforward concept. Now let's do something a little more interesting. Absolute value of negative 10."}, {"video_title": "Absolute value of integers   Negative numbers and absolute value   Pre-Algebra   Khan Academy.mp3", "Sentence": "So the absolute value of 5 is just 5. Now, I think you already get to see this is a pretty straightforward concept. Now let's do something a little more interesting. Absolute value of negative 10. Or the absolute value of x when x is equal to negative 10. So let's just put negative 10 in for x. This is the distance that negative 10 is from 0."}, {"video_title": "Absolute value of integers   Negative numbers and absolute value   Pre-Algebra   Khan Academy.mp3", "Sentence": "Absolute value of negative 10. Or the absolute value of x when x is equal to negative 10. So let's just put negative 10 in for x. This is the distance that negative 10 is from 0. So let's just go negative 1, negative 2, negative 3, negative 4, negative 5, negative 6, negative 7, negative 8, negative 9, negative 10. I should extend the number line more. So this right here is negative 10."}, {"video_title": "Absolute value of integers   Negative numbers and absolute value   Pre-Algebra   Khan Academy.mp3", "Sentence": "This is the distance that negative 10 is from 0. So let's just go negative 1, negative 2, negative 3, negative 4, negative 5, negative 6, negative 7, negative 8, negative 9, negative 10. I should extend the number line more. So this right here is negative 10. So how far is it away from 0? Well, it's 10 to the left of 0. So you put a 10 here."}, {"video_title": "Absolute value of integers   Negative numbers and absolute value   Pre-Algebra   Khan Academy.mp3", "Sentence": "So this right here is negative 10. So how far is it away from 0? Well, it's 10 to the left of 0. So you put a 10 here. And so in general, absolute value will always be a positive quantity. And when we're thinking about just absolute values of just numbers, it's just going to be really the positive version of that number. Let's do one more."}, {"video_title": "Absolute value of integers   Negative numbers and absolute value   Pre-Algebra   Khan Academy.mp3", "Sentence": "So you put a 10 here. And so in general, absolute value will always be a positive quantity. And when we're thinking about just absolute values of just numbers, it's just going to be really the positive version of that number. Let's do one more. Well, they tell us to do one more. The absolute value of x when x is equal to negative 12. So we have the absolute value of negative 12."}, {"video_title": "Absolute value of integers   Negative numbers and absolute value   Pre-Algebra   Khan Academy.mp3", "Sentence": "Let's do one more. Well, they tell us to do one more. The absolute value of x when x is equal to negative 12. So we have the absolute value of negative 12. We don't even have to look at the number line. It's just going to be the positive version of negative 12. It's just going to be equal to 12."}, {"video_title": "Absolute value of integers   Negative numbers and absolute value   Pre-Algebra   Khan Academy.mp3", "Sentence": "So we have the absolute value of negative 12. We don't even have to look at the number line. It's just going to be the positive version of negative 12. It's just going to be equal to 12. And this is just saying that negative 12 is 12 away from 0. And we could draw it over here. This negative 11, negative 12 is right over here."}, {"video_title": "Adding decimals example 1   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "And I encourage you to pause the video and try to do it on your own. So I'm assuming you have tried to do it on your own. And now let's see how we could actually tackle this. Now, one thing I want to point out, some of you might see these numbers all lined up and immediately want to say, hey, 7 plus 1 is 8, and 8 plus 3 is 11, carry the 1, et cetera, et cetera. And if you did that, you would be making a mistake. Because you see right over here, these decimals aren't lined up. Here, if you did that, you would be adding the 7,000s to 100s."}, {"video_title": "Adding decimals example 1   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "Now, one thing I want to point out, some of you might see these numbers all lined up and immediately want to say, hey, 7 plus 1 is 8, and 8 plus 3 is 11, carry the 1, et cetera, et cetera. And if you did that, you would be making a mistake. Because you see right over here, these decimals aren't lined up. Here, if you did that, you would be adding the 7,000s to 100s. You would be adding 0 tenths to 5 ones. You would be adding 9 to 1 tenths, or essentially this is a 10 right over here. So the place is would be all mixed up."}, {"video_title": "Adding decimals example 1   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "Here, if you did that, you would be adding the 7,000s to 100s. You would be adding 0 tenths to 5 ones. You would be adding 9 to 1 tenths, or essentially this is a 10 right over here. So the place is would be all mixed up. So what you need to do is actually align the decimals so that your place values are aligned. So what you want to do is you want to align things up. So we could write 9.087."}, {"video_title": "Adding decimals example 1   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "So the place is would be all mixed up. So what you need to do is actually align the decimals so that your place values are aligned. So what you want to do is you want to align things up. So we could write 9.087. And then we want to align the decimals. So let's align the decimal. This is what has to match up."}, {"video_title": "Adding decimals example 1   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "So we could write 9.087. And then we want to align the decimals. So let's align the decimal. This is what has to match up. And this is going to be 15.31. And this should hopefully make sense to you as well. This is 9 point something plus 15 point something."}, {"video_title": "Adding decimals example 1   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "This is what has to match up. And this is going to be 15.31. And this should hopefully make sense to you as well. This is 9 point something plus 15 point something. So it's going to be, if you add 9 to 15, you get it will be 24 point something, give or take a little bit. So when you see that here, here you have a 9 plus the 15. So you've lined up the appropriate place values."}, {"video_title": "Adding decimals example 1   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "This is 9 point something plus 15 point something. So it's going to be, if you add 9 to 15, you get it will be 24 point something, give or take a little bit. So when you see that here, here you have a 9 plus the 15. So you've lined up the appropriate place values. And now we are ready to add. It's a good idea to start with the smallest place value. So if you have any extra in a certain place, you can bring something into the next place value."}, {"video_title": "Adding decimals example 1   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "So you've lined up the appropriate place values. And now we are ready to add. It's a good idea to start with the smallest place value. So if you have any extra in a certain place, you can bring something into the next place value. So here you say 7 plus, well, this is 7 thousandths. It's in the thousandths place. And you might want to, you say, well, what do I add it to?"}, {"video_title": "Adding decimals example 1   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "So if you have any extra in a certain place, you can bring something into the next place value. So here you say 7 plus, well, this is 7 thousandths. It's in the thousandths place. And you might want to, you say, well, what do I add it to? There's no thousandths right over here. And you're right. There are no thousandths."}, {"video_title": "Adding decimals example 1   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "And you might want to, you say, well, what do I add it to? There's no thousandths right over here. And you're right. There are no thousandths. So we could literally write 0 thousandths. So 7 thousandths plus 0 thousandths is 7 thousandths. 8 hundredths plus 1 hundredth is 9 hundredth."}, {"video_title": "Adding decimals example 1   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "There are no thousandths. So we could literally write 0 thousandths. So 7 thousandths plus 0 thousandths is 7 thousandths. 8 hundredths plus 1 hundredth is 9 hundredth. 0 plus 0 tenths plus 3 tenths is 3 tenths. We got our decimal. Then you have 9 ones plus 5 ones is 14 ones."}, {"video_title": "Adding decimals example 1   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "8 hundredths plus 1 hundredth is 9 hundredth. 0 plus 0 tenths plus 3 tenths is 3 tenths. We got our decimal. Then you have 9 ones plus 5 ones is 14 ones. Well, 14 ones is the same thing as 4 ones and 1 tenth. So we'll carry that 1 right over there. So this is just 1 tenth plus 4 ones is 14."}, {"video_title": "Comparing features of quadratic functions   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "So we're asked which function has the greater y-intercept. So the y-intercept is the y-coordinate when x is equal to zero. So f of zero, when x is equal to zero, the function is equal to, let's see, f of zero is going to be equal to zero minus zero plus four, is going to be equal to four. So this function right over here, it has a y-intercept of four. So it would intersect the y-axis right over there, while the function that we're comparing it to, g of x, we're looking at its graph, y is equal to g of x, its y-intercept is right over here at y is equal to three. So which function has a greater y-intercept? Well, it's going to be f of x. F of x has a greater y-intercept than g of x does."}, {"video_title": "Comparing features of quadratic functions   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "So this function right over here, it has a y-intercept of four. So it would intersect the y-axis right over there, while the function that we're comparing it to, g of x, we're looking at its graph, y is equal to g of x, its y-intercept is right over here at y is equal to three. So which function has a greater y-intercept? Well, it's going to be f of x. F of x has a greater y-intercept than g of x does. Let's do a few more of these where we're comparing different functions, one of them that's in a visual, has a visual depiction, and one of them where we're just given the equation. How many roots do the functions have in common? Well, g of x, we can see their roots."}, {"video_title": "Comparing features of quadratic functions   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "Well, it's going to be f of x. F of x has a greater y-intercept than g of x does. Let's do a few more of these where we're comparing different functions, one of them that's in a visual, has a visual depiction, and one of them where we're just given the equation. How many roots do the functions have in common? Well, g of x, we can see their roots. The roots are x equals negative one, and x is equal to two. So these two functions, at most, are going to have two roots in common, because this g of x only has two roots. There's a couple of ways we could tackle it."}, {"video_title": "Comparing features of quadratic functions   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "Well, g of x, we can see their roots. The roots are x equals negative one, and x is equal to two. So these two functions, at most, are going to have two roots in common, because this g of x only has two roots. There's a couple of ways we could tackle it. We could just try to find f's roots, or we could plug in either one of these values and see if it makes the function equal to zero. I'll do the first way. I'll try to factor this."}, {"video_title": "Comparing features of quadratic functions   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "There's a couple of ways we could tackle it. We could just try to find f's roots, or we could plug in either one of these values and see if it makes the function equal to zero. I'll do the first way. I'll try to factor this. Let's see, what two numbers, if I add them, I get one, because that's the coefficient here, or implicitly there. If I take their product, I get negative six. Well, they're going to have to have different signs since their product is negative."}, {"video_title": "Comparing features of quadratic functions   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "I'll try to factor this. Let's see, what two numbers, if I add them, I get one, because that's the coefficient here, or implicitly there. If I take their product, I get negative six. Well, they're going to have to have different signs since their product is negative. Let's see, negative three and positive two. No, actually, the other way around, because it's positive one. So positive three and negative two."}, {"video_title": "Comparing features of quadratic functions   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "Well, they're going to have to have different signs since their product is negative. Let's see, negative three and positive two. No, actually, the other way around, because it's positive one. So positive three and negative two. This is equal to x plus three times x minus two. So f of x is going to have zeros when x is equal to negative three, x is equal to negative three, or x is equal to two. These are the two zeros."}, {"video_title": "Comparing features of quadratic functions   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "So positive three and negative two. This is equal to x plus three times x minus two. So f of x is going to have zeros when x is equal to negative three, x is equal to negative three, or x is equal to two. These are the two zeros. If x is equal to negative three, this expression becomes zero. Zero times anything is zero. If x equals two, this expression becomes zero, and zero times anything is zero."}, {"video_title": "Comparing features of quadratic functions   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "These are the two zeros. If x is equal to negative three, this expression becomes zero. Zero times anything is zero. If x equals two, this expression becomes zero, and zero times anything is zero. So f of negative three is zero, and f of two is zero. These are the zeros of that function. So let's see, which of these are in common?"}, {"video_title": "Comparing features of quadratic functions   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "If x equals two, this expression becomes zero, and zero times anything is zero. So f of negative three is zero, and f of two is zero. These are the zeros of that function. So let's see, which of these are in common? Well, negative three is out here. That's not in common. x equals two is in common."}, {"video_title": "Comparing features of quadratic functions   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "So let's see, which of these are in common? Well, negative three is out here. That's not in common. x equals two is in common. So they only have one common zero right over there. So how many roots do the functions have in common? One."}, {"video_title": "Comparing features of quadratic functions   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "x equals two is in common. So they only have one common zero right over there. So how many roots do the functions have in common? One. All right, let's do one more of these. And they ask us, do the functions have the same concavity? And the way I think, or one way to think about concavity is whether it's opening upwards or opening downwards."}, {"video_title": "Comparing features of quadratic functions   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "One. All right, let's do one more of these. And they ask us, do the functions have the same concavity? And the way I think, or one way to think about concavity is whether it's opening upwards or opening downwards. So this is often viewed as concave upwards, and this is viewed as concave downwards. Concave, concave downwards. And the key realization is, well, you know, if you just look at this blue, if you look at g of x right over here, it is concave downwards."}, {"video_title": "Comparing features of quadratic functions   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "And the way I think, or one way to think about concavity is whether it's opening upwards or opening downwards. So this is often viewed as concave upwards, and this is viewed as concave downwards. Concave, concave downwards. And the key realization is, well, you know, if you just look at this blue, if you look at g of x right over here, it is concave downwards. So the question is, would this be concave downwards or upwards? And the key here is the coefficient on the second degree term, on the x squared term. If the coefficient is positive, you're going to be concave upwards, because as x gets suitably far away from zero, this term is going to overpower everything else, and it's going to become positive."}, {"video_title": "Comparing features of quadratic functions   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "And the key realization is, well, you know, if you just look at this blue, if you look at g of x right over here, it is concave downwards. So the question is, would this be concave downwards or upwards? And the key here is the coefficient on the second degree term, on the x squared term. If the coefficient is positive, you're going to be concave upwards, because as x gets suitably far away from zero, this term is going to overpower everything else, and it's going to become positive. So as x gets further and further away, or not even further away from zero, as x gets further and further away from the vertex, as x gets further and further away from the vertex, this term dominates everything else, and we get more and more positive values. And so that's why, if your coefficient is positive, you're going to have concave upwards, a concave upwards graph. And so if this is concave upwards, this one is clearly concave downwards."}, {"video_title": "Comparing features of quadratic functions   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "If the coefficient is positive, you're going to be concave upwards, because as x gets suitably far away from zero, this term is going to overpower everything else, and it's going to become positive. So as x gets further and further away, or not even further away from zero, as x gets further and further away from the vertex, as x gets further and further away from the vertex, this term dominates everything else, and we get more and more positive values. And so that's why, if your coefficient is positive, you're going to have concave upwards, a concave upwards graph. And so if this is concave upwards, this one is clearly concave downwards. They do not have the same concavity. So no. If this was negative four x squared minus 108, then it would be concave downwards, and we would say yes."}, {"video_title": "How to determine the domain of a modeling function   Functions   Algebra I   Khan Academy (2).mp3", "Sentence": "It says, Mason stands on the fifth step of a vertical ladder. The ladder has 15 steps, and the height difference between consecutive steps is 0.5 meters. He is thinking about moving up, down, or staying put. So let me draw this ladder that Mason is on. So it's a vertical ladder. So that's one side of the ladder. This is the other side of the ladder."}, {"video_title": "How to determine the domain of a modeling function   Functions   Algebra I   Khan Academy (2).mp3", "Sentence": "So let me draw this ladder that Mason is on. So it's a vertical ladder. So that's one side of the ladder. This is the other side of the ladder. And it has 15 steps. Let me see if I can draw that. So this is the first one, two, three, four, actually I'm gonna run out of space."}, {"video_title": "How to determine the domain of a modeling function   Functions   Algebra I   Khan Academy (2).mp3", "Sentence": "This is the other side of the ladder. And it has 15 steps. Let me see if I can draw that. So this is the first one, two, three, four, actually I'm gonna run out of space. I need to make them closer together. It's gonna be one, two, three, four, five, six, seven, eight, nine, 10, 11, 12, 13, 14, and 15, 15 steps. Let me make sure it's even."}, {"video_title": "How to determine the domain of a modeling function   Functions   Algebra I   Khan Academy (2).mp3", "Sentence": "So this is the first one, two, three, four, actually I'm gonna run out of space. I need to make them closer together. It's gonna be one, two, three, four, five, six, seven, eight, nine, 10, 11, 12, 13, 14, and 15, 15 steps. Let me make sure it's even. So the top and the bottom. And the distance between each of these, I guess you could say steps, or the rungs of the ladder, are half a meter. So this distance right over here is 0.5 meters."}, {"video_title": "How to determine the domain of a modeling function   Functions   Algebra I   Khan Academy (2).mp3", "Sentence": "Let me make sure it's even. So the top and the bottom. And the distance between each of these, I guess you could say steps, or the rungs of the ladder, are half a meter. So this distance right over here is 0.5 meters. And it says that he's on the fifth step of this vertical ladder. So he's on the fifth step. So one, two, three, four, five."}, {"video_title": "How to determine the domain of a modeling function   Functions   Algebra I   Khan Academy (2).mp3", "Sentence": "So this distance right over here is 0.5 meters. And it says that he's on the fifth step of this vertical ladder. So he's on the fifth step. So one, two, three, four, five. So this is where he is right now. He's on this fifth step. And he's thinking about moving up or down or staying put."}, {"video_title": "How to determine the domain of a modeling function   Functions   Algebra I   Khan Academy (2).mp3", "Sentence": "So one, two, three, four, five. So this is where he is right now. He's on this fifth step. And he's thinking about moving up or down or staying put. Let h of n denote the height above the ground, h, of Mason's feet, measured in meters, after moving n steps. If Mason went down the ladder, n is negative. Alright, h of n. So when he is, note the height above the ground after moving n steps."}, {"video_title": "How to determine the domain of a modeling function   Functions   Algebra I   Khan Academy (2).mp3", "Sentence": "And he's thinking about moving up or down or staying put. Let h of n denote the height above the ground, h, of Mason's feet, measured in meters, after moving n steps. If Mason went down the ladder, n is negative. Alright, h of n. So when he is, note the height above the ground after moving n steps. So let's just make sure we understand this. If I were to say h of zero, what is that going to be? Well, h of zero means that he's moved zero steps."}, {"video_title": "How to determine the domain of a modeling function   Functions   Algebra I   Khan Academy (2).mp3", "Sentence": "Alright, h of n. So when he is, note the height above the ground after moving n steps. So let's just make sure we understand this. If I were to say h of zero, what is that going to be? Well, h of zero means that he's moved zero steps. He's moved zero steps, he's still going to be on this fifth step of the ladder. And so how high is he going to be? So if he's on the fifth step of a vertical ladder, so he is going to be, so I'm assuming that there's 0.5 from the ground."}, {"video_title": "How to determine the domain of a modeling function   Functions   Algebra I   Khan Academy (2).mp3", "Sentence": "Well, h of zero means that he's moved zero steps. He's moved zero steps, he's still going to be on this fifth step of the ladder. And so how high is he going to be? So if he's on the fifth step of a vertical ladder, so he is going to be, so I'm assuming that there's 0.5 from the ground. So this is the ground right over here. So he is one, two, three, four, five steps. Each of them is half a meter."}, {"video_title": "How to determine the domain of a modeling function   Functions   Algebra I   Khan Academy (2).mp3", "Sentence": "So if he's on the fifth step of a vertical ladder, so he is going to be, so I'm assuming that there's 0.5 from the ground. So this is the ground right over here. So he is one, two, three, four, five steps. Each of them is half a meter. So five times 0.5 is going to give us 2.5 meters. So h of zero is 2.5 meters. If I said h of one, that means he goes up, h of one means he goes up one step."}, {"video_title": "How to determine the domain of a modeling function   Functions   Algebra I   Khan Academy (2).mp3", "Sentence": "Each of them is half a meter. So five times 0.5 is going to give us 2.5 meters. So h of zero is 2.5 meters. If I said h of one, that means he goes up, h of one means he goes up one step. So here, n would be equal to one. So if he goes up one step, h of one, he's going to be half a meter higher. So it's going to be equal to three meters."}, {"video_title": "How to determine the domain of a modeling function   Functions   Algebra I   Khan Academy (2).mp3", "Sentence": "If I said h of one, that means he goes up, h of one means he goes up one step. So here, n would be equal to one. So if he goes up one step, h of one, he's going to be half a meter higher. So it's going to be equal to three meters. So we could keep doing that for a bunch of different inputs. So let me write that. That's going to be equal to three meters."}, {"video_title": "How to determine the domain of a modeling function   Functions   Algebra I   Khan Academy (2).mp3", "Sentence": "So it's going to be equal to three meters. So we could keep doing that for a bunch of different inputs. So let me write that. That's going to be equal to three meters. But anyway, that's not what they're asking us about. They're saying which number type is more appropriate for the domain of the function? So the domain, just as a review, that's the set of numbers that we can input into the function and get a valid output."}, {"video_title": "How to determine the domain of a modeling function   Functions   Algebra I   Khan Academy (2).mp3", "Sentence": "That's going to be equal to three meters. But anyway, that's not what they're asking us about. They're saying which number type is more appropriate for the domain of the function? So the domain, just as a review, that's the set of numbers that we can input into the function and get a valid output. And it's clear here, let's see, we have to pick between integers or real numbers. Well, n, which is our input, that's the number of steps he goes up or down. So it could be positive or negative, but we're not going to talk about half steps."}, {"video_title": "How to determine the domain of a modeling function   Functions   Algebra I   Khan Academy (2).mp3", "Sentence": "So the domain, just as a review, that's the set of numbers that we can input into the function and get a valid output. And it's clear here, let's see, we have to pick between integers or real numbers. Well, n, which is our input, that's the number of steps he goes up or down. So it could be positive or negative, but we're not going to talk about half steps. Then he'll put his foot in the air right over here. He has to take integer-valued steps up or down. Well, I guess he's taking integer-valued steps."}, {"video_title": "How to determine the domain of a modeling function   Functions   Algebra I   Khan Academy (2).mp3", "Sentence": "So it could be positive or negative, but we're not going to talk about half steps. Then he'll put his foot in the air right over here. He has to take integer-valued steps up or down. Well, I guess he's taking integer-valued steps. If it's positive, it's up. If it's negative, it's down. If it's zero, that means he's staying put."}, {"video_title": "How to determine the domain of a modeling function   Functions   Algebra I   Khan Academy (2).mp3", "Sentence": "Well, I guess he's taking integer-valued steps. If it's positive, it's up. If it's negative, it's down. If it's zero, that means he's staying put. If n is zero, that means he's staying put. It's not real numbers. He can't take, he can't move pi steps from where he is."}, {"video_title": "How to determine the domain of a modeling function   Functions   Algebra I   Khan Academy (2).mp3", "Sentence": "If it's zero, that means he's staying put. If n is zero, that means he's staying put. It's not real numbers. He can't take, he can't move pi steps from where he is. He can't move square root of two steps from where he is. He can't even move 0.25 steps. Then he'd put his foot in the air."}, {"video_title": "How to determine the domain of a modeling function   Functions   Algebra I   Khan Academy (2).mp3", "Sentence": "He can't take, he can't move pi steps from where he is. He can't move square root of two steps from where he is. He can't even move 0.25 steps. Then he'd put his foot in the air. So this is definitely going to be about the integers, not the real numbers. This function right over here, the valid inputs, I want to be able to input an integer. In fact, it's not even all integers because he can't go down an arbitrary amount."}, {"video_title": "How to determine the domain of a modeling function   Functions   Algebra I   Khan Academy (2).mp3", "Sentence": "Then he'd put his foot in the air. So this is definitely going to be about the integers, not the real numbers. This function right over here, the valid inputs, I want to be able to input an integer. In fact, it's not even all integers because he can't go down an arbitrary amount. In fact, he can't go up an arbitrary amount either. But it's going to be, the domain is going to be a subset of integers. And then they say define the interval of the domain."}, {"video_title": "How to determine the domain of a modeling function   Functions   Algebra I   Khan Academy (2).mp3", "Sentence": "In fact, it's not even all integers because he can't go down an arbitrary amount. In fact, he can't go up an arbitrary amount either. But it's going to be, the domain is going to be a subset of integers. And then they say define the interval of the domain. We have these little toggles here to think about, to define the interval of the domain. Let's see. The lowest value for n, he can go as far as one, two, three, four, five steps down."}, {"video_title": "How to determine the domain of a modeling function   Functions   Algebra I   Khan Academy (2).mp3", "Sentence": "And then they say define the interval of the domain. We have these little toggles here to think about, to define the interval of the domain. Let's see. The lowest value for n, he can go as far as one, two, three, four, five steps down. In that case, n would be equal to negative five. Then the highest value for n is if he takes one, two, three, four, five, six, seven, eight, nine, 10 steps up. That would be n is equal to 10."}, {"video_title": "How to determine the domain of a modeling function   Functions   Algebra I   Khan Academy (2).mp3", "Sentence": "The lowest value for n, he can go as far as one, two, three, four, five steps down. In that case, n would be equal to negative five. Then the highest value for n is if he takes one, two, three, four, five, six, seven, eight, nine, 10 steps up. That would be n is equal to 10. The interval of the domain, and actually I'm just using, I just copied and pasted this onto my scratch pad. He can, n can be as low as negative five and as high as 10. And it can include them as well."}, {"video_title": "How to determine the domain of a modeling function   Functions   Algebra I   Khan Academy (2).mp3", "Sentence": "That would be n is equal to 10. The interval of the domain, and actually I'm just using, I just copied and pasted this onto my scratch pad. He can, n can be as low as negative five and as high as 10. And it can include them as well. I'm going to use brackets. My domain would include negative five. If it didn't include negative five, I'd put a parenthesis."}, {"video_title": "How to determine the domain of a modeling function   Functions   Algebra I   Khan Academy (2).mp3", "Sentence": "And it can include them as well. I'm going to use brackets. My domain would include negative five. If it didn't include negative five, I'd put a parenthesis. But I could put brackets here and I could put brackets there as well. And I can actually, let me, just for fun, let me actually input it into the actual exercise. I'm saying integers and I'm as low, n can, I can go down five steps and I can go up 10 steps."}, {"video_title": "Number of solutions to linear equations ex 3   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Whereas to use the drop down to form a linear equation with infinitely many solutions. So an equation with infinitely many solutions essentially has the same thing on both sides no matter what x you pick. So first my brain just wants to simplify this left hand side a little bit. And then think about how I can engineer the right hand side so it's going to be the same as the left no matter what x I pick. So right over here if I distribute the 4 over x minus 2 I get 4x minus 8 and then I'm adding x to that. And that's of course going to be equal to 5x plus blank and I get to pick what my blank is. And so 4x plus x is 5x and of course we still have our minus 8 and that's going to be equal to 5x plus blank."}, {"video_title": "Number of solutions to linear equations ex 3   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "And then think about how I can engineer the right hand side so it's going to be the same as the left no matter what x I pick. So right over here if I distribute the 4 over x minus 2 I get 4x minus 8 and then I'm adding x to that. And that's of course going to be equal to 5x plus blank and I get to pick what my blank is. And so 4x plus x is 5x and of course we still have our minus 8 and that's going to be equal to 5x plus blank. So what could we make that blank? So this is true for any x we pick. Well over here we have 5 times an x minus 8."}, {"video_title": "Number of solutions to linear equations ex 3   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "And so 4x plus x is 5x and of course we still have our minus 8 and that's going to be equal to 5x plus blank. So what could we make that blank? So this is true for any x we pick. Well over here we have 5 times an x minus 8. Well if we make this a minus 8 or if we subtract 8 here or if we make this a negative 8 this is going to be true for any x. So if we make this a negative 8 this is going to be true for any x you pick. You give me any x you multiply it by 5 and subtract 8."}, {"video_title": "Number of solutions to linear equations ex 3   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Well over here we have 5 times an x minus 8. Well if we make this a minus 8 or if we subtract 8 here or if we make this a negative 8 this is going to be true for any x. So if we make this a negative 8 this is going to be true for any x you pick. You give me any x you multiply it by 5 and subtract 8. That's of course going to be that same x multiplied by 5 and subtracting 8. And if you were to try to somehow solve this equation, subtract 5x from both sides, you would get negative 8 is equal to negative 8 which is absolutely true for absolutely any x that you pick. So let's go, let me actually fill this in on the exercise."}, {"video_title": "Constructing linear and exponential functions from data   Algebra II   Khan Academy.mp3", "Sentence": "For a linear function, f of x is equal to mx plus b. And an exponential function, g of x is equal to a times r to the x. Write the equation for each function. And so they give us, for each x value, what f of x is and what g of x is. And we need to figure out the equation for each function and type them in over here. So I copied and pasted this problem on my little scratch pad. So let's first think about the linear function."}, {"video_title": "Constructing linear and exponential functions from data   Algebra II   Khan Academy.mp3", "Sentence": "And so they give us, for each x value, what f of x is and what g of x is. And we need to figure out the equation for each function and type them in over here. So I copied and pasted this problem on my little scratch pad. So let's first think about the linear function. And to figure out the equation of a line or a linear function right over here, you really just need two points. And I always like to use the situation when x equals 0, because that makes it very clear what the y-intercept is going to be. So for example, we can say that f of 0 is going to be equal to m times 0 plus b."}, {"video_title": "Constructing linear and exponential functions from data   Algebra II   Khan Academy.mp3", "Sentence": "So let's first think about the linear function. And to figure out the equation of a line or a linear function right over here, you really just need two points. And I always like to use the situation when x equals 0, because that makes it very clear what the y-intercept is going to be. So for example, we can say that f of 0 is going to be equal to m times 0 plus b. Or this is just going to be equal to b. And they tell us that f of 0 is equal to 5. b is equal to 5. So we immediately know that this b right over here is equal to 5."}, {"video_title": "Constructing linear and exponential functions from data   Algebra II   Khan Academy.mp3", "Sentence": "So for example, we can say that f of 0 is going to be equal to m times 0 plus b. Or this is just going to be equal to b. And they tell us that f of 0 is equal to 5. b is equal to 5. So we immediately know that this b right over here is equal to 5. Now we just have to figure out the m. We have to figure out the slope of this line. So just as a little bit of a refresher on slope, the slope of this line is going to be our change in y or our change in our function, I guess we could say, if we say that this is y is equal to f of x, over our change in x. And actually, let me write it that way."}, {"video_title": "Constructing linear and exponential functions from data   Algebra II   Khan Academy.mp3", "Sentence": "So we immediately know that this b right over here is equal to 5. Now we just have to figure out the m. We have to figure out the slope of this line. So just as a little bit of a refresher on slope, the slope of this line is going to be our change in y or our change in our function, I guess we could say, if we say that this is y is equal to f of x, over our change in x. And actually, let me write it that way. We could write this as our change in our function over our change in x, if you want to look at it that way. So let's look at this first change in x. When x goes from 0 to 1."}, {"video_title": "Constructing linear and exponential functions from data   Algebra II   Khan Academy.mp3", "Sentence": "And actually, let me write it that way. We could write this as our change in our function over our change in x, if you want to look at it that way. So let's look at this first change in x. When x goes from 0 to 1. So we finish at 1. We started at 0. And f of x finishes at 7 and started at 5."}, {"video_title": "Constructing linear and exponential functions from data   Algebra II   Khan Academy.mp3", "Sentence": "When x goes from 0 to 1. So we finish at 1. We started at 0. And f of x finishes at 7 and started at 5. So when x is 1, f of x is 7. When x is 0, f of x is 5. And we get a change in our function of 2 when x changes by 1."}, {"video_title": "Constructing linear and exponential functions from data   Algebra II   Khan Academy.mp3", "Sentence": "And f of x finishes at 7 and started at 5. So when x is 1, f of x is 7. When x is 0, f of x is 5. And we get a change in our function of 2 when x changes by 1. So our m is equal to 2. And you see that. When x increases by 1, our function increases by 2."}, {"video_title": "Constructing linear and exponential functions from data   Algebra II   Khan Academy.mp3", "Sentence": "And we get a change in our function of 2 when x changes by 1. So our m is equal to 2. And you see that. When x increases by 1, our function increases by 2. So now we know the equation for f of x. f of x is going to be equal to 2 times 2x plus b, or 5. So we figured out what f of x is. Now we need to figure out what g of x is."}, {"video_title": "Constructing linear and exponential functions from data   Algebra II   Khan Academy.mp3", "Sentence": "When x increases by 1, our function increases by 2. So now we know the equation for f of x. f of x is going to be equal to 2 times 2x plus b, or 5. So we figured out what f of x is. Now we need to figure out what g of x is. So g of x is an exponential function. And there's really two things that we need to figure out. We need to figure out what a is."}, {"video_title": "Constructing linear and exponential functions from data   Algebra II   Khan Academy.mp3", "Sentence": "Now we need to figure out what g of x is. So g of x is an exponential function. And there's really two things that we need to figure out. We need to figure out what a is. And we need to figure out what r is. And let me just rewrite that. So we know that g of x, maybe I'll do it down here."}, {"video_title": "Constructing linear and exponential functions from data   Algebra II   Khan Academy.mp3", "Sentence": "We need to figure out what a is. And we need to figure out what r is. And let me just rewrite that. So we know that g of x, maybe I'll do it down here. g of x is equal to a times r to the x power. And if we know what g of 0 is, that's a pretty useful thing. Because r to the 0th power, regardless of what r is, where I guess we could assume that r is not equal to 0, that leads to people can debate what 0 to the 0 power is."}, {"video_title": "Constructing linear and exponential functions from data   Algebra II   Khan Academy.mp3", "Sentence": "So we know that g of x, maybe I'll do it down here. g of x is equal to a times r to the x power. And if we know what g of 0 is, that's a pretty useful thing. Because r to the 0th power, regardless of what r is, where I guess we could assume that r is not equal to 0, that leads to people can debate what 0 to the 0 power is. But if r is any non-zero number, we know that you raise that to the 0 power, you get 1. And so that essentially gives us a. So let's just write that down."}, {"video_title": "Constructing linear and exponential functions from data   Algebra II   Khan Academy.mp3", "Sentence": "Because r to the 0th power, regardless of what r is, where I guess we could assume that r is not equal to 0, that leads to people can debate what 0 to the 0 power is. But if r is any non-zero number, we know that you raise that to the 0 power, you get 1. And so that essentially gives us a. So let's just write that down. g of 0 is a times r to the 0 power, which is just going to be equal to a times 1, or a. And they tell us what g of 0 is. g of 0 is equal to 3."}, {"video_title": "Constructing linear and exponential functions from data   Algebra II   Khan Academy.mp3", "Sentence": "So let's just write that down. g of 0 is a times r to the 0 power, which is just going to be equal to a times 1, or a. And they tell us what g of 0 is. g of 0 is equal to 3. So we know that a is equal to 3. So so far, we know that our g of x can be written as 3 times r to the x power. So now we can just use any one of the other values they gave us to solve for r. For example, they tell us that g of 1 is equal to 2."}, {"video_title": "Constructing linear and exponential functions from data   Algebra II   Khan Academy.mp3", "Sentence": "g of 0 is equal to 3. So we know that a is equal to 3. So so far, we know that our g of x can be written as 3 times r to the x power. So now we can just use any one of the other values they gave us to solve for r. For example, they tell us that g of 1 is equal to 2. So let's write that down. g of 1, which would be 3 times r to the 1st power, or just 3. Let me just write."}, {"video_title": "Constructing linear and exponential functions from data   Algebra II   Khan Academy.mp3", "Sentence": "So now we can just use any one of the other values they gave us to solve for r. For example, they tell us that g of 1 is equal to 2. So let's write that down. g of 1, which would be 3 times r to the 1st power, or just 3. Let me just write. It could be 3 times r to the 1st power, or we could just write that as 3 times r. They tell us that g of 1 is equal to 2. So we get 3 times r is equal to 2. Or we get that r is equal to 2 thirds."}, {"video_title": "Constructing linear and exponential functions from data   Algebra II   Khan Academy.mp3", "Sentence": "Let me just write. It could be 3 times r to the 1st power, or we could just write that as 3 times r. They tell us that g of 1 is equal to 2. So we get 3 times r is equal to 2. Or we get that r is equal to 2 thirds. Divide both sides of this equation by 3. So r is 2 thirds, and we're done. g of x is equal to 3 times 2 thirds to the x power."}, {"video_title": "Constructing linear and exponential functions from data   Algebra II   Khan Academy.mp3", "Sentence": "Or we get that r is equal to 2 thirds. Divide both sides of this equation by 3. So r is 2 thirds, and we're done. g of x is equal to 3 times 2 thirds to the x power. We could write it that way if you want, any which way. So 3 times 2 thirds to the x power, and f of x is 2x plus 5. So let's actually just type that in."}, {"video_title": "Constructing linear and exponential functions from data   Algebra II   Khan Academy.mp3", "Sentence": "g of x is equal to 3 times 2 thirds to the x power. We could write it that way if you want, any which way. So 3 times 2 thirds to the x power, and f of x is 2x plus 5. So let's actually just type that in. So f of x is 2x plus 5. f of x is 2x plus 5. And we can verify that that's the expression that we want. And g of x is 3 times 2 over 3 to the x power."}, {"video_title": "Constructing linear and exponential functions from data   Algebra II   Khan Academy.mp3", "Sentence": "So let's actually just type that in. So f of x is 2x plus 5. f of x is 2x plus 5. And we can verify that that's the expression that we want. And g of x is 3 times 2 over 3 to the x power. Let me just verify that that's what I did there. I have a short memory. All right, yeah, that looks right."}, {"video_title": "Solve by completing the square Integer solutions   Algebra I   Khan Academy.mp3", "Sentence": "So we're given this equation here. What I want you to do is pause this video and see if you can solve it. What x values satisfy the equation? All right, now let's work through this together. So one technique could be just, let's just try to complete the square here on the left-hand side. So to do that, let me write it this way, x squared minus eight x, and then I have, I'll write the plus one out here, is equal to 85. Now, if I wanna complete the square, I just have to think, what can I add to both sides of this equation that could make this part of the left-hand expression a perfect square?"}, {"video_title": "Solve by completing the square Integer solutions   Algebra I   Khan Academy.mp3", "Sentence": "All right, now let's work through this together. So one technique could be just, let's just try to complete the square here on the left-hand side. So to do that, let me write it this way, x squared minus eight x, and then I have, I'll write the plus one out here, is equal to 85. Now, if I wanna complete the square, I just have to think, what can I add to both sides of this equation that could make this part of the left-hand expression a perfect square? Well, if I look at this negative eight coefficient on the first degree term, I could say, okay, let me take half of negative eight, that would be negative four, and then negative four squared is going to be positive 16. So I'm gonna add a positive 16 on the left-hand side. And if I want, I could then subtract a 16 from the left-hand side, or I could add a 16 on the right-hand side."}, {"video_title": "Solve by completing the square Integer solutions   Algebra I   Khan Academy.mp3", "Sentence": "Now, if I wanna complete the square, I just have to think, what can I add to both sides of this equation that could make this part of the left-hand expression a perfect square? Well, if I look at this negative eight coefficient on the first degree term, I could say, okay, let me take half of negative eight, that would be negative four, and then negative four squared is going to be positive 16. So I'm gonna add a positive 16 on the left-hand side. And if I want, I could then subtract a 16 from the left-hand side, or I could add a 16 on the right-hand side. Notice, I've just done the same thing to both sides of this equation. And why was that useful? Well, now what I've just put in parentheses is a perfect square."}, {"video_title": "Solve by completing the square Integer solutions   Algebra I   Khan Academy.mp3", "Sentence": "And if I want, I could then subtract a 16 from the left-hand side, or I could add a 16 on the right-hand side. Notice, I've just done the same thing to both sides of this equation. And why was that useful? Well, now what I've just put in parentheses is a perfect square. This is the same thing as x minus four squared. It was by design. We looked at that negative eight, half of that is negative four, you square it, you get 16, and you can verify x minus four times x minus four is indeed equal to this."}, {"video_title": "Solve by completing the square Integer solutions   Algebra I   Khan Academy.mp3", "Sentence": "Well, now what I've just put in parentheses is a perfect square. This is the same thing as x minus four squared. It was by design. We looked at that negative eight, half of that is negative four, you square it, you get 16, and you can verify x minus four times x minus four is indeed equal to this. And then we have plus one is going to be equal to, what's 85 plus 16? That is 101. And now we wanna get rid of this one on the left-hand side, and the easiest way we can do that is subtract one from both sides."}, {"video_title": "Solve by completing the square Integer solutions   Algebra I   Khan Academy.mp3", "Sentence": "We looked at that negative eight, half of that is negative four, you square it, you get 16, and you can verify x minus four times x minus four is indeed equal to this. And then we have plus one is going to be equal to, what's 85 plus 16? That is 101. And now we wanna get rid of this one on the left-hand side, and the easiest way we can do that is subtract one from both sides. That way we'll just isolate that x minus four squared. And we are left with x minus four squared, four squared, these cancel out, is going to be equal to 100. Now if something squared is equal to 100, that means that the something is equal to the positive or the negative square root of 100, or that that something, x minus four, is equal to positive or negative 10, positive or negative 10."}, {"video_title": "Solve by completing the square Integer solutions   Algebra I   Khan Academy.mp3", "Sentence": "And now we wanna get rid of this one on the left-hand side, and the easiest way we can do that is subtract one from both sides. That way we'll just isolate that x minus four squared. And we are left with x minus four squared, four squared, these cancel out, is going to be equal to 100. Now if something squared is equal to 100, that means that the something is equal to the positive or the negative square root of 100, or that that something, x minus four, is equal to positive or negative 10, positive or negative 10. All I did is took the plus or minus square root of 100. And this makes sense. If I took positive 10 squared, I'll get 100."}, {"video_title": "Solve by completing the square Integer solutions   Algebra I   Khan Academy.mp3", "Sentence": "Now if something squared is equal to 100, that means that the something is equal to the positive or the negative square root of 100, or that that something, x minus four, is equal to positive or negative 10, positive or negative 10. All I did is took the plus or minus square root of 100. And this makes sense. If I took positive 10 squared, I'll get 100. If I take negative 10 squared, I get 100. So x minus four could be either one of those. And now I just add four to both sides of this equation."}, {"video_title": "Solve by completing the square Integer solutions   Algebra I   Khan Academy.mp3", "Sentence": "If I took positive 10 squared, I'll get 100. If I take negative 10 squared, I get 100. So x minus four could be either one of those. And now I just add four to both sides of this equation. And then what do I get? I get x is equal to four plus or minus 10. Or another way of thinking about it, I could write it as x is equal to, four plus 10 is 14, and then four minus 10 is equal to negative six."}, {"video_title": "Solve by completing the square Integer solutions   Algebra I   Khan Academy.mp3", "Sentence": "And now I just add four to both sides of this equation. And then what do I get? I get x is equal to four plus or minus 10. Or another way of thinking about it, I could write it as x is equal to, four plus 10 is 14, and then four minus 10 is equal to negative six. So these are two ways to solve it. But there's other ways to solve this equation. We could, right from the get-go, try to subtract 85 from both sides."}, {"video_title": "Solve by completing the square Integer solutions   Algebra I   Khan Academy.mp3", "Sentence": "Or another way of thinking about it, I could write it as x is equal to, four plus 10 is 14, and then four minus 10 is equal to negative six. So these are two ways to solve it. But there's other ways to solve this equation. We could, right from the get-go, try to subtract 85 from both sides. Some people feel more comfortable solving quadratics if they have the quadratic expression be equal to zero. And if you did that, you would get x squared minus eight x minus 84 is equal to zero. All I did is I subtracted 85 from both sides of this equation to get this right over here."}, {"video_title": "Solve by completing the square Integer solutions   Algebra I   Khan Academy.mp3", "Sentence": "We could, right from the get-go, try to subtract 85 from both sides. Some people feel more comfortable solving quadratics if they have the quadratic expression be equal to zero. And if you did that, you would get x squared minus eight x minus 84 is equal to zero. All I did is I subtracted 85 from both sides of this equation to get this right over here. Now this one, we can approach in two different ways. We can complete the square again, or we could just try to factor. If we complete the square, we're going to see something very similar to this."}, {"video_title": "Solve by completing the square Integer solutions   Algebra I   Khan Academy.mp3", "Sentence": "All I did is I subtracted 85 from both sides of this equation to get this right over here. Now this one, we can approach in two different ways. We can complete the square again, or we could just try to factor. If we complete the square, we're going to see something very similar to this. Actually, let me just do that really fast. If I look at this part right over there, I could say x squared minus eight x. And then once again, half of negative eight is negative four."}, {"video_title": "Solve by completing the square Integer solutions   Algebra I   Khan Academy.mp3", "Sentence": "If we complete the square, we're going to see something very similar to this. Actually, let me just do that really fast. If I look at this part right over there, I could say x squared minus eight x. And then once again, half of negative eight is negative four. That squared is plus 16. And then I'd have minus 84. So let me do that in that blue color so we can keep track."}, {"video_title": "Solve by completing the square Integer solutions   Algebra I   Khan Academy.mp3", "Sentence": "And then once again, half of negative eight is negative four. That squared is plus 16. And then I'd have minus 84. So let me do that in that blue color so we can keep track. Minus 84. And then if I added 16 on the left-hand side, I could either add that to the right-hand side so both sides have 16 added to it, or if I want to maintain the equality, I could just subtract 16 from the left-hand side. So I've added 16, subtracted 16."}, {"video_title": "Solve by completing the square Integer solutions   Algebra I   Khan Academy.mp3", "Sentence": "So let me do that in that blue color so we can keep track. Minus 84. And then if I added 16 on the left-hand side, I could either add that to the right-hand side so both sides have 16 added to it, or if I want to maintain the equality, I could just subtract 16 from the left-hand side. So I've added 16, subtracted 16. I haven't changed the left-hand side's value. And then that would be equal to zero. This part right over here, this is x minus four squared."}, {"video_title": "Solve by completing the square Integer solutions   Algebra I   Khan Academy.mp3", "Sentence": "So I've added 16, subtracted 16. I haven't changed the left-hand side's value. And then that would be equal to zero. This part right over here, this is x minus four squared. This part right over here is minus 100 is equal to zero. And then you add 100 to both sides of this and you get exactly this step right over here. Now, another way that we could have approached it without completing the square, we could have said x squared minus eight x minus 84 is equal to zero."}, {"video_title": "Solve by completing the square Integer solutions   Algebra I   Khan Academy.mp3", "Sentence": "This part right over here, this is x minus four squared. This part right over here is minus 100 is equal to zero. And then you add 100 to both sides of this and you get exactly this step right over here. Now, another way that we could have approached it without completing the square, we could have said x squared minus eight x minus 84 is equal to zero. And think about what two numbers, if I multiply them, I get negative 84. So they'd have to have different signs since when I take their product, I get a negative number. And when I add them together, I get negative eight."}, {"video_title": "Solve by completing the square Integer solutions   Algebra I   Khan Academy.mp3", "Sentence": "Now, another way that we could have approached it without completing the square, we could have said x squared minus eight x minus 84 is equal to zero. And think about what two numbers, if I multiply them, I get negative 84. So they'd have to have different signs since when I take their product, I get a negative number. And when I add them together, I get negative eight. And there we could just look at the factorization of negative 84, of 84 generally. It could be two times, let's just think about 84. 84 could be two times 42."}, {"video_title": "Solve by completing the square Integer solutions   Algebra I   Khan Academy.mp3", "Sentence": "And when I add them together, I get negative eight. And there we could just look at the factorization of negative 84, of 84 generally. It could be two times, let's just think about 84. 84 could be two times 42. And obviously one of them would have to be negative, one of them would have to be positive in order to get to negative 84. But the difference between these two numbers, if one was positive and one is negative, is a lot more than eight. So that doesn't work."}, {"video_title": "Solve by completing the square Integer solutions   Algebra I   Khan Academy.mp3", "Sentence": "84 could be two times 42. And obviously one of them would have to be negative, one of them would have to be positive in order to get to negative 84. But the difference between these two numbers, if one was positive and one is negative, is a lot more than eight. So that doesn't work. So let's try, let's see, I'll do a few in my head. Three times 28, but still that difference is way more than eight. Four times, four times, let's see, four times 21, now that difference between four and 21 is still larger than eight."}, {"video_title": "Solve by completing the square Integer solutions   Algebra I   Khan Academy.mp3", "Sentence": "So that doesn't work. So let's try, let's see, I'll do a few in my head. Three times 28, but still that difference is way more than eight. Four times, four times, let's see, four times 21, now that difference between four and 21 is still larger than eight. Let's see, five doesn't go to it. Six times 14, that's interesting now. Okay, so let's think about this."}, {"video_title": "Solve by completing the square Integer solutions   Algebra I   Khan Academy.mp3", "Sentence": "Four times, four times, let's see, four times 21, now that difference between four and 21 is still larger than eight. Let's see, five doesn't go to it. Six times 14, that's interesting now. Okay, so let's think about this. So six times 14 is equal to 84. One of them has to be negative. And since when we take the sum of the two numbers, we get a negative number, that means the larger one is negative."}, {"video_title": "Solve by completing the square Integer solutions   Algebra I   Khan Academy.mp3", "Sentence": "Okay, so let's think about this. So six times 14 is equal to 84. One of them has to be negative. And since when we take the sum of the two numbers, we get a negative number, that means the larger one is negative. So let's see, six times negative 14 is negative 84. Six plus negative 14 is indeed equal to negative eight. So we can factor this as x plus six times x minus 14 is equal to zero."}, {"video_title": "Solve by completing the square Integer solutions   Algebra I   Khan Academy.mp3", "Sentence": "And since when we take the sum of the two numbers, we get a negative number, that means the larger one is negative. So let's see, six times negative 14 is negative 84. Six plus negative 14 is indeed equal to negative eight. So we can factor this as x plus six times x minus 14 is equal to zero. And so the product of two things is equal to zero. That means if either of them is equal to zero, that would make the entire expression equal to zero. So we could say x plus six is equal to zero or x minus 14 is equal to zero."}, {"video_title": "Subtracting decimals example 2   Arithmetic operations   6th grade   Khan Academy.mp3", "Sentence": "So let's try to calculate 39.1 minus 0.794. And so pause the video and try this on your own. All right, I'm assuming you've given a go at it. So now let's work through it together. So I'm going to rewrite this as 39.1 minus, I'm going to line up the decimals so that I have the right place values below the right place values, minus, this 0 is in the ones place, so I'll put it in the ones place, 0.794. And now we're ready to subtract. Now, how do we subtract 4 from nothingness here and 9 from nothingness here?"}, {"video_title": "Subtracting decimals example 2   Arithmetic operations   6th grade   Khan Academy.mp3", "Sentence": "So now let's work through it together. So I'm going to rewrite this as 39.1 minus, I'm going to line up the decimals so that I have the right place values below the right place values, minus, this 0 is in the ones place, so I'll put it in the ones place, 0.794. And now we're ready to subtract. Now, how do we subtract 4 from nothingness here and 9 from nothingness here? Well, the same thing as nothing is a 0. And so now we can start to think about how to subtract. Well, we still have the problem."}, {"video_title": "Subtracting decimals example 2   Arithmetic operations   6th grade   Khan Academy.mp3", "Sentence": "Now, how do we subtract 4 from nothingness here and 9 from nothingness here? Well, the same thing as nothing is a 0. And so now we can start to think about how to subtract. Well, we still have the problem. Well, we're trying to subtract 4 from 0, so we're trying to subtract 9 from 0. So what we could do is take this 1 tenth and try to regroup it into the hundredths place and the thousandths place. So let's think about this."}, {"video_title": "Subtracting decimals example 2   Arithmetic operations   6th grade   Khan Academy.mp3", "Sentence": "Well, we still have the problem. Well, we're trying to subtract 4 from 0, so we're trying to subtract 9 from 0. So what we could do is take this 1 tenth and try to regroup it into the hundredths place and the thousandths place. So let's think about this. If we make this, if we make, actually, that's not actually going to solve our problem. Well, we could do it, but then we're going to have 0 tenths and we're still going to have a problem here. So actually, let me go to the ones place."}, {"video_title": "Subtracting decimals example 2   Arithmetic operations   6th grade   Khan Academy.mp3", "Sentence": "So let's think about this. If we make this, if we make, actually, that's not actually going to solve our problem. Well, we could do it, but then we're going to have 0 tenths and we're still going to have a problem here. So actually, let me go to the ones place. So let me get rid of a ones. So that's 8 ones, which is going to be 10 tenths. So that's going to now, we're going to have 11 tenths."}, {"video_title": "Subtracting decimals example 2   Arithmetic operations   6th grade   Khan Academy.mp3", "Sentence": "So actually, let me go to the ones place. So let me get rid of a ones. So that's 8 ones, which is going to be 10 tenths. So that's going to now, we're going to have 11 tenths. 10 tenths from here plus 1 is 11 tenths. Now let's take one of those tenths so that we have 10 tenths and give it to the hundredths. So that's going to be 10 hundredths."}, {"video_title": "Subtracting decimals example 2   Arithmetic operations   6th grade   Khan Academy.mp3", "Sentence": "So that's going to now, we're going to have 11 tenths. 10 tenths from here plus 1 is 11 tenths. Now let's take one of those tenths so that we have 10 tenths and give it to the hundredths. So that's going to be 10 hundredths. And now let's take one of those hundredths, so now we have 9 hundredths, and give it to the thousandths. So that's going to be 10 thousandths. Now we're ready to subtract."}, {"video_title": "Subtracting decimals example 2   Arithmetic operations   6th grade   Khan Academy.mp3", "Sentence": "So that's going to be 10 hundredths. And now let's take one of those hundredths, so now we have 9 hundredths, and give it to the thousandths. So that's going to be 10 thousandths. Now we're ready to subtract. So 10, let me do this in yellow. 10 minus 4 is 6. 9 minus 9 is 0."}, {"video_title": "Subtracting decimals example 2   Arithmetic operations   6th grade   Khan Academy.mp3", "Sentence": "Now we're ready to subtract. So 10, let me do this in yellow. 10 minus 4 is 6. 9 minus 9 is 0. 10 minus 7 is 3. We have our decimal point. 8 minus 0 is 8."}, {"video_title": "Raising a number to the 0 and 1st power   Pre-Algebra   Khan Academy.mp3", "Sentence": "What I want to do in this video is think about exponents in a slightly different way that will be useful for different contexts, and also go through a lot more examples. So in the last video, we saw that taking something to an exponent means multiplying that number that many times. So if I had the number negative 2, and I want to raise it to the third power, this literally means taking three negative 2's. So negative 2, negative 2, and negative 2, and then multiplying them. And then multiplying them. So what's this going to be? Well, let's see."}, {"video_title": "Raising a number to the 0 and 1st power   Pre-Algebra   Khan Academy.mp3", "Sentence": "So negative 2, negative 2, and negative 2, and then multiplying them. And then multiplying them. So what's this going to be? Well, let's see. Negative 2 times negative 2 is positive 4. And then positive 4 times negative 2 is negative 8. So this would be equal to negative 8."}, {"video_title": "Raising a number to the 0 and 1st power   Pre-Algebra   Khan Academy.mp3", "Sentence": "Well, let's see. Negative 2 times negative 2 is positive 4. And then positive 4 times negative 2 is negative 8. So this would be equal to negative 8. Now, another way of thinking about exponents, instead of saying you're just taking three negative 2's and multiplying, and this is a completely reasonable way of viewing it, you could also view it as this is the number of times you're going to multiply this number times 1. So you could completely view this as being equal to. So you're going to start with a 1, and you're going to multiply 1 times negative 2 three times."}, {"video_title": "Raising a number to the 0 and 1st power   Pre-Algebra   Khan Academy.mp3", "Sentence": "So this would be equal to negative 8. Now, another way of thinking about exponents, instead of saying you're just taking three negative 2's and multiplying, and this is a completely reasonable way of viewing it, you could also view it as this is the number of times you're going to multiply this number times 1. So you could completely view this as being equal to. So you're going to start with a 1, and you're going to multiply 1 times negative 2 three times. So this is times negative 2, times negative 2, times times negative 2. So clearly, these are the same number. Here, we just took this, and we're just multiplying it by 1."}, {"video_title": "Raising a number to the 0 and 1st power   Pre-Algebra   Khan Academy.mp3", "Sentence": "So you're going to start with a 1, and you're going to multiply 1 times negative 2 three times. So this is times negative 2, times negative 2, times times negative 2. So clearly, these are the same number. Here, we just took this, and we're just multiplying it by 1. So you're still going to get negative 8. And this might be a slightly more useful idea to get an intuition for exponents, especially when you start taking things to the 1 or 0 power. So let's think about that a little bit."}, {"video_title": "Raising a number to the 0 and 1st power   Pre-Algebra   Khan Academy.mp3", "Sentence": "Here, we just took this, and we're just multiplying it by 1. So you're still going to get negative 8. And this might be a slightly more useful idea to get an intuition for exponents, especially when you start taking things to the 1 or 0 power. So let's think about that a little bit. What is positive 2 based on this definition to the 0 power going to be equal to? Well, we just said, this says how many times are you going to multiply 1 times this number? So this literally says, I'm going to take a 1, and I'm going to multiply it by 2, 0 times."}, {"video_title": "Raising a number to the 0 and 1st power   Pre-Algebra   Khan Academy.mp3", "Sentence": "So let's think about that a little bit. What is positive 2 based on this definition to the 0 power going to be equal to? Well, we just said, this says how many times are you going to multiply 1 times this number? So this literally says, I'm going to take a 1, and I'm going to multiply it by 2, 0 times. Well, if I'm going to multiply it by 2, 0 times, that means I'm just left with the 1. So 2 to the 0 power is going to be equal to 1. And actually, any non-zero number to the 0 power is 1 by that same rationale."}, {"video_title": "Raising a number to the 0 and 1st power   Pre-Algebra   Khan Academy.mp3", "Sentence": "So this literally says, I'm going to take a 1, and I'm going to multiply it by 2, 0 times. Well, if I'm going to multiply it by 2, 0 times, that means I'm just left with the 1. So 2 to the 0 power is going to be equal to 1. And actually, any non-zero number to the 0 power is 1 by that same rationale. And I'll make another video that will also give a little bit more intuition on there. That might seem very counterintuitive, but it's based on one way of thinking about it is thinking of an exponent as this. And this will also make sense if we start thinking of what 2 to the first power is."}, {"video_title": "Raising a number to the 0 and 1st power   Pre-Algebra   Khan Academy.mp3", "Sentence": "And actually, any non-zero number to the 0 power is 1 by that same rationale. And I'll make another video that will also give a little bit more intuition on there. That might seem very counterintuitive, but it's based on one way of thinking about it is thinking of an exponent as this. And this will also make sense if we start thinking of what 2 to the first power is. So 2 to the first power. So let's go to this definition we just gave of the exponent. We always start with a 1, and we multiply it by the 2 one time."}, {"video_title": "Raising a number to the 0 and 1st power   Pre-Algebra   Khan Academy.mp3", "Sentence": "And this will also make sense if we start thinking of what 2 to the first power is. So 2 to the first power. So let's go to this definition we just gave of the exponent. We always start with a 1, and we multiply it by the 2 one time. So 2 is going to be 1. We're only going to multiply it by the 2. I'll use this for multiplication."}, {"video_title": "Raising a number to the 0 and 1st power   Pre-Algebra   Khan Academy.mp3", "Sentence": "We always start with a 1, and we multiply it by the 2 one time. So 2 is going to be 1. We're only going to multiply it by the 2. I'll use this for multiplication. I'll use the dot. We're only going to multiply it by 2 one time. So 1 times 2, well, that's clearly just going to be equal to 2."}, {"video_title": "Raising a number to the 0 and 1st power   Pre-Algebra   Khan Academy.mp3", "Sentence": "I'll use this for multiplication. I'll use the dot. We're only going to multiply it by 2 one time. So 1 times 2, well, that's clearly just going to be equal to 2. And any number to the first power is just going to be equal to that number. And then we can go from there, and you'll, of course, see the pattern. If we say what 2 squared is, well, based on this definition, we start with a 1, and we multiply it by 2 two times."}, {"video_title": "Raising a number to the 0 and 1st power   Pre-Algebra   Khan Academy.mp3", "Sentence": "So 1 times 2, well, that's clearly just going to be equal to 2. And any number to the first power is just going to be equal to that number. And then we can go from there, and you'll, of course, see the pattern. If we say what 2 squared is, well, based on this definition, we start with a 1, and we multiply it by 2 two times. So times 2 is going to be equal to 4. And we've seen this before. You go to 2 to the third power."}, {"video_title": "Raising a number to the 0 and 1st power   Pre-Algebra   Khan Academy.mp3", "Sentence": "If we say what 2 squared is, well, based on this definition, we start with a 1, and we multiply it by 2 two times. So times 2 is going to be equal to 4. And we've seen this before. You go to 2 to the third power. You start with a 1. And then multiply it by 2 three times. So times 2, times 2, times 2."}, {"video_title": "Raising a number to the 0 and 1st power   Pre-Algebra   Khan Academy.mp3", "Sentence": "You go to 2 to the third power. You start with a 1. And then multiply it by 2 three times. So times 2, times 2, times 2. This is going to give us positive 8. And you probably see a pattern here. Every time we multiply by 2, or every time, I should say, we raise 2 to one more power, we are multiplying by 2."}, {"video_title": "Raising a number to the 0 and 1st power   Pre-Algebra   Khan Academy.mp3", "Sentence": "So times 2, times 2, times 2. This is going to give us positive 8. And you probably see a pattern here. Every time we multiply by 2, or every time, I should say, we raise 2 to one more power, we are multiplying by 2. Notice this. To go from 2 to the 0 to 2 to the 1, we multiplied by 2. I'll use a little x for the multiplication symbol now."}, {"video_title": "Raising a number to the 0 and 1st power   Pre-Algebra   Khan Academy.mp3", "Sentence": "Every time we multiply by 2, or every time, I should say, we raise 2 to one more power, we are multiplying by 2. Notice this. To go from 2 to the 0 to 2 to the 1, we multiplied by 2. I'll use a little x for the multiplication symbol now. A little cross. And then to go from 2 to the first power to 2 to the second power, we multiply by 2 again. And that makes complete sense, because this is literally telling us how many times are we going to take this number, and how many times are we going to take 1 and multiply it by this number."}, {"video_title": "Raising a number to the 0 and 1st power   Pre-Algebra   Khan Academy.mp3", "Sentence": "I'll use a little x for the multiplication symbol now. A little cross. And then to go from 2 to the first power to 2 to the second power, we multiply by 2 again. And that makes complete sense, because this is literally telling us how many times are we going to take this number, and how many times are we going to take 1 and multiply it by this number. And so when you go from 2 to the second power to 2 to the third, you're multiplying by 2 one more time. And this is another intuition of why something to the 0 power is equal to 1. If you were to go backwards, if say we didn't know what 2 to the 0 power is, and we were just trying to figure out what would make sense."}, {"video_title": "Raising a number to the 0 and 1st power   Pre-Algebra   Khan Academy.mp3", "Sentence": "And that makes complete sense, because this is literally telling us how many times are we going to take this number, and how many times are we going to take 1 and multiply it by this number. And so when you go from 2 to the second power to 2 to the third, you're multiplying by 2 one more time. And this is another intuition of why something to the 0 power is equal to 1. If you were to go backwards, if say we didn't know what 2 to the 0 power is, and we were just trying to figure out what would make sense. Well, when we go from 2 to the third power to 2 to the second, we'd be dividing by 2. We're going from 8 to 4. Then we divide by 2 again to go from 2 to the second to 2 to the first."}, {"video_title": "Comparing exponent expressions.mp3", "Sentence": "So we are asked to order the expressions from least to greatest. And this is from the exercises on Khan Academy. And if we're doing it on Khan Academy, we would drag these little tiles around from least to greatest. Least on the left, greatest on the right. I can't drag it around, because this is just a picture. So I'm gonna evaluate each of these, and then I'm gonna rewrite them from least to greatest. So let's start with two to the third minus two to the first."}, {"video_title": "Comparing exponent expressions.mp3", "Sentence": "Least on the left, greatest on the right. I can't drag it around, because this is just a picture. So I'm gonna evaluate each of these, and then I'm gonna rewrite them from least to greatest. So let's start with two to the third minus two to the first. What is that going to be? Two to the third minus two to the first. And if you feel really confident, just pause this video and try to figure out the whole thing."}, {"video_title": "Comparing exponent expressions.mp3", "Sentence": "So let's start with two to the third minus two to the first. What is that going to be? Two to the third minus two to the first. And if you feel really confident, just pause this video and try to figure out the whole thing. Order them from least to greatest. Well, two to the third, that is two times two times two. And then two to the first, well, that's just two."}, {"video_title": "Comparing exponent expressions.mp3", "Sentence": "And if you feel really confident, just pause this video and try to figure out the whole thing. Order them from least to greatest. Well, two to the third, that is two times two times two. And then two to the first, well, that's just two. So two times two is four, times two is eight, minus two. This is going to be equal to six. So this expression right over here could be evaluated as being equal to six."}, {"video_title": "Comparing exponent expressions.mp3", "Sentence": "And then two to the first, well, that's just two. So two times two is four, times two is eight, minus two. This is going to be equal to six. So this expression right over here could be evaluated as being equal to six. Now, what about this right over here? What is this equal to? Well, let's see."}, {"video_title": "Comparing exponent expressions.mp3", "Sentence": "So this expression right over here could be evaluated as being equal to six. Now, what about this right over here? What is this equal to? Well, let's see. We have two squared plus three to the zero. Two squared is two times two. And anything to the zero power is going to be equal to one."}, {"video_title": "Comparing exponent expressions.mp3", "Sentence": "Well, let's see. We have two squared plus three to the zero. Two squared is two times two. And anything to the zero power is going to be equal to one. It's an interesting thing to think about what zero to the zeroth power should be, but that'll be a topic for another video. But here we have three to the zeroth power, which is clearly equal to one. And so we have two times two plus one."}, {"video_title": "Comparing exponent expressions.mp3", "Sentence": "And anything to the zero power is going to be equal to one. It's an interesting thing to think about what zero to the zeroth power should be, but that'll be a topic for another video. But here we have three to the zeroth power, which is clearly equal to one. And so we have two times two plus one. This is four plus one, which is equal to five. So the second tile is equal to five. And then three squared."}, {"video_title": "Comparing exponent expressions.mp3", "Sentence": "And so we have two times two plus one. This is four plus one, which is equal to five. So the second tile is equal to five. And then three squared. Well, three squared, that's just three times three. Three times three is equal to nine. So if I were to order them from least to greatest, the smallest of these is two squared plus three to the zeroth power."}, {"video_title": "Comparing exponent expressions.mp3", "Sentence": "And then three squared. Well, three squared, that's just three times three. Three times three is equal to nine. So if I were to order them from least to greatest, the smallest of these is two squared plus three to the zeroth power. That one is equal to five, so I'd put that on the left. Then we have this thing that's equal to six. Two to the third power minus two to the first power."}, {"video_title": "Comparing exponent expressions.mp3", "Sentence": "So if I were to order them from least to greatest, the smallest of these is two squared plus three to the zeroth power. That one is equal to five, so I'd put that on the left. Then we have this thing that's equal to six. Two to the third power minus two to the first power. And then the largest value here is three squared. So we would put that tile, three squared, we will put that tile on the right. And we're done."}, {"video_title": "Labeling parts of a circle   Perimeter, area, and volume   Geometry   Khan Academy.mp3", "Sentence": "It won't be that well drawn of a circle, but I think you get the idea. So that is my circle. I'm going to label the center over here. So I'll do the center. I'll call it C. So that is my center. I'll draw an arrow there. That is the center of the circle."}, {"video_title": "Labeling parts of a circle   Perimeter, area, and volume   Geometry   Khan Academy.mp3", "Sentence": "So I'll do the center. I'll call it C. So that is my center. I'll draw an arrow there. That is the center of the circle. And actually, the circle itself is the set of all points that are a fixed distance away from that center. And that fixed distance away, that they're all from that center, that is the radius. So let me draw the radius."}, {"video_title": "Labeling parts of a circle   Perimeter, area, and volume   Geometry   Khan Academy.mp3", "Sentence": "That is the center of the circle. And actually, the circle itself is the set of all points that are a fixed distance away from that center. And that fixed distance away, that they're all from that center, that is the radius. So let me draw the radius. So this distance right over here is the radius. That is the radius. And that's going to be the same as this distance, which is the same as that distance."}, {"video_title": "Labeling parts of a circle   Perimeter, area, and volume   Geometry   Khan Academy.mp3", "Sentence": "So let me draw the radius. So this distance right over here is the radius. That is the radius. And that's going to be the same as this distance, which is the same as that distance. I can draw multiple radii. All of these are radii. The distance between the center and any point on the circle."}, {"video_title": "Labeling parts of a circle   Perimeter, area, and volume   Geometry   Khan Academy.mp3", "Sentence": "And that's going to be the same as this distance, which is the same as that distance. I can draw multiple radii. All of these are radii. The distance between the center and any point on the circle. Now a diameter just goes straight across the circle, going through the center. From one side of the circle to the other side, going through the center. And it's essentially two radii put together."}, {"video_title": "Labeling parts of a circle   Perimeter, area, and volume   Geometry   Khan Academy.mp3", "Sentence": "The distance between the center and any point on the circle. Now a diameter just goes straight across the circle, going through the center. From one side of the circle to the other side, going through the center. And it's essentially two radii put together. So for example, this would be a diameter. That would be a diameter. You have one radii, then another radii, all one line, going from one side of the circle to the other, going through the center."}, {"video_title": "Labeling parts of a circle   Perimeter, area, and volume   Geometry   Khan Academy.mp3", "Sentence": "And it's essentially two radii put together. So for example, this would be a diameter. That would be a diameter. You have one radii, then another radii, all one line, going from one side of the circle to the other, going through the center. So that is a diameter. That is a diameter. And I could have drawn it other ways."}, {"video_title": "Labeling parts of a circle   Perimeter, area, and volume   Geometry   Khan Academy.mp3", "Sentence": "You have one radii, then another radii, all one line, going from one side of the circle to the other, going through the center. So that is a diameter. That is a diameter. And I could have drawn it other ways. I could have drawn it like this. That would be another diameter. But they're going to have the exact same length."}, {"video_title": "Labeling parts of a circle   Perimeter, area, and volume   Geometry   Khan Academy.mp3", "Sentence": "And I could have drawn it other ways. I could have drawn it like this. That would be another diameter. But they're going to have the exact same length. And then finally, we have to think about the circumference. And the circumference is really just how far you have to go to go around the circle. Or if you put a string on this circle, how long would that string have to be?"}, {"video_title": "Labeling parts of a circle   Perimeter, area, and volume   Geometry   Khan Academy.mp3", "Sentence": "But they're going to have the exact same length. And then finally, we have to think about the circumference. And the circumference is really just how far you have to go to go around the circle. Or if you put a string on this circle, how long would that string have to be? So what I'm tracing out in blue right now, the length of what I'm tracing out, is the circumference. So right over here, that is the circumference. And we're done."}, {"video_title": "Introduction to exponents   Pre-Algebra   Khan Academy.mp3", "Sentence": "So it could be 2 plus 2 plus 2. Notice this is 1, 2, 3 2's. And when you add those 2's, you get 6. What we're going to introduce you to in this video is the idea of repeated multiplication, a new operation that really can be viewed as repeated multiplication. And that's the operation of taking an exponent. And it sounds very fancy, but we'll see with a few examples that it's not too bad. So now let's take the idea of 2 to the third power, which is how we would write this down in the appropriate colors."}, {"video_title": "Introduction to exponents   Pre-Algebra   Khan Academy.mp3", "Sentence": "What we're going to introduce you to in this video is the idea of repeated multiplication, a new operation that really can be viewed as repeated multiplication. And that's the operation of taking an exponent. And it sounds very fancy, but we'll see with a few examples that it's not too bad. So now let's take the idea of 2 to the third power, which is how we would write this down in the appropriate colors. So 2 to the third power. So you might be tempted to say, hey, maybe this is 2 times 3, which would be 6. But remember, I just said this is repeated multiplication."}, {"video_title": "Introduction to exponents   Pre-Algebra   Khan Academy.mp3", "Sentence": "So now let's take the idea of 2 to the third power, which is how we would write this down in the appropriate colors. So 2 to the third power. So you might be tempted to say, hey, maybe this is 2 times 3, which would be 6. But remember, I just said this is repeated multiplication. So if I have 2 to the third power, this literally means multiplying three 2's. So this would be equal to not 2 plus 2 plus 2, but 2 times. I'll use a little dot for multiplication."}, {"video_title": "Introduction to exponents   Pre-Algebra   Khan Academy.mp3", "Sentence": "But remember, I just said this is repeated multiplication. So if I have 2 to the third power, this literally means multiplying three 2's. So this would be equal to not 2 plus 2 plus 2, but 2 times. I'll use a little dot for multiplication. 2 times 2 times 2. Well, what's 2 times 2 times 2? Well, that is equal to 8."}, {"video_title": "Introduction to exponents   Pre-Algebra   Khan Academy.mp3", "Sentence": "I'll use a little dot for multiplication. 2 times 2 times 2. Well, what's 2 times 2 times 2? Well, that is equal to 8. So that is equal to 8. So 2 to the third power is equal to 8. Let's try a few more examples here."}, {"video_title": "Introduction to exponents   Pre-Algebra   Khan Academy.mp3", "Sentence": "Well, that is equal to 8. So that is equal to 8. So 2 to the third power is equal to 8. Let's try a few more examples here. What is 3 to the second power going to be equal to? I'll let you think about that for a second. I encourage you to pause the video."}, {"video_title": "Introduction to exponents   Pre-Algebra   Khan Academy.mp3", "Sentence": "Let's try a few more examples here. What is 3 to the second power going to be equal to? I'll let you think about that for a second. I encourage you to pause the video. So let's think it through. This literally means multiplying two 2 3's. So let's multiply 3."}, {"video_title": "Introduction to exponents   Pre-Algebra   Khan Academy.mp3", "Sentence": "I encourage you to pause the video. So let's think it through. This literally means multiplying two 2 3's. So let's multiply 3. Let me do that in yellow color. Let's multiply 3 times 3. So this is going to be equal to 9."}, {"video_title": "Introduction to exponents   Pre-Algebra   Khan Academy.mp3", "Sentence": "So let's multiply 3. Let me do that in yellow color. Let's multiply 3 times 3. So this is going to be equal to 9. Let's do a few more examples. What is, say, 5 to the fourth power? And what you'll see here is this number is going to get large very, very, very fast."}, {"video_title": "Introduction to exponents   Pre-Algebra   Khan Academy.mp3", "Sentence": "So this is going to be equal to 9. Let's do a few more examples. What is, say, 5 to the fourth power? And what you'll see here is this number is going to get large very, very, very fast. So 5 to the fourth power is going to be equal to multiplying 4 5's. So 5 times 5 times 5 times 5. Notice we have 1, 2, 3, 4 5's."}, {"video_title": "Introduction to exponents   Pre-Algebra   Khan Academy.mp3", "Sentence": "And what you'll see here is this number is going to get large very, very, very fast. So 5 to the fourth power is going to be equal to multiplying 4 5's. So 5 times 5 times 5 times 5. Notice we have 1, 2, 3, 4 5's. And we are multiplying them. We're not adding it. This is not 5 times 4."}, {"video_title": "Introduction to exponents   Pre-Algebra   Khan Academy.mp3", "Sentence": "Notice we have 1, 2, 3, 4 5's. And we are multiplying them. We're not adding it. This is not 5 times 4. This is not 20. This is 5 times 5 times 5 times 5. So what is this going to be?"}, {"video_title": "Introduction to exponents   Pre-Algebra   Khan Academy.mp3", "Sentence": "This is not 5 times 4. This is not 20. This is 5 times 5 times 5 times 5. So what is this going to be? Well, 5 times 5 is 25. 25 times 5 is 125. 125 times 5 is 625."}, {"video_title": "Why aren't we using the multiplication sign   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "But it can be anything from an x to a y, a z, an a, a b. And so oftentimes, we'll start using Greek letters like theta. But you can really use any symbol to say, hey, this is going to vary. It can take on multiple values. But out of all of these, the one that's most typically used in algebra, or really in all of mathematics, is the variable x, although all of these are used to some degree. But given that x is used so heavily, it does introduce a slight problem. And the problem is it looks a lot like the multiplication symbol or the one that we use in arithmetic."}, {"video_title": "Why aren't we using the multiplication sign   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "It can take on multiple values. But out of all of these, the one that's most typically used in algebra, or really in all of mathematics, is the variable x, although all of these are used to some degree. But given that x is used so heavily, it does introduce a slight problem. And the problem is it looks a lot like the multiplication symbol or the one that we use in arithmetic. So in arithmetic, if I want to write 2 times 3, I literally write 2 times 3. But now that we're starting to use variables and if I want to write 2 times x, well, if I use this as the multiplication symbol, it would be 2 times x. And the times symbol and the x look awfully similar."}, {"video_title": "Why aren't we using the multiplication sign   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And the problem is it looks a lot like the multiplication symbol or the one that we use in arithmetic. So in arithmetic, if I want to write 2 times 3, I literally write 2 times 3. But now that we're starting to use variables and if I want to write 2 times x, well, if I use this as the multiplication symbol, it would be 2 times x. And the times symbol and the x look awfully similar. And if I'm not really careful with my penmanship, it can get very confusing. Is this 2xx? Is this 2 times times something?"}, {"video_title": "Why aren't we using the multiplication sign   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And the times symbol and the x look awfully similar. And if I'm not really careful with my penmanship, it can get very confusing. Is this 2xx? Is this 2 times times something? What exactly is going on here? And because this is confusing, this right over here is extremely confusing and it can be misinterpreted, we tend to not use this multiplication symbol when we are doing algebra. Instead of that, to represent multiplication, we have several options."}, {"video_title": "Why aren't we using the multiplication sign   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Is this 2 times times something? What exactly is going on here? And because this is confusing, this right over here is extremely confusing and it can be misinterpreted, we tend to not use this multiplication symbol when we are doing algebra. Instead of that, to represent multiplication, we have several options. Instead of writing 2 times x this way, we could write 2 dot x. And this dot, I want to be very clear, this is not a decimal. This is written a little bit higher."}, {"video_title": "Why aren't we using the multiplication sign   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Instead of that, to represent multiplication, we have several options. Instead of writing 2 times x this way, we could write 2 dot x. And this dot, I want to be very clear, this is not a decimal. This is written a little bit higher. And we write this so we don't get confusion between this and one of these variables right here. But this really can be interpreted as 2 times x. So for example, if someone says 2 dot x, what is 2 dot x when x is equal to 3?"}, {"video_title": "Why aren't we using the multiplication sign   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "This is written a little bit higher. And we write this so we don't get confusion between this and one of these variables right here. But this really can be interpreted as 2 times x. So for example, if someone says 2 dot x, what is 2 dot x when x is equal to 3? Well, this would be the same thing as 2 times 3 when x is equal to 3. Another way that you could write it is you could write 2 and then you could write the x in parentheses right next to it. This is also interpreted as 2 times x."}, {"video_title": "Why aren't we using the multiplication sign   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "So for example, if someone says 2 dot x, what is 2 dot x when x is equal to 3? Well, this would be the same thing as 2 times 3 when x is equal to 3. Another way that you could write it is you could write 2 and then you could write the x in parentheses right next to it. This is also interpreted as 2 times x. Once again, so if in this situation x were 7, this would be 2 times 7 or 14. And then the most traditional way of doing it is to just write the x right after the 2. And sometimes this will be read as 2x."}, {"video_title": "Why aren't we using the multiplication sign   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "This is also interpreted as 2 times x. Once again, so if in this situation x were 7, this would be 2 times 7 or 14. And then the most traditional way of doing it is to just write the x right after the 2. And sometimes this will be read as 2x. But this literally does mean 2 times x. And you might say, well, how come we didn't always do that? Well, it would be literally confusing if we did it over here."}, {"video_title": "Why aren't we using the multiplication sign   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And sometimes this will be read as 2x. But this literally does mean 2 times x. And you might say, well, how come we didn't always do that? Well, it would be literally confusing if we did it over here. If we just wrote, instead of writing 2 times 3, we just wrote 2, 3, well, that looks like 23. This doesn't look like 2 times 3. That's why we never did it."}, {"video_title": "Why aren't we using the multiplication sign   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Well, it would be literally confusing if we did it over here. If we just wrote, instead of writing 2 times 3, we just wrote 2, 3, well, that looks like 23. This doesn't look like 2 times 3. That's why we never did it. But here, since we're using a letter now, it's clear that this isn't part of that number. This isn't 20-something. This is 2 times this variable x."}, {"video_title": "Why aren't we using the multiplication sign   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "That's why we never did it. But here, since we're using a letter now, it's clear that this isn't part of that number. This isn't 20-something. This is 2 times this variable x. So all of these are really the same expression, 2 times x, 2 times x, and 2 times x. And so with that out of the way, let's try a few worked examples, a few practice problems. And this will hopefully prepare you for the next exercise where you get a lot of chance to practice this."}, {"video_title": "Why aren't we using the multiplication sign   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "This is 2 times this variable x. So all of these are really the same expression, 2 times x, 2 times x, and 2 times x. And so with that out of the way, let's try a few worked examples, a few practice problems. And this will hopefully prepare you for the next exercise where you get a lot of chance to practice this. So if I were to say, what is 10 minus 3y? And what does this equal when y is equal to 2? Well, every time you see the y, you'd want that 2 there."}, {"video_title": "Why aren't we using the multiplication sign   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And this will hopefully prepare you for the next exercise where you get a lot of chance to practice this. So if I were to say, what is 10 minus 3y? And what does this equal when y is equal to 2? Well, every time you see the y, you'd want that 2 there. So this is when y is equal to 2. Let's set that y equal to 2. This is the same thing as 10 minus 3 times 2."}, {"video_title": "Why aren't we using the multiplication sign   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Well, every time you see the y, you'd want that 2 there. So this is when y is equal to 2. Let's set that y equal to 2. This is the same thing as 10 minus 3 times 2. You do the multiplication first. Multiplication takes precedence in order of operations. So 3 times 2 is 6."}, {"video_title": "Why aren't we using the multiplication sign   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "This is the same thing as 10 minus 3 times 2. You do the multiplication first. Multiplication takes precedence in order of operations. So 3 times 2 is 6. 10 minus 6 is equal to 4. Let's do another one. Let's say that we had 7x minus 4."}, {"video_title": "Why aren't we using the multiplication sign   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "So 3 times 2 is 6. 10 minus 6 is equal to 4. Let's do another one. Let's say that we had 7x minus 4. And let me do it in that same green color. 7x minus 4. And we want to evaluate that when x is equal to 3."}, {"video_title": "Why aren't we using the multiplication sign   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Let's say that we had 7x minus 4. And let me do it in that same green color. 7x minus 4. And we want to evaluate that when x is equal to 3. Where will we see the x? We want to put a 3 there. So this is the same thing as 7 times 3."}, {"video_title": "Why aren't we using the multiplication sign   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And we want to evaluate that when x is equal to 3. Where will we see the x? We want to put a 3 there. So this is the same thing as 7 times 3. And I'll actually use this notation, because I can use that with numbers. 7 times 3 minus 4. And once again, multiplication takes precedence by order of operations over addition or subtraction."}, {"video_title": "Why aren't we using the multiplication sign   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "So this is the same thing as 7 times 3. And I'll actually use this notation, because I can use that with numbers. 7 times 3 minus 4. And once again, multiplication takes precedence by order of operations over addition or subtraction. So we want to do the multiplying first. 7 times 3 is 21. 21 minus 4 is equal to 17."}, {"video_title": "Why aren't we using the multiplication sign   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And once again, multiplication takes precedence by order of operations over addition or subtraction. So we want to do the multiplying first. 7 times 3 is 21. 21 minus 4 is equal to 17. So hopefully that gives you a little bit of background. I really encourage you to try the next exercise. It'll give you a lot of practice on being able to evaluate expressions like this."}, {"video_title": "Recognize functions from tables   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "So a good place to start is just think about what a functional relationship means. Now there's definitely a relationship. They say, hey, if you're Joel, you're 5' 6\". If you're Nathan, you're 4' 11\". If you're Stuart, you're 5' 11\". That is a relationship. Now in order for it to be a functional relationship, for every instance or every example of the independent variable, you can only have one example of the value of the function for it."}, {"video_title": "Recognize functions from tables   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "If you're Nathan, you're 4' 11\". If you're Stuart, you're 5' 11\". That is a relationship. Now in order for it to be a functional relationship, for every instance or every example of the independent variable, you can only have one example of the value of the function for it. So if this is a height function, so if you say height is the function, in order for this to be a functional relationship, no matter whose name you put inside of the height function, you need to only be able to get one value. If there were two values associated with one person's name, it would not be a functional relationship. So if I were to ask you, what is the height of Nathan?"}, {"video_title": "Recognize functions from tables   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "Now in order for it to be a functional relationship, for every instance or every example of the independent variable, you can only have one example of the value of the function for it. So if this is a height function, so if you say height is the function, in order for this to be a functional relationship, no matter whose name you put inside of the height function, you need to only be able to get one value. If there were two values associated with one person's name, it would not be a functional relationship. So if I were to ask you, what is the height of Nathan? Well, you look at the table and say, well, Nathan's height is 4' 11\". Nathan is 4' 11\". There are not two heights for Nathan."}, {"video_title": "Recognize functions from tables   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "So if I were to ask you, what is the height of Nathan? Well, you look at the table and say, well, Nathan's height is 4' 11\". Nathan is 4' 11\". There are not two heights for Nathan. There's only one height. And for any one of these people that we can input into the function, there's only one height associated with them. So it is a functional relationship."}, {"video_title": "Recognize functions from tables   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "There are not two heights for Nathan. There's only one height. And for any one of these people that we can input into the function, there's only one height associated with them. So it is a functional relationship. And we can even see that on a graph. Let me graph that out for you. So let's graph out."}, {"video_title": "Recognize functions from tables   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "So it is a functional relationship. And we can even see that on a graph. Let me graph that out for you. So let's graph out. So let's see. The highest height here is 6' 1\". So if we start off with, well, we don't have to start."}, {"video_title": "Recognize functions from tables   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "So let's graph out. So let's see. The highest height here is 6' 1\". So if we start off with, well, we don't have to start. Well, let me just start at 1 foot, 2 feet, 3 feet, 4 feet, 5 feet, 5 feet, and 6 feet. And then if I were to plot the different names, the different people that I could put into our height function, we have, I'll just put the first letters of their names. We have Joel."}, {"video_title": "Recognize functions from tables   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "So if we start off with, well, we don't have to start. Well, let me just start at 1 foot, 2 feet, 3 feet, 4 feet, 5 feet, 5 feet, and 6 feet. And then if I were to plot the different names, the different people that I could put into our height function, we have, I'll just put the first letters of their names. We have Joel. We have Nathan. We have Stuart. We have LJ."}, {"video_title": "Recognize functions from tables   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "We have Joel. We have Nathan. We have Stuart. We have LJ. And then we have Tariq right there. So let's plot them. So you have Joel."}, {"video_title": "Recognize functions from tables   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "We have LJ. And then we have Tariq right there. So let's plot them. So you have Joel. Joel's height is 5' 6\". So 5' 6 is right about there. Then you have Nathan."}, {"video_title": "Recognize functions from tables   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "So you have Joel. Joel's height is 5' 6\". So 5' 6 is right about there. Then you have Nathan. Nathan's height, let me do it in a different color. Nathan's height is 4' 11\". So Nathan's height is 4' 11\"."}, {"video_title": "Recognize functions from tables   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "Then you have Nathan. Nathan's height, let me do it in a different color. Nathan's height is 4' 11\". So Nathan's height is 4' 11\". We will plot him right over there. Then you have Stuart. Stuart's height is 5' 11\"."}, {"video_title": "Recognize functions from tables   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "So Nathan's height is 4' 11\". We will plot him right over there. Then you have Stuart. Stuart's height is 5' 11\". So Stuart's height is 5' 11\". He is pretty close to 6 feet. So Stuart's height is 5' 11\"."}, {"video_title": "Recognize functions from tables   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "Stuart's height is 5' 11\". So Stuart's height is 5' 11\". He is pretty close to 6 feet. So Stuart's height is 5' 11\". Then you have LJ. LJ's height is 5' 6\". So you have two people with a height of 5' 6\"."}, {"video_title": "Recognize functions from tables   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "So Stuart's height is 5' 11\". Then you have LJ. LJ's height is 5' 6\". So you have two people with a height of 5' 6\". But that's OK. As long as for each person, you only have one height. And then finally, Tariq is 6' 1\". He's the tallest guy here."}, {"video_title": "Recognize functions from tables   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "So you have two people with a height of 5' 6\". But that's OK. As long as for each person, you only have one height. And then finally, Tariq is 6' 1\". He's the tallest guy here. Tariq is right up here at 6' 1\". So notice, for any one of the inputs into our function, we only have one value. So this is a functional relationship."}, {"video_title": "Recognize functions from tables   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "He's the tallest guy here. Tariq is right up here at 6' 1\". So notice, for any one of the inputs into our function, we only have one value. So this is a functional relationship. Now, you might say, OK, well, isn't everything a functional relationship? No. If I gave you the situation, if I also wrote here, let's say the table was like this, and I also wrote that Stuart is 5' 3\"."}, {"video_title": "Recognize functions from tables   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "So this is a functional relationship. Now, you might say, OK, well, isn't everything a functional relationship? No. If I gave you the situation, if I also wrote here, let's say the table was like this, and I also wrote that Stuart is 5' 3\". If this was our table, then we would no longer have a functional relationship. Because for the input of Stuart, we would have two different values. If we were to graph this, we have Stuart here at 5' 11\"."}, {"video_title": "Recognize functions from tables   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "If I gave you the situation, if I also wrote here, let's say the table was like this, and I also wrote that Stuart is 5' 3\". If this was our table, then we would no longer have a functional relationship. Because for the input of Stuart, we would have two different values. If we were to graph this, we have Stuart here at 5' 11\". And then all of a sudden, we would also have Stuart at 5' 3\". Now, this doesn't make a lot of sense. So we would plot it right over here."}, {"video_title": "Recognize functions from tables   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "If we were to graph this, we have Stuart here at 5' 11\". And then all of a sudden, we would also have Stuart at 5' 3\". Now, this doesn't make a lot of sense. So we would plot it right over here. So for Stuart, you would have two values. And so this wouldn't be a valid functional relationship. Because you wouldn't know what value to give if you were to take the height of Stuart."}, {"video_title": "Recognize functions from tables   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "So we would plot it right over here. So for Stuart, you would have two values. And so this wouldn't be a valid functional relationship. Because you wouldn't know what value to give if you were to take the height of Stuart. In order for this to be a function, there can only be one value for this. You don't know in this situation, when I add this, whether it's 5' 3\", or 5' 11\". Now, this wasn't the case."}, {"video_title": "Recognize functions from tables   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "Because you wouldn't know what value to give if you were to take the height of Stuart. In order for this to be a function, there can only be one value for this. You don't know in this situation, when I add this, whether it's 5' 3\", or 5' 11\". Now, this wasn't the case. So that isn't there. And so we know that the height of Stuart is 5' 11\". And this is a functional relationship."}, {"video_title": "Recognize functions from tables   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "Now, this wasn't the case. So that isn't there. And so we know that the height of Stuart is 5' 11\". And this is a functional relationship. I think to some level, it might be confusing. Because it's such a simple idea. Each of these values can only have one height associated with it."}, {"video_title": "Multiplying binomials area model   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "So I have this big rectangle here that's divided into four smaller rectangles. So what I want to do is I want to express the area of this larger rectangle, and I want to do it two ways. The first way I want to express it as the product of two binomials, and then I want to express it as a trinomial. So let's think about this a little bit. So one way to say, well look, the height of this larger rectangle from here to here, we see that that distance is x, and then from here to here it's two. So the entire height right over here, the entire height right over here is going to be x plus two. So the height is x plus two, and what's the width?"}, {"video_title": "Multiplying binomials area model   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "So let's think about this a little bit. So one way to say, well look, the height of this larger rectangle from here to here, we see that that distance is x, and then from here to here it's two. So the entire height right over here, the entire height right over here is going to be x plus two. So the height is x plus two, and what's the width? Well, the width is, we go from there to there is x, and then from there to there is three. So the entire width is x plus three, x plus three. So just like that, I've expressed the area of the entire rectangle, and it's the product of two binomials."}, {"video_title": "Multiplying binomials area model   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "So the height is x plus two, and what's the width? Well, the width is, we go from there to there is x, and then from there to there is three. So the entire width is x plus three, x plus three. So just like that, I've expressed the area of the entire rectangle, and it's the product of two binomials. But now let's express it as a trinomial. Well, to do that, we can break down the larger area into the areas of each of these smaller rectangles. So what's the area of this purple rectangle right over here?"}, {"video_title": "Multiplying binomials area model   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "So just like that, I've expressed the area of the entire rectangle, and it's the product of two binomials. But now let's express it as a trinomial. Well, to do that, we can break down the larger area into the areas of each of these smaller rectangles. So what's the area of this purple rectangle right over here? Well, the purple rectangle, its height is x, and its width is x, so its area is x squared. Let me write that, that's x squared. What's the area of this yellow rectangle?"}, {"video_title": "Multiplying binomials area model   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "So what's the area of this purple rectangle right over here? Well, the purple rectangle, its height is x, and its width is x, so its area is x squared. Let me write that, that's x squared. What's the area of this yellow rectangle? Well, its height is x, same height as right over here. Its height is x, and its width is three. So it's gonna be x times three, or three x."}, {"video_title": "Multiplying binomials area model   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "What's the area of this yellow rectangle? Well, its height is x, same height as right over here. Its height is x, and its width is three. So it's gonna be x times three, or three x. It'll have an area of three x. So that area is three x. If we're summing up the area of the entire thing, this would be plus three x."}, {"video_title": "Multiplying binomials area model   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "So it's gonna be x times three, or three x. It'll have an area of three x. So that area is three x. If we're summing up the area of the entire thing, this would be plus three x. So this expression right over here, that's the area of this purple region plus the area of this yellow region. And then we can move on to this green region. What's the area going to be here?"}, {"video_title": "Multiplying binomials area model   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "If we're summing up the area of the entire thing, this would be plus three x. So this expression right over here, that's the area of this purple region plus the area of this yellow region. And then we can move on to this green region. What's the area going to be here? Well, the height is two, and the width is x. So multiplying height times width is gonna be two times x. And we can just add that, plus two times x."}, {"video_title": "Multiplying binomials area model   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "What's the area going to be here? Well, the height is two, and the width is x. So multiplying height times width is gonna be two times x. And we can just add that, plus two times x. And then finally, this little gray box here, its height is two, we see that right over there. Its height is two, and its width is three. We see it right over there."}, {"video_title": "Multiplying binomials area model   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "And we can just add that, plus two times x. And then finally, this little gray box here, its height is two, we see that right over there. Its height is two, and its width is three. We see it right over there. So it has an area of six, two times three. So plus six, and you might say, well, this isn't a trinomial. This has four terms right over here."}, {"video_title": "Multiplying binomials area model   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "We see it right over there. So it has an area of six, two times three. So plus six, and you might say, well, this isn't a trinomial. This has four terms right over here. But you might notice that we can add, that we can add these middle two terms. Three x plus two x. I have three x's, and I add two x's to that, I'm gonna have five x's. So this entire thing simplifies to x squared, x squared, plus five x, plus six, plus six."}, {"video_title": "Multiplying binomials area model   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "This has four terms right over here. But you might notice that we can add, that we can add these middle two terms. Three x plus two x. I have three x's, and I add two x's to that, I'm gonna have five x's. So this entire thing simplifies to x squared, x squared, plus five x, plus six, plus six. So this and this are two ways of expressing the area, so they're going to be equal. And that makes sense, because if you multiply it out, these binomials, and simplify it, you would get this trinomial. We could do that really fast."}, {"video_title": "Multiplying binomials area model   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "So this entire thing simplifies to x squared, x squared, plus five x, plus six, plus six. So this and this are two ways of expressing the area, so they're going to be equal. And that makes sense, because if you multiply it out, these binomials, and simplify it, you would get this trinomial. We could do that really fast. You multiply the x times the x. Actually, let me do it in the same colors. You multiply, you multiply the x times the x, you get the x squared."}, {"video_title": "Multiplying binomials area model   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "We could do that really fast. You multiply the x times the x. Actually, let me do it in the same colors. You multiply, you multiply the x times the x, you get the x squared. You multiply this x times the three, you get your three x. You multiply, you multiply the two times the x, you get your two x. And then you multiply the two times the three, and you get your six."}, {"video_title": "Introduction to graphing systems of linear inequalities   Algebra II   Khan Academy.mp3", "Sentence": "graph the solution set for this system. It's a system of inequalities. We have y is greater than x minus 8, and y is less than 5 minus x. Let's graph the solution set for each of these inequalities. And then essentially where they overlap is the solution set for the system, the set of numbers or the set of coordinates that satisfy both. So let me draw a coordinate axis here. So that is my x-axis, and then I have my y-axis."}, {"video_title": "Introduction to graphing systems of linear inequalities   Algebra II   Khan Academy.mp3", "Sentence": "Let's graph the solution set for each of these inequalities. And then essentially where they overlap is the solution set for the system, the set of numbers or the set of coordinates that satisfy both. So let me draw a coordinate axis here. So that is my x-axis, and then I have my y-axis. That is my y-axis. And now let me draw the boundary line, the boundary for this first inequality. So the boundary line is going to look like y is equal to x minus 8, but it's not going to include it because it's only greater than x minus 8."}, {"video_title": "Introduction to graphing systems of linear inequalities   Algebra II   Khan Academy.mp3", "Sentence": "So that is my x-axis, and then I have my y-axis. That is my y-axis. And now let me draw the boundary line, the boundary for this first inequality. So the boundary line is going to look like y is equal to x minus 8, but it's not going to include it because it's only greater than x minus 8. But let's just graph x minus 8. So the y-intercept here is negative 8. When x is 0, y is going to be negative 8."}, {"video_title": "Introduction to graphing systems of linear inequalities   Algebra II   Khan Academy.mp3", "Sentence": "So the boundary line is going to look like y is equal to x minus 8, but it's not going to include it because it's only greater than x minus 8. But let's just graph x minus 8. So the y-intercept here is negative 8. When x is 0, y is going to be negative 8. So let's go negative 1, negative 2, 3, 4, 5, 6, 7, 8. So that is negative 8. So the point 0, negative 8 is on the line."}, {"video_title": "Introduction to graphing systems of linear inequalities   Algebra II   Khan Academy.mp3", "Sentence": "When x is 0, y is going to be negative 8. So let's go negative 1, negative 2, 3, 4, 5, 6, 7, 8. So that is negative 8. So the point 0, negative 8 is on the line. And then it has a slope of 1. You don't see it right there, but I could write it as 1x. So the slope here is going to be 1."}, {"video_title": "Introduction to graphing systems of linear inequalities   Algebra II   Khan Academy.mp3", "Sentence": "So the point 0, negative 8 is on the line. And then it has a slope of 1. You don't see it right there, but I could write it as 1x. So the slope here is going to be 1. I could just draw a line that goes straight up, or you could even say that it'll intersect if y is equal to 0. If y were equal to 0, x would be equal to 8. So 1, 2, 3, 4, 5, 6, 7, 8."}, {"video_title": "Introduction to graphing systems of linear inequalities   Algebra II   Khan Academy.mp3", "Sentence": "So the slope here is going to be 1. I could just draw a line that goes straight up, or you could even say that it'll intersect if y is equal to 0. If y were equal to 0, x would be equal to 8. So 1, 2, 3, 4, 5, 6, 7, 8. And so this is x is equal to 8. It has a slope of 1. For every time you move to the right one, you're going to move up 1."}, {"video_title": "Introduction to graphing systems of linear inequalities   Algebra II   Khan Academy.mp3", "Sentence": "So 1, 2, 3, 4, 5, 6, 7, 8. And so this is x is equal to 8. It has a slope of 1. For every time you move to the right one, you're going to move up 1. So the line is going to look something like this. And actually, let me not draw it as a solid line. If I did it as a solid line, that would actually be this equation right here."}, {"video_title": "Introduction to graphing systems of linear inequalities   Algebra II   Khan Academy.mp3", "Sentence": "For every time you move to the right one, you're going to move up 1. So the line is going to look something like this. And actually, let me not draw it as a solid line. If I did it as a solid line, that would actually be this equation right here. But we're not going to include that line. We care about the y values that are greater than that line. So what we want to do is do a dotted line to show that that's just the boundary, that we're not including that in our solution set."}, {"video_title": "Introduction to graphing systems of linear inequalities   Algebra II   Khan Academy.mp3", "Sentence": "If I did it as a solid line, that would actually be this equation right here. But we're not going to include that line. We care about the y values that are greater than that line. So what we want to do is do a dotted line to show that that's just the boundary, that we're not including that in our solution set. Let me do this in a new color. So this will be the color for that line, or for that inequality, I should say. So that is the boundary line."}, {"video_title": "Introduction to graphing systems of linear inequalities   Algebra II   Khan Academy.mp3", "Sentence": "So what we want to do is do a dotted line to show that that's just the boundary, that we're not including that in our solution set. Let me do this in a new color. So this will be the color for that line, or for that inequality, I should say. So that is the boundary line. And this says y is greater than x minus 8. So you pick an x, and then x minus 8 would get us on the boundary line, and then y is greater than that. So it's all the y values above the line for any given x."}, {"video_title": "Introduction to graphing systems of linear inequalities   Algebra II   Khan Academy.mp3", "Sentence": "So that is the boundary line. And this says y is greater than x minus 8. So you pick an x, and then x minus 8 would get us on the boundary line, and then y is greater than that. So it's all the y values above the line for any given x. So it'll be this region above the line right over here. And if that confuses you, in general, I like to just think, oh, greater than is going to be above the line. If it's less than, it's going to be below a line."}, {"video_title": "Introduction to graphing systems of linear inequalities   Algebra II   Khan Academy.mp3", "Sentence": "So it's all the y values above the line for any given x. So it'll be this region above the line right over here. And if that confuses you, in general, I like to just think, oh, greater than is going to be above the line. If it's less than, it's going to be below a line. But if you want to make sure, you can just test on either side of this line. So you could try the point 0, 0, which should be in our solution set. And if you say 0 is greater than 0 minus 8, or 0 is greater than negative 8, that works."}, {"video_title": "Introduction to graphing systems of linear inequalities   Algebra II   Khan Academy.mp3", "Sentence": "If it's less than, it's going to be below a line. But if you want to make sure, you can just test on either side of this line. So you could try the point 0, 0, which should be in our solution set. And if you say 0 is greater than 0 minus 8, or 0 is greater than negative 8, that works. So this definitely should be part of the solution set. And then you could try something out here like 10, 0 and see that it doesn't work. Because you would have 10 minus 8, which would be 2."}, {"video_title": "Introduction to graphing systems of linear inequalities   Algebra II   Khan Academy.mp3", "Sentence": "And if you say 0 is greater than 0 minus 8, or 0 is greater than negative 8, that works. So this definitely should be part of the solution set. And then you could try something out here like 10, 0 and see that it doesn't work. Because you would have 10 minus 8, which would be 2. And then you'd have 0, and 0 is not greater than 2. So when you test something out here, you also see that it won't work. But in general, I like to just say, hey, look."}, {"video_title": "Introduction to graphing systems of linear inequalities   Algebra II   Khan Academy.mp3", "Sentence": "Because you would have 10 minus 8, which would be 2. And then you'd have 0, and 0 is not greater than 2. So when you test something out here, you also see that it won't work. But in general, I like to just say, hey, look. This is the boundary line, and we're greater than the boundary line for any given x. Now let's do this one over here. Let's do this one."}, {"video_title": "Introduction to graphing systems of linear inequalities   Algebra II   Khan Academy.mp3", "Sentence": "But in general, I like to just say, hey, look. This is the boundary line, and we're greater than the boundary line for any given x. Now let's do this one over here. Let's do this one. The boundary line for it is going to be y is equal to 5 minus x. So the boundary line is y is equal to 5 minus x. So once again, if x is equal to 0, y is 5."}, {"video_title": "Introduction to graphing systems of linear inequalities   Algebra II   Khan Academy.mp3", "Sentence": "Let's do this one. The boundary line for it is going to be y is equal to 5 minus x. So the boundary line is y is equal to 5 minus x. So once again, if x is equal to 0, y is 5. So 1, 2, 3, 4, 5. And then it has a slope of negative 1. We could write this as y is equal to negative 1x plus 5."}, {"video_title": "Introduction to graphing systems of linear inequalities   Algebra II   Khan Academy.mp3", "Sentence": "So once again, if x is equal to 0, y is 5. So 1, 2, 3, 4, 5. And then it has a slope of negative 1. We could write this as y is equal to negative 1x plus 5. That's a little bit more traditional. So once again, y-intercept at 5. It has a slope of negative 1."}, {"video_title": "Introduction to graphing systems of linear inequalities   Algebra II   Khan Academy.mp3", "Sentence": "We could write this as y is equal to negative 1x plus 5. That's a little bit more traditional. So once again, y-intercept at 5. It has a slope of negative 1. Or another way to think about it, when y is 0, x will be equal to 5. So 1, 2, 3, 4, 5. So every time we move to the right one, we go down 1."}, {"video_title": "Introduction to graphing systems of linear inequalities   Algebra II   Khan Academy.mp3", "Sentence": "It has a slope of negative 1. Or another way to think about it, when y is 0, x will be equal to 5. So 1, 2, 3, 4, 5. So every time we move to the right one, we go down 1. Because we have a negative 1 slope. So it will look like this. And once again, I want to do a dotted line."}, {"video_title": "Introduction to graphing systems of linear inequalities   Algebra II   Khan Academy.mp3", "Sentence": "So every time we move to the right one, we go down 1. Because we have a negative 1 slope. So it will look like this. And once again, I want to do a dotted line. So that is our dotted line. And I'm doing a dotted line because it says y is less than 5 minus x. If it was y is equal to 5 minus x, I would have included the line, if it was y is less than or equal to 5 minus x, I also would have made this line solid."}, {"video_title": "Introduction to graphing systems of linear inequalities   Algebra II   Khan Academy.mp3", "Sentence": "And once again, I want to do a dotted line. So that is our dotted line. And I'm doing a dotted line because it says y is less than 5 minus x. If it was y is equal to 5 minus x, I would have included the line, if it was y is less than or equal to 5 minus x, I also would have made this line solid. But it's only less than. So for any x value, this is what 5 minus x. 5 minus x will sit on that boundary line."}, {"video_title": "Introduction to graphing systems of linear inequalities   Algebra II   Khan Academy.mp3", "Sentence": "If it was y is equal to 5 minus x, I would have included the line, if it was y is less than or equal to 5 minus x, I also would have made this line solid. But it's only less than. So for any x value, this is what 5 minus x. 5 minus x will sit on that boundary line. But we care about the y values that are less than that. So we want everything that is below the line. And once again, you can test on either side of the line."}, {"video_title": "Introduction to graphing systems of linear inequalities   Algebra II   Khan Academy.mp3", "Sentence": "5 minus x will sit on that boundary line. But we care about the y values that are less than that. So we want everything that is below the line. And once again, you can test on either side of the line. 0, 0 should work for this second inequality right here. 0 is indeed less than 5 minus 0. 0 is less than 5."}, {"video_title": "Introduction to graphing systems of linear inequalities   Algebra II   Khan Academy.mp3", "Sentence": "And once again, you can test on either side of the line. 0, 0 should work for this second inequality right here. 0 is indeed less than 5 minus 0. 0 is less than 5. And you could try something like 0, 10 and see that it doesn't work. Because if you had 10 is less than 5 minus 0, that doesn't work. So it is everything below the line like that."}, {"video_title": "Introduction to graphing systems of linear inequalities   Algebra II   Khan Academy.mp3", "Sentence": "0 is less than 5. And you could try something like 0, 10 and see that it doesn't work. Because if you had 10 is less than 5 minus 0, that doesn't work. So it is everything below the line like that. And like we said, the solution set for this system are all of the x's and y's, all of the coordinates that satisfy both of them. So all the shaded in purple satisfies the second inequality. All the shaded in green satisfies the first inequality."}, {"video_title": "Introduction to graphing systems of linear inequalities   Algebra II   Khan Academy.mp3", "Sentence": "So it is everything below the line like that. And like we said, the solution set for this system are all of the x's and y's, all of the coordinates that satisfy both of them. So all the shaded in purple satisfies the second inequality. All the shaded in green satisfies the first inequality. So the stuff that satisfies both of them is their overlap. So it's all of this region in blue. Hopefully this isn't making it too messy."}, {"video_title": "Introduction to graphing systems of linear inequalities   Algebra II   Khan Academy.mp3", "Sentence": "All the shaded in green satisfies the first inequality. So the stuff that satisfies both of them is their overlap. So it's all of this region in blue. Hopefully this isn't making it too messy. All of this region in blue where the two overlap below the magenta dotted line on the left-hand side and above the green magenta line. That's only where they overlap. So it's only this region over here."}, {"video_title": "Introduction to cube roots   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "For example, if I were to tell you that seven squared is equal to 49, that's equivalent to saying that seven is equal to the square root of 49. The square root essentially unwinds taking the square of something. In fact, we could write it like this. We could write the square root of 49. So this is whatever number times itself is equal to 49. If I multiply that number times itself, if I square it, well I'm gonna get 49. And that's going to be true for any number, not just 49."}, {"video_title": "Introduction to cube roots   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "We could write the square root of 49. So this is whatever number times itself is equal to 49. If I multiply that number times itself, if I square it, well I'm gonna get 49. And that's going to be true for any number, not just 49. If I write this, if I write the square root of x, and if I were to square it, that's going to be equal to x. And that's going to be true for any x for which we can evaluate the square root, evaluate the principal root. Now, typically, and as you advance in math, you're going to see that this will change."}, {"video_title": "Introduction to cube roots   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "And that's going to be true for any number, not just 49. If I write this, if I write the square root of x, and if I were to square it, that's going to be equal to x. And that's going to be true for any x for which we can evaluate the square root, evaluate the principal root. Now, typically, and as you advance in math, you're going to see that this will change. But typically, you say, okay, if I'm gonna take the square root of something, x has to be non-negative. X has to be non-negative. This is going to change once we start thinking about imaginary and complex numbers."}, {"video_title": "Introduction to cube roots   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "Now, typically, and as you advance in math, you're going to see that this will change. But typically, you say, okay, if I'm gonna take the square root of something, x has to be non-negative. X has to be non-negative. This is going to change once we start thinking about imaginary and complex numbers. But typically, for the principal square root, we assume that whatever's under the radical, whatever's under here, is going to be non-negative. Because it's hard to square a number, at least the numbers that we know about, it's hard to square them and get a negative number. So for this thing to be defined, for it to make sense, it's typical to say that, okay, we need to put a non-negative number in here."}, {"video_title": "Introduction to cube roots   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "This is going to change once we start thinking about imaginary and complex numbers. But typically, for the principal square root, we assume that whatever's under the radical, whatever's under here, is going to be non-negative. Because it's hard to square a number, at least the numbers that we know about, it's hard to square them and get a negative number. So for this thing to be defined, for it to make sense, it's typical to say that, okay, we need to put a non-negative number in here. But anyway, the focus of this video is not on the square root. It's really just to review things so we can start thinking about the cube root. And as you can imagine, where does the whole notion of taking a square of something or a square root come from?"}, {"video_title": "Introduction to cube roots   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "So for this thing to be defined, for it to make sense, it's typical to say that, okay, we need to put a non-negative number in here. But anyway, the focus of this video is not on the square root. It's really just to review things so we can start thinking about the cube root. And as you can imagine, where does the whole notion of taking a square of something or a square root come from? Well, it comes from the notion of finding the area of a square. If I have a square like this, and if this side is seven, well, if it's a square, all the sides are going to be seven. And if I wanted to find the area of this, it would be seven times seven, or seven squared."}, {"video_title": "Introduction to cube roots   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "And as you can imagine, where does the whole notion of taking a square of something or a square root come from? Well, it comes from the notion of finding the area of a square. If I have a square like this, and if this side is seven, well, if it's a square, all the sides are going to be seven. And if I wanted to find the area of this, it would be seven times seven, or seven squared. That would be the area of this. Or if I were to say, well, what is, if I have a square, if I have, and that doesn't look like a perfect square, but you get the idea, all the sides are the same length. If I have a square with area x, if the area here is x, what are the lengths of the sides going to be?"}, {"video_title": "Introduction to cube roots   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "And if I wanted to find the area of this, it would be seven times seven, or seven squared. That would be the area of this. Or if I were to say, well, what is, if I have a square, if I have, and that doesn't look like a perfect square, but you get the idea, all the sides are the same length. If I have a square with area x, if the area here is x, what are the lengths of the sides going to be? Well, it's going to be square root of x. All of the sides are going to be square root of x. So it's going to be square root of x by square root of x."}, {"video_title": "Introduction to cube roots   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "If I have a square with area x, if the area here is x, what are the lengths of the sides going to be? Well, it's going to be square root of x. All of the sides are going to be square root of x. So it's going to be square root of x by square root of x. And this side's going to be square root of x as well, and that's going to be square root of x as well. So that's where the term square root comes from, where the square comes from. Now, what do you think of cube root?"}, {"video_title": "Introduction to cube roots   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "So it's going to be square root of x by square root of x. And this side's going to be square root of x as well, and that's going to be square root of x as well. So that's where the term square root comes from, where the square comes from. Now, what do you think of cube root? Well, same idea. If I have a cube, my best attempt at drawing a cube really fast. If I have a cube, and a cube, all of its dimensions have the same length, so this is a two by two by two cube, what's the volume over here?"}, {"video_title": "Introduction to cube roots   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "Now, what do you think of cube root? Well, same idea. If I have a cube, my best attempt at drawing a cube really fast. If I have a cube, and a cube, all of its dimensions have the same length, so this is a two by two by two cube, what's the volume over here? Well, the volume is going to be two times two times two, which is two to the third power, or two cubed. This is two cubed. That's why they use the word cubed, because this would be the volume of a cube where each of its sides have length two, and this, of course, is going to be equal to eight."}, {"video_title": "Introduction to cube roots   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "If I have a cube, and a cube, all of its dimensions have the same length, so this is a two by two by two cube, what's the volume over here? Well, the volume is going to be two times two times two, which is two to the third power, or two cubed. This is two cubed. That's why they use the word cubed, because this would be the volume of a cube where each of its sides have length two, and this, of course, is going to be equal to eight. But what if we went the other way around? What if we started with the cube, or what if we started with its volume? What if we started with a cube's volume, and let's say the volume here is eight cubic units, so volume is equal to eight, and we wanted to find the lengths of the sides, so we wanted to figure out what x, we wanted to figure out what x is, because that's x, that's x, and that's x."}, {"video_title": "Introduction to cube roots   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "That's why they use the word cubed, because this would be the volume of a cube where each of its sides have length two, and this, of course, is going to be equal to eight. But what if we went the other way around? What if we started with the cube, or what if we started with its volume? What if we started with a cube's volume, and let's say the volume here is eight cubic units, so volume is equal to eight, and we wanted to find the lengths of the sides, so we wanted to figure out what x, we wanted to figure out what x is, because that's x, that's x, and that's x. It's a cube, so all its dimensions have the same length. Well, there's two ways that we could express this. We could say that x times x times x, or x to the third power, is equal to eight, or we could use the cube root symbol, which is a radical with a little three in the right place, or we could write that x is equal to, it's going to look very similar to the square root, this would be the square root of eight, but to make it clear that we're talking about the cube root of eight, we would write a little three over there."}, {"video_title": "Introduction to cube roots   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "What if we started with a cube's volume, and let's say the volume here is eight cubic units, so volume is equal to eight, and we wanted to find the lengths of the sides, so we wanted to figure out what x, we wanted to figure out what x is, because that's x, that's x, and that's x. It's a cube, so all its dimensions have the same length. Well, there's two ways that we could express this. We could say that x times x times x, or x to the third power, is equal to eight, or we could use the cube root symbol, which is a radical with a little three in the right place, or we could write that x is equal to, it's going to look very similar to the square root, this would be the square root of eight, but to make it clear that we're talking about the cube root of eight, we would write a little three over there. In theory, for square root, you could put a little two over here, but that'd be redundant. If there's no number here, people just assume that it's the square root, but if you're figuring out the cube root, or sometimes people say the third root, well, then you have to say, well, you have to put this little three right over here and this little notch in the radical symbol right over here, and so this is saying x is going to be some number that if I cube it, I get eight. So with that out of the way, let's do some examples."}, {"video_title": "Introduction to cube roots   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "We could say that x times x times x, or x to the third power, is equal to eight, or we could use the cube root symbol, which is a radical with a little three in the right place, or we could write that x is equal to, it's going to look very similar to the square root, this would be the square root of eight, but to make it clear that we're talking about the cube root of eight, we would write a little three over there. In theory, for square root, you could put a little two over here, but that'd be redundant. If there's no number here, people just assume that it's the square root, but if you're figuring out the cube root, or sometimes people say the third root, well, then you have to say, well, you have to put this little three right over here and this little notch in the radical symbol right over here, and so this is saying x is going to be some number that if I cube it, I get eight. So with that out of the way, let's do some examples. Let's say that I have, let's say that I want to calculate the cube root, the cube root of 27. What's that going to be? Well, if I say that this is going to be equal to x, this is equivalent to saying that x to the third, or that 27 is equal to x to the third power."}, {"video_title": "Introduction to cube roots   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "So with that out of the way, let's do some examples. Let's say that I have, let's say that I want to calculate the cube root, the cube root of 27. What's that going to be? Well, if I say that this is going to be equal to x, this is equivalent to saying that x to the third, or that 27 is equal to x to the third power. So what is x going to be? Well, x times x times x is equal to 27. Well, the number I can think of is three."}, {"video_title": "Introduction to cube roots   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "Well, if I say that this is going to be equal to x, this is equivalent to saying that x to the third, or that 27 is equal to x to the third power. So what is x going to be? Well, x times x times x is equal to 27. Well, the number I can think of is three. So we would say that x, let me scroll down a little bit, x is equal to three. Now let me ask you a question. Can we write, can we write something like, can we pick a new color, the cube root of, let me write negative 64."}, {"video_title": "Introduction to cube roots   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "Well, the number I can think of is three. So we would say that x, let me scroll down a little bit, x is equal to three. Now let me ask you a question. Can we write, can we write something like, can we pick a new color, the cube root of, let me write negative 64. I already talked about that. If we're talking the square root, it's fairly typical that, hey, you put a negative number in there, at least until we learn about imaginary numbers, we don't know what to do with it. But can we do something with this?"}, {"video_title": "Introduction to cube roots   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "Can we write, can we write something like, can we pick a new color, the cube root of, let me write negative 64. I already talked about that. If we're talking the square root, it's fairly typical that, hey, you put a negative number in there, at least until we learn about imaginary numbers, we don't know what to do with it. But can we do something with this? Well, if I multiply, if I cube something, can I get a negative number? Sure, if I take, so if I say this is equal to x, this is the same thing as saying that negative 64 is equal to x to the third power. Well, what could x be?"}, {"video_title": "Introduction to cube roots   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "But can we do something with this? Well, if I multiply, if I cube something, can I get a negative number? Sure, if I take, so if I say this is equal to x, this is the same thing as saying that negative 64 is equal to x to the third power. Well, what could x be? Well, what happens if you take negative four times negative four times negative four? Negative four times negative four is positive 16, but then times negative four is negative 64, is equal to negative 64. So what could x be here?"}, {"video_title": "Introduction to cube roots   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "Well, what could x be? Well, what happens if you take negative four times negative four times negative four? Negative four times negative four is positive 16, but then times negative four is negative 64, is equal to negative 64. So what could x be here? Well, x could be equal to negative four. x could be equal to negative four. So based on the math that we know so far, you actually can take the cube root of a negative number."}, {"video_title": "Introduction to cube roots   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "So what could x be here? Well, x could be equal to negative four. x could be equal to negative four. So based on the math that we know so far, you actually can take the cube root of a negative number. And just so you know, you don't have to stop there. You could take a fourth root, and in this case you would have a four here, a fifth root, a sixth root, a seventh root of numbers, and we'll talk about that in, well, later in your mathematical career. But most of what you're gonna see is actually going to be square root, and every now and then you're going to see a cube root."}, {"video_title": "Introduction to cube roots   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "So based on the math that we know so far, you actually can take the cube root of a negative number. And just so you know, you don't have to stop there. You could take a fourth root, and in this case you would have a four here, a fifth root, a sixth root, a seventh root of numbers, and we'll talk about that in, well, later in your mathematical career. But most of what you're gonna see is actually going to be square root, and every now and then you're going to see a cube root. Now you might be saying, well, hey, look, you know, you just knew that three to the third power is 27. You took the cube root, you get x. Is there any simple way to do this?"}, {"video_title": "Introduction to cube roots   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "But most of what you're gonna see is actually going to be square root, and every now and then you're going to see a cube root. Now you might be saying, well, hey, look, you know, you just knew that three to the third power is 27. You took the cube root, you get x. Is there any simple way to do this? And like, you know, if I give you an arbitrary number, if I were to just say, I don't know, if I were to say cube root of, cube root of 125. And the simple answer is, well, you know, the easiest way to actually figure this out is actually just to do a factorization, in particular a prime factorization of this thing right over here, and then you would figure it out. So you would say, okay, well, 125 is five times 25, which is five times five."}, {"video_title": "Introduction to cube roots   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "Is there any simple way to do this? And like, you know, if I give you an arbitrary number, if I were to just say, I don't know, if I were to say cube root of, cube root of 125. And the simple answer is, well, you know, the easiest way to actually figure this out is actually just to do a factorization, in particular a prime factorization of this thing right over here, and then you would figure it out. So you would say, okay, well, 125 is five times 25, which is five times five. All right, so this is the same thing as the cube root of five to the third power, which of course is going to be equal to, which is going to be equal to five. If you have a much larger number here, yes, there's no very simple way to compute what a cube root or a fourth root or a fifth root might be. And even square root can get quite difficult."}, {"video_title": "Factoring quadratics as (x+a)(x+b)   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "And what I'd like to do in this video is I'd like to factor it as the product of two binomials. Or to put it another way, I wanna write it as the product x plus a, that's one binomial, times x plus b, where we need to figure out what a and b are going to be. So I encourage you to pause the video and see if you can figure out what a and b need to be. Can we rewrite this expression as a product of two binomials where we know what a and b are? So let's work through this together now. And I'll highlight a and b in different colors. So I'll put a in yellow and I'll put b in magenta."}, {"video_title": "Factoring quadratics as (x+a)(x+b)   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "Can we rewrite this expression as a product of two binomials where we know what a and b are? So let's work through this together now. And I'll highlight a and b in different colors. So I'll put a in yellow and I'll put b in magenta. So if you, one way to think about it, so let's just multiply these two binomials using a and b. And we've done this in previous videos. You might wanna review multiplying binomials if any of this looks strange to you."}, {"video_title": "Factoring quadratics as (x+a)(x+b)   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "So I'll put a in yellow and I'll put b in magenta. So if you, one way to think about it, so let's just multiply these two binomials using a and b. And we've done this in previous videos. You might wanna review multiplying binomials if any of this looks strange to you. But if you were to multiply what we have on the right-hand side out, it would be equal to, you're going to have the x times the x, which is going to be x squared. Then you are going to have the a times the x, which is ax. And then you're gonna have the b times the x, which is bx."}, {"video_title": "Factoring quadratics as (x+a)(x+b)   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "You might wanna review multiplying binomials if any of this looks strange to you. But if you were to multiply what we have on the right-hand side out, it would be equal to, you're going to have the x times the x, which is going to be x squared. Then you are going to have the a times the x, which is ax. And then you're gonna have the b times the x, which is bx. Actually, I'm not gonna skip any steps here just to see it this time. But this is all review, or it should be review. So then we have, so we did x times x to get x squared."}, {"video_title": "Factoring quadratics as (x+a)(x+b)   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "And then you're gonna have the b times the x, which is bx. Actually, I'm not gonna skip any steps here just to see it this time. But this is all review, or it should be review. So then we have, so we did x times x to get x squared. Then we have a times x to get ax, to get ax. And then we're gonna have x times b. So we're multiplying each term times every other term."}, {"video_title": "Factoring quadratics as (x+a)(x+b)   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "So then we have, so we did x times x to get x squared. Then we have a times x to get ax, to get ax. And then we're gonna have x times b. So we're multiplying each term times every other term. So then we have x times b to get bx. So plus bx, bx. And then finally we have plus the a times the b, which is of course going to be ab."}, {"video_title": "Factoring quadratics as (x+a)(x+b)   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "So we're multiplying each term times every other term. So then we have x times b to get bx. So plus bx, bx. And then finally we have plus the a times the b, which is of course going to be ab. And now we can simplify this. And you might have been able to go straight to this if you are familiar with multiplying binomials. This would be x squared plus, we can add these two coefficients because they're both on the first degree terms."}, {"video_title": "Factoring quadratics as (x+a)(x+b)   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "And then finally we have plus the a times the b, which is of course going to be ab. And now we can simplify this. And you might have been able to go straight to this if you are familiar with multiplying binomials. This would be x squared plus, we can add these two coefficients because they're both on the first degree terms. They're both multiplied by x. If I have ax's and I add bx's to that, I'm gonna have a plus bx's. So let me write that down."}, {"video_title": "Factoring quadratics as (x+a)(x+b)   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "This would be x squared plus, we can add these two coefficients because they're both on the first degree terms. They're both multiplied by x. If I have ax's and I add bx's to that, I'm gonna have a plus bx's. So let me write that down. a plus bx's. And then finally I have the plus, let me do that blue color. Finally I have it as plus ab."}, {"video_title": "Factoring quadratics as (x+a)(x+b)   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "So let me write that down. a plus bx's. And then finally I have the plus, let me do that blue color. Finally I have it as plus ab. Plus ab. And now we can use this, and now we can use this to think about what a and b need to be. If we do a little bit of pattern matching, we see we have an x squared there, we have an x squared there."}, {"video_title": "Factoring quadratics as (x+a)(x+b)   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "Finally I have it as plus ab. Plus ab. And now we can use this, and now we can use this to think about what a and b need to be. If we do a little bit of pattern matching, we see we have an x squared there, we have an x squared there. We have something, we have something times x. In this case it's a negative three times x. And here we have something times x."}, {"video_title": "Factoring quadratics as (x+a)(x+b)   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "If we do a little bit of pattern matching, we see we have an x squared there, we have an x squared there. We have something, we have something times x. In this case it's a negative three times x. And here we have something times x. So one way to think about it is that a plus b needs to be equal to negative three. They need to add up to be this coefficient. So let me write that down."}, {"video_title": "Factoring quadratics as (x+a)(x+b)   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "And here we have something times x. So one way to think about it is that a plus b needs to be equal to negative three. They need to add up to be this coefficient. So let me write that down. So we have a plus b, a plus b, needs to be equal to negative three. Needs to be equal to negative three. And we're not done yet."}, {"video_title": "Factoring quadratics as (x+a)(x+b)   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "So let me write that down. So we have a plus b, a plus b, needs to be equal to negative three. Needs to be equal to negative three. And we're not done yet. We finally look at this last term, we have a times b. Well a times b needs to be equal to negative 10. So let's write that down."}, {"video_title": "Factoring quadratics as (x+a)(x+b)   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "And we're not done yet. We finally look at this last term, we have a times b. Well a times b needs to be equal to negative 10. So let's write that down. So we have a times b needs to be equal to, needs to be equal to negative 10. And in general, whenever you're factoring something, a quadratic expression that has a one on the second degree term, so it has a one coefficient on x squared, or you don't even see it, but it's implicitly there, you could write this as one x squared. Way to factor it is say, well can I come up with two numbers that add up to the coefficient on the first degree term?"}, {"video_title": "Factoring quadratics as (x+a)(x+b)   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "So let's write that down. So we have a times b needs to be equal to, needs to be equal to negative 10. And in general, whenever you're factoring something, a quadratic expression that has a one on the second degree term, so it has a one coefficient on x squared, or you don't even see it, but it's implicitly there, you could write this as one x squared. Way to factor it is say, well can I come up with two numbers that add up to the coefficient on the first degree term? So two numbers that add up to negative three. And if I multiply those same two numbers, I'm going to get negative 10. So two numbers that add up to negative three to add up to the coefficient here."}, {"video_title": "Factoring quadratics as (x+a)(x+b)   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "Way to factor it is say, well can I come up with two numbers that add up to the coefficient on the first degree term? So two numbers that add up to negative three. And if I multiply those same two numbers, I'm going to get negative 10. So two numbers that add up to negative three to add up to the coefficient here. And now when I multiply it, I get the constant term. I get this right over here. Two numbers when I multiply, I get negative 10."}, {"video_title": "Factoring quadratics as (x+a)(x+b)   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "So two numbers that add up to negative three to add up to the coefficient here. And now when I multiply it, I get the constant term. I get this right over here. Two numbers when I multiply, I get negative 10. Well what could those numbers be? Well, since when you multiply them, we get a negative number, we know that they're going to have different signs. And so let's see how we could think about it."}, {"video_title": "Factoring quadratics as (x+a)(x+b)   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "Two numbers when I multiply, I get negative 10. Well what could those numbers be? Well, since when you multiply them, we get a negative number, we know that they're going to have different signs. And so let's see how we could think about it. And since when we add them, we get a negative number, we know that the negative number must be the larger one. So if I were to just factor 10, 10, I could, 10, you could view that as one times 10, one times 10, or two times five. And two and five are interesting because if one of them are negative, their difference is three."}, {"video_title": "Factoring quadratics as (x+a)(x+b)   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "And so let's see how we could think about it. And since when we add them, we get a negative number, we know that the negative number must be the larger one. So if I were to just factor 10, 10, I could, 10, you could view that as one times 10, one times 10, or two times five. And two and five are interesting because if one of them are negative, their difference is three. So if one is negative, so let's see, if we're talking about negative 10, negative 10, you could say negative two times five. And when you multiply them, you do get negative 10. But if you add these two, you're going to get positive three."}, {"video_title": "Factoring quadratics as (x+a)(x+b)   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "And two and five are interesting because if one of them are negative, their difference is three. So if one is negative, so let's see, if we're talking about negative 10, negative 10, you could say negative two times five. And when you multiply them, you do get negative 10. But if you add these two, you're going to get positive three. But what if you went positive two times negative five? Now this is interesting because still when you multiply them you get negative 10. And when you add them, two plus negative five is going to be negative three."}, {"video_title": "Factoring quadratics as (x+a)(x+b)   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "But if you add these two, you're going to get positive three. But what if you went positive two times negative five? Now this is interesting because still when you multiply them you get negative 10. And when you add them, two plus negative five is going to be negative three. So we have just figured out, we have just figured out our two numbers. We could say that A, and we could say that A is two, or we could say that B is two, but I'll just say that A is two. So A is equal to two, and B is equal to negative five."}, {"video_title": "Factoring quadratics as (x+a)(x+b)   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "And when you add them, two plus negative five is going to be negative three. So we have just figured out, we have just figured out our two numbers. We could say that A, and we could say that A is two, or we could say that B is two, but I'll just say that A is two. So A is equal to two, and B is equal to negative five. B is equal to negative five. And so our original expression, we can rewrite as, so we can rewrite x squared minus three x minus 10. We can say that that is going to be equal to x plus two, x plus two times, times x, instead of saying plus negative five, which we could say, we could just say, see let me write that down."}, {"video_title": "Factoring quadratics as (x+a)(x+b)   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "So A is equal to two, and B is equal to negative five. B is equal to negative five. And so our original expression, we can rewrite as, so we can rewrite x squared minus three x minus 10. We can say that that is going to be equal to x plus two, x plus two times, times x, instead of saying plus negative five, which we could say, we could just say, see let me write that down. I could write just plus negative five right over there because that's our B. I could just write x minus, x minus five, and we're done. We've just factored it as a product of two binomials. Now, I did it fairly involved, mainly so you see where all this came from, but in the future, whenever you see a quadratic expression, and you have a one coefficient on the second degree term right over here, you could say, all right, well I need to figure out two numbers that add up to the coefficient on the first degree term, on the x term, and those same two numbers, when I multiply them, need to be equal to this constant term, need to be equal to negative 10."}, {"video_title": "Factoring quadratics as (x+a)(x+b)   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "We can say that that is going to be equal to x plus two, x plus two times, times x, instead of saying plus negative five, which we could say, we could just say, see let me write that down. I could write just plus negative five right over there because that's our B. I could just write x minus, x minus five, and we're done. We've just factored it as a product of two binomials. Now, I did it fairly involved, mainly so you see where all this came from, but in the future, whenever you see a quadratic expression, and you have a one coefficient on the second degree term right over here, you could say, all right, well I need to figure out two numbers that add up to the coefficient on the first degree term, on the x term, and those same two numbers, when I multiply them, need to be equal to this constant term, need to be equal to negative 10. You say, okay, well let's see. They're gonna be different signs because when I multiply them, I get a negative number. The negative one's gonna be the larger one since when I add them, I got a negative number."}, {"video_title": "Factoring quadratics as (x+a)(x+b)   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "Now, I did it fairly involved, mainly so you see where all this came from, but in the future, whenever you see a quadratic expression, and you have a one coefficient on the second degree term right over here, you could say, all right, well I need to figure out two numbers that add up to the coefficient on the first degree term, on the x term, and those same two numbers, when I multiply them, need to be equal to this constant term, need to be equal to negative 10. You say, okay, well let's see. They're gonna be different signs because when I multiply them, I get a negative number. The negative one's gonna be the larger one since when I add them, I got a negative number. So let's see. Five and two seem interesting. Well, negative five and positive two."}, {"video_title": "Algebraic expressions with fraction division   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "So let's deal with some algebraic expressions that involve multiplying fractions. So let's say that I had A over B, A over B times, times C over D. What is this going to be? I encourage you to pause the video and try to figure it out on your own. Well, when you multiply fractions, you just multiply the numerators and multiply the denominators. So the numerators here, A, C, you're just gonna multiply those out. So it's gonna be A times C, which we could just write as A, C. That's just A times C. All of that over the denominator, B times D. B times, B times D. Instead of multiplying, what would have happened if we were dividing? So if we had A over B, A over B divided by, divided by C over D. What would this be?"}, {"video_title": "Algebraic expressions with fraction division   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Well, when you multiply fractions, you just multiply the numerators and multiply the denominators. So the numerators here, A, C, you're just gonna multiply those out. So it's gonna be A times C, which we could just write as A, C. That's just A times C. All of that over the denominator, B times D. B times, B times D. Instead of multiplying, what would have happened if we were dividing? So if we had A over B, A over B divided by, divided by C over D. What would this be? And once again, I encourage you, encourage you to pause the video and figure it out on your own. Well, when you divide by a fraction, it is equivalent to multiplying by its reciprocal. So this is going to be the same thing as A over B, A over B times, times the reciprocal of this."}, {"video_title": "Algebraic expressions with fraction division   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "So if we had A over B, A over B divided by, divided by C over D. What would this be? And once again, I encourage you, encourage you to pause the video and figure it out on your own. Well, when you divide by a fraction, it is equivalent to multiplying by its reciprocal. So this is going to be the same thing as A over B, A over B times, times the reciprocal of this. So times D over, let me do that same color just so I don't confuse you, that D was purple, times D over C, times D over C. And then it reduces to a problem like this. You know what, I shouldn't even use this multiplication symbol now that we're in algebra because you might confuse that with an X. Let me write that as times, times D, D over C, times D over C. And what are you left with?"}, {"video_title": "Algebraic expressions with fraction division   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "So this is going to be the same thing as A over B, A over B times, times the reciprocal of this. So times D over, let me do that same color just so I don't confuse you, that D was purple, times D over C, times D over C. And then it reduces to a problem like this. You know what, I shouldn't even use this multiplication symbol now that we're in algebra because you might confuse that with an X. Let me write that as times, times D, D over C, times D over C. And what are you left with? Well, the numerator, you're gonna have A times D. So it's A D, A, A D over, over B C, over B times C. Now let's do one that's maybe a little bit more involved and see if you can pull it off. So let's say that I had, let's say that I had, I don't know, let me write it as one over A minus one over B, all of that over, all of that over C. And let's say, let's just, let's also divide that by one over D. So this is a more involved expression than what we've seen so far, but I think we have all the tools to tackle it. So I encourage you to pause the video and see if you can simplify this, if you can actually carry out these operations and come up with a one fraction that represents this."}, {"video_title": "Algebraic expressions with fraction division   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Let me write that as times, times D, D over C, times D over C. And what are you left with? Well, the numerator, you're gonna have A times D. So it's A D, A, A D over, over B C, over B times C. Now let's do one that's maybe a little bit more involved and see if you can pull it off. So let's say that I had, let's say that I had, I don't know, let me write it as one over A minus one over B, all of that over, all of that over C. And let's say, let's just, let's also divide that by one over D. So this is a more involved expression than what we've seen so far, but I think we have all the tools to tackle it. So I encourage you to pause the video and see if you can simplify this, if you can actually carry out these operations and come up with a one fraction that represents this. All right, so let's work through it step by step. So one over A minus one over B. Let me work through just that part by itself."}, {"video_title": "Algebraic expressions with fraction division   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "So I encourage you to pause the video and see if you can simplify this, if you can actually carry out these operations and come up with a one fraction that represents this. All right, so let's work through it step by step. So one over A minus one over B. Let me work through just that part by itself. So one over A minus one over B. We know how to tackle that. We can find a common denominator."}, {"video_title": "Algebraic expressions with fraction division   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Let me work through just that part by itself. So one over A minus one over B. We know how to tackle that. We can find a common denominator. Let me write it up here. So one over A minus one over B is going to be equal to, we can multiply one over A times B over B. So it's going to be B over BA."}, {"video_title": "Algebraic expressions with fraction division   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "We can find a common denominator. Let me write it up here. So one over A minus one over B is going to be equal to, we can multiply one over A times B over B. So it's going to be B over BA. Notice I haven't changed its value. I just multiplied it times one, B over B, minus, well I'm gonna multiply the numerator and denominator here by A. Minus A over AB, or I could write that as BA."}, {"video_title": "Algebraic expressions with fraction division   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "So it's going to be B over BA. Notice I haven't changed its value. I just multiplied it times one, B over B, minus, well I'm gonna multiply the numerator and denominator here by A. Minus A over AB, or I could write that as BA. And the whole reason why I did that is to have the same denominator. So this is going to be equal to B minus A over, I could write it as BA or AB. So this is going to be equal to, this is going to be equal to, this numerator right over here, B minus A over AB."}, {"video_title": "Algebraic expressions with fraction division   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Minus A over AB, or I could write that as BA. And the whole reason why I did that is to have the same denominator. So this is going to be equal to B minus A over, I could write it as BA or AB. So this is going to be equal to, this is going to be equal to, this numerator right over here, B minus A over AB. And then if I'm dividing it by C, that's the same thing as multiplying by the reciprocal of C. So if I'm dividing it by C, that's the same thing as multiplication. That's the same thing as multiplying times one over C. And if I am, and I'll just keep going here, if I'm dividing by one over D, notice this is the same thing as division right over here. If I'm dividing by C, that's the same thing as multiplying by the reciprocal of C. And then finally, I'm dividing by one over D, that's the same thing as multiplying by the reciprocal of one over D. So the reciprocal of one over D is D over D over one."}, {"video_title": "Algebraic expressions with fraction division   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "So this is going to be equal to, this is going to be equal to, this numerator right over here, B minus A over AB. And then if I'm dividing it by C, that's the same thing as multiplying by the reciprocal of C. So if I'm dividing it by C, that's the same thing as multiplication. That's the same thing as multiplying times one over C. And if I am, and I'll just keep going here, if I'm dividing by one over D, notice this is the same thing as division right over here. If I'm dividing by C, that's the same thing as multiplying by the reciprocal of C. And then finally, I'm dividing by one over D, that's the same thing as multiplying by the reciprocal of one over D. So the reciprocal of one over D is D over D over one. And so what does this result with? Well, in the numerator I have B minus A times one times D. So we could write this as D times B minus A. Times B minus A."}, {"video_title": "Algebraic expressions with fraction division   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "If I'm dividing by C, that's the same thing as multiplying by the reciprocal of C. And then finally, I'm dividing by one over D, that's the same thing as multiplying by the reciprocal of one over D. So the reciprocal of one over D is D over D over one. And so what does this result with? Well, in the numerator I have B minus A times one times D. So we could write this as D times B minus A. Times B minus A. And then in the denominator I have A, B, C. A, B, and C. And then finally, we can use the distributive property here. We can distribute this D. And we're gonna be left with, we're gonna be left with, we deserve a minor drum roll at this point. We could write this as D times B, D times B minus D, whoops, I wanna do that same green color so you really see how it got distributed."}, {"video_title": "Example 4 Factoring quadratics as a perfect square of a difference (a-b)^2   Khan Academy.mp3", "Sentence": "Factor 25x squared minus 30x plus 9. And so we have a leading coefficient that's not a 1, and it doesn't look like there are any common factors. Both 25 and 30 are divisible by 5, but 9 isn't divisible by 5. So we could factor this by grouping, but if we look a little bit more carefully here, see something interesting. 25 is a perfect square, and so 25x squared is a perfect square. It's the square of 5x. And then 9 is also a perfect square."}, {"video_title": "Example 4 Factoring quadratics as a perfect square of a difference (a-b)^2   Khan Academy.mp3", "Sentence": "So we could factor this by grouping, but if we look a little bit more carefully here, see something interesting. 25 is a perfect square, and so 25x squared is a perfect square. It's the square of 5x. And then 9 is also a perfect square. It's the square of 3, or actually it could be the square of negative 3. This could also be the square of negative 5x. So maybe, just maybe, this could be a perfect square."}, {"video_title": "Example 4 Factoring quadratics as a perfect square of a difference (a-b)^2   Khan Academy.mp3", "Sentence": "And then 9 is also a perfect square. It's the square of 3, or actually it could be the square of negative 3. This could also be the square of negative 5x. So maybe, just maybe, this could be a perfect square. So let's just think about what happens when we take the perfect square of a binomial, especially when the coefficient on the x term is not a 1. So if we have ax plus b squared, what will this look like when we expand this into a trinomial? Well, this is the same thing as ax plus b times ax plus b, which is the same thing as ax times ax."}, {"video_title": "Example 4 Factoring quadratics as a perfect square of a difference (a-b)^2   Khan Academy.mp3", "Sentence": "So maybe, just maybe, this could be a perfect square. So let's just think about what happens when we take the perfect square of a binomial, especially when the coefficient on the x term is not a 1. So if we have ax plus b squared, what will this look like when we expand this into a trinomial? Well, this is the same thing as ax plus b times ax plus b, which is the same thing as ax times ax. ax times ax is a squared x squared plus ax times b, which is abx, plus b times ax, which is another. You could call it bax or abx, plus b times b, so plus b squared. So this is equal to a squared x squared plus, these two are the same term, plus 2abx plus b squared."}, {"video_title": "Example 4 Factoring quadratics as a perfect square of a difference (a-b)^2   Khan Academy.mp3", "Sentence": "Well, this is the same thing as ax plus b times ax plus b, which is the same thing as ax times ax. ax times ax is a squared x squared plus ax times b, which is abx, plus b times ax, which is another. You could call it bax or abx, plus b times b, so plus b squared. So this is equal to a squared x squared plus, these two are the same term, plus 2abx plus b squared. So this is what happens when you square a binomial. Now, this pattern seems to work out pretty good. Let me rewrite our problem right below it."}, {"video_title": "Example 4 Factoring quadratics as a perfect square of a difference (a-b)^2   Khan Academy.mp3", "Sentence": "So this is equal to a squared x squared plus, these two are the same term, plus 2abx plus b squared. So this is what happens when you square a binomial. Now, this pattern seems to work out pretty good. Let me rewrite our problem right below it. We have 25x squared minus 30x plus 9. So if this is a perfect square, then that means that the a squared part right over here is 25, and then that means that the b squared part, and that means that the, let me do this in a different color, the b squared part is 9. So that tells us that a could be plus or minus 5, and that b could be plus or minus 3."}, {"video_title": "Example 4 Factoring quadratics as a perfect square of a difference (a-b)^2   Khan Academy.mp3", "Sentence": "Let me rewrite our problem right below it. We have 25x squared minus 30x plus 9. So if this is a perfect square, then that means that the a squared part right over here is 25, and then that means that the b squared part, and that means that the, let me do this in a different color, the b squared part is 9. So that tells us that a could be plus or minus 5, and that b could be plus or minus 3. Now let's see if this gels with this middle term. If these, for this middle term to work out, for this middle term to work out, I'm trying to look for good colors, 2ab, this part right over here, 2ab needs to be equal to negative 30. Or another way, let me write it over here, 2ab needs to be equal to negative 30, or if we divide both sides by 2, ab needs to be equal to negative 15."}, {"video_title": "Example 4 Factoring quadratics as a perfect square of a difference (a-b)^2   Khan Academy.mp3", "Sentence": "So that tells us that a could be plus or minus 5, and that b could be plus or minus 3. Now let's see if this gels with this middle term. If these, for this middle term to work out, for this middle term to work out, I'm trying to look for good colors, 2ab, this part right over here, 2ab needs to be equal to negative 30. Or another way, let me write it over here, 2ab needs to be equal to negative 30, or if we divide both sides by 2, ab needs to be equal to negative 15. So that tells us, since the product is negative, one has to be positive and one has to be negative. Now, lucky for us, the product of 5 and 3 is 15. So if we make one of them positive and one of them negative, we'll get up to negative 15."}, {"video_title": "Example 4 Factoring quadratics as a perfect square of a difference (a-b)^2   Khan Academy.mp3", "Sentence": "Or another way, let me write it over here, 2ab needs to be equal to negative 30, or if we divide both sides by 2, ab needs to be equal to negative 15. So that tells us, since the product is negative, one has to be positive and one has to be negative. Now, lucky for us, the product of 5 and 3 is 15. So if we make one of them positive and one of them negative, we'll get up to negative 15. So it looks like things are going to work out. So we could select a is equal to positive 5, and b is equal to negative 3. Those would work out to ab being equal to negative 15."}, {"video_title": "Example 4 Factoring quadratics as a perfect square of a difference (a-b)^2   Khan Academy.mp3", "Sentence": "So if we make one of them positive and one of them negative, we'll get up to negative 15. So it looks like things are going to work out. So we could select a is equal to positive 5, and b is equal to negative 3. Those would work out to ab being equal to negative 15. Or we could make a is equal to negative 5, and b is equal to positive 3. So either of these will work. So if we factor this out, this could be either, a is, let's do this first one, it could either be a is 5, b is negative 3."}, {"video_title": "Example 4 Factoring quadratics as a perfect square of a difference (a-b)^2   Khan Academy.mp3", "Sentence": "Those would work out to ab being equal to negative 15. Or we could make a is equal to negative 5, and b is equal to positive 3. So either of these will work. So if we factor this out, this could be either, a is, let's do this first one, it could either be a is 5, b is negative 3. So this could either be 5x minus 3 squared. a is 5, b is negative 3. It could be that."}, {"video_title": "Example 4 Factoring quadratics as a perfect square of a difference (a-b)^2   Khan Academy.mp3", "Sentence": "So if we factor this out, this could be either, a is, let's do this first one, it could either be a is 5, b is negative 3. So this could either be 5x minus 3 squared. a is 5, b is negative 3. It could be that. Or we could switch the signs on the two terms. Or a could be negative 5, and b could be positive 3. Or it could be negative 5x plus 3 squared."}, {"video_title": "Example 4 Factoring quadratics as a perfect square of a difference (a-b)^2   Khan Academy.mp3", "Sentence": "It could be that. Or we could switch the signs on the two terms. Or a could be negative 5, and b could be positive 3. Or it could be negative 5x plus 3 squared. So either of these are possible ways to factor this term out here. And you're going to say, wait, how does this work out? How can both of these multiply to the same thing?"}, {"video_title": "Example 4 Factoring quadratics as a perfect square of a difference (a-b)^2   Khan Academy.mp3", "Sentence": "Or it could be negative 5x plus 3 squared. So either of these are possible ways to factor this term out here. And you're going to say, wait, how does this work out? How can both of these multiply to the same thing? Well, this term, remember, this negative 5x plus 3, we could factor out a negative 1. So this right here is the same thing as negative 1 times 5x minus 3, the whole thing squared. And that's the same thing as negative 1 squared times 5x minus 3 squared."}, {"video_title": "Example 4 Factoring quadratics as a perfect square of a difference (a-b)^2   Khan Academy.mp3", "Sentence": "How can both of these multiply to the same thing? Well, this term, remember, this negative 5x plus 3, we could factor out a negative 1. So this right here is the same thing as negative 1 times 5x minus 3, the whole thing squared. And that's the same thing as negative 1 squared times 5x minus 3 squared. And negative 1 squared is clearly equal to 1. So that's why this and this are the same thing. This comes out to the same thing as 5x minus 3 squared, which is the same thing as that over there."}, {"video_title": "Checking solutions to systems of equations example   Algebra I   Khan Academy.mp3", "Sentence": "And they give us, the first equation is x plus two y is equal to 13. Second equation is three x minus y is equal to negative 11. In order for negative one comma seven to be a solution for the system, it needs to satisfy both equations. Or another way of thinking about it, x equals seven and y, or sorry, x is equal to negative one. This is the x coordinate. X equals negative one and y is equal to seven need to satisfy both of these equations in order for it to be a solution. So let's try it out."}, {"video_title": "Checking solutions to systems of equations example   Algebra I   Khan Academy.mp3", "Sentence": "Or another way of thinking about it, x equals seven and y, or sorry, x is equal to negative one. This is the x coordinate. X equals negative one and y is equal to seven need to satisfy both of these equations in order for it to be a solution. So let's try it out. Let's try it out with the first equation. So we have x plus two y is equal to 13. So if we're thinking about that, we're testing to see if when x is equal to negative one and y is equal to seven, will x plus two y equal 13?"}, {"video_title": "Checking solutions to systems of equations example   Algebra I   Khan Academy.mp3", "Sentence": "So let's try it out. Let's try it out with the first equation. So we have x plus two y is equal to 13. So if we're thinking about that, we're testing to see if when x is equal to negative one and y is equal to seven, will x plus two y equal 13? So we have negative one plus two times seven. Y should be seven. This needs to be equal to 13."}, {"video_title": "Checking solutions to systems of equations example   Algebra I   Khan Academy.mp3", "Sentence": "So if we're thinking about that, we're testing to see if when x is equal to negative one and y is equal to seven, will x plus two y equal 13? So we have negative one plus two times seven. Y should be seven. This needs to be equal to 13. And I'll put a question mark there because we don't know whether it does. So this is the same thing as negative one plus two times seven plus 14. That does indeed equal 13."}, {"video_title": "Checking solutions to systems of equations example   Algebra I   Khan Academy.mp3", "Sentence": "This needs to be equal to 13. And I'll put a question mark there because we don't know whether it does. So this is the same thing as negative one plus two times seven plus 14. That does indeed equal 13. This is negative one plus 14. This is 13. So 13 does definitely equal 13."}, {"video_title": "Checking solutions to systems of equations example   Algebra I   Khan Academy.mp3", "Sentence": "That does indeed equal 13. This is negative one plus 14. This is 13. So 13 does definitely equal 13. So this point, it does at least satisfy this first equation. This point does sit on the graph of this first equation or on the line of this first equation. Now let's look at the second equation."}, {"video_title": "Checking solutions to systems of equations example   Algebra I   Khan Academy.mp3", "Sentence": "So 13 does definitely equal 13. So this point, it does at least satisfy this first equation. This point does sit on the graph of this first equation or on the line of this first equation. Now let's look at the second equation. I'll do that one in blue. We have three times negative one, three times negative one minus y, so minus seven, needs to be equal to negative 11. And I'll put a question mark here because we don't know whether it's true or not."}, {"video_title": "Checking solutions to systems of equations example   Algebra I   Khan Academy.mp3", "Sentence": "Now let's look at the second equation. I'll do that one in blue. We have three times negative one, three times negative one minus y, so minus seven, needs to be equal to negative 11. And I'll put a question mark here because we don't know whether it's true or not. So let's see. We have three times negative one is negative three. And then we have minus seven needs to be equal to negative 11."}, {"video_title": "Checking solutions to systems of equations example   Algebra I   Khan Academy.mp3", "Sentence": "And I'll put a question mark here because we don't know whether it's true or not. So let's see. We have three times negative one is negative three. And then we have minus seven needs to be equal to negative 11. I'll put the question mark there. Negative three minus seven, that's negative 10. So we get negative 10 equaling negative 11."}, {"video_title": "Checking solutions to systems of equations example   Algebra I   Khan Academy.mp3", "Sentence": "And then we have minus seven needs to be equal to negative 11. I'll put the question mark there. Negative three minus seven, that's negative 10. So we get negative 10 equaling negative 11. No, negative 10 does not equal a negative 11. So x equaling negative one and y equaling seven does not satisfy this second equation. So it does not sit on its graph."}, {"video_title": "Checking solutions to systems of equations example   Algebra I   Khan Academy.mp3", "Sentence": "So we get negative 10 equaling negative 11. No, negative 10 does not equal a negative 11. So x equaling negative one and y equaling seven does not satisfy this second equation. So it does not sit on its graph. So this over here is not a solution for the system. So the answer is no. It satisfies the first equation, but it doesn't satisfy the second."}, {"video_title": "Similar triangle basics   Similarity   Geometry   Khan Academy.mp3", "Sentence": "But there does seem to be something interesting about the relationship between these two triangles. One, all of their corresponding angles are the same. So the angle right here, angle BAC, is congruent to angle YXZ. Angle BCA is congruent to angle YZX. And angle ABC is congruent to angle XYZ. So all of their angles, their corresponding angles, are the same. And we also see that the sides are just scaled up versions of each other."}, {"video_title": "Similar triangle basics   Similarity   Geometry   Khan Academy.mp3", "Sentence": "Angle BCA is congruent to angle YZX. And angle ABC is congruent to angle XYZ. So all of their angles, their corresponding angles, are the same. And we also see that the sides are just scaled up versions of each other. So to go from the length of XZ to AC, we can multiply by 3. We multiplied by 3 there. To go from XY, the length of XY, to the length of AB, which is the corresponding side, we are multiplying by 3."}, {"video_title": "Similar triangle basics   Similarity   Geometry   Khan Academy.mp3", "Sentence": "And we also see that the sides are just scaled up versions of each other. So to go from the length of XZ to AC, we can multiply by 3. We multiplied by 3 there. To go from XY, the length of XY, to the length of AB, which is the corresponding side, we are multiplying by 3. We had to multiply by 3. And then to go from the length of YZ to the length of BC, we also multiplied by 3. So essentially, triangle ABC is just a scaled up version of triangle XYZ."}, {"video_title": "Similar triangle basics   Similarity   Geometry   Khan Academy.mp3", "Sentence": "To go from XY, the length of XY, to the length of AB, which is the corresponding side, we are multiplying by 3. We had to multiply by 3. And then to go from the length of YZ to the length of BC, we also multiplied by 3. So essentially, triangle ABC is just a scaled up version of triangle XYZ. If they were the same scale, they would be the exact same triangles. But one is just a bigger, a blown up version of the other one. Or this is a miniaturized version of that one over there."}, {"video_title": "Similar triangle basics   Similarity   Geometry   Khan Academy.mp3", "Sentence": "So essentially, triangle ABC is just a scaled up version of triangle XYZ. If they were the same scale, they would be the exact same triangles. But one is just a bigger, a blown up version of the other one. Or this is a miniaturized version of that one over there. If you just multiply all the sides by 3, you get to this triangle. And so we can't call them congruent. But this does seem to be a bit of a special relationship."}, {"video_title": "Similar triangle basics   Similarity   Geometry   Khan Academy.mp3", "Sentence": "Or this is a miniaturized version of that one over there. If you just multiply all the sides by 3, you get to this triangle. And so we can't call them congruent. But this does seem to be a bit of a special relationship. So we call this special relationship similarity. So we can write that triangle ABC is similar to triangle. And we want to make sure we get the corresponding sides right."}, {"video_title": "Similar triangle basics   Similarity   Geometry   Khan Academy.mp3", "Sentence": "But this does seem to be a bit of a special relationship. So we call this special relationship similarity. So we can write that triangle ABC is similar to triangle. And we want to make sure we get the corresponding sides right. ABC is going to be similar to XYZ. And so based on what we just saw, there's actually kind of three ideas here. And they're all equivalent ways of thinking about similarity."}, {"video_title": "Similar triangle basics   Similarity   Geometry   Khan Academy.mp3", "Sentence": "And we want to make sure we get the corresponding sides right. ABC is going to be similar to XYZ. And so based on what we just saw, there's actually kind of three ideas here. And they're all equivalent ways of thinking about similarity. One way to think about it is that one is a scaled up version of the other. So scaled up or down version of the other. When we talked about congruency, they had to be exactly the same."}, {"video_title": "Similar triangle basics   Similarity   Geometry   Khan Academy.mp3", "Sentence": "And they're all equivalent ways of thinking about similarity. One way to think about it is that one is a scaled up version of the other. So scaled up or down version of the other. When we talked about congruency, they had to be exactly the same. You could rotate it. You could shift it. You could flip it."}, {"video_title": "Similar triangle basics   Similarity   Geometry   Khan Academy.mp3", "Sentence": "When we talked about congruency, they had to be exactly the same. You could rotate it. You could shift it. You could flip it. But when you do all those things, they would have to essentially be identical. With similarity, you can rotate it. You can shift it."}, {"video_title": "Similar triangle basics   Similarity   Geometry   Khan Academy.mp3", "Sentence": "You could flip it. But when you do all those things, they would have to essentially be identical. With similarity, you can rotate it. You can shift it. You can flip it. And you can also scale it up and down in order for something to be similar. So for example, if you say if something is congruent, if let me say triangle CDE, if we know that triangle CDE is congruent to triangle FGH, then we definitely know that they are similar."}, {"video_title": "Similar triangle basics   Similarity   Geometry   Khan Academy.mp3", "Sentence": "You can shift it. You can flip it. And you can also scale it up and down in order for something to be similar. So for example, if you say if something is congruent, if let me say triangle CDE, if we know that triangle CDE is congruent to triangle FGH, then we definitely know that they are similar. They're scaled up by a factor of 1. Then we know for a fact that CDE is also similar to triangle FGH. But we can't say it the other way around."}, {"video_title": "Similar triangle basics   Similarity   Geometry   Khan Academy.mp3", "Sentence": "So for example, if you say if something is congruent, if let me say triangle CDE, if we know that triangle CDE is congruent to triangle FGH, then we definitely know that they are similar. They're scaled up by a factor of 1. Then we know for a fact that CDE is also similar to triangle FGH. But we can't say it the other way around. If triangle ABC is similar to XYZ, we can't say that it's necessarily congruent. And we see for this particular example, they definitely are not congruent. So this is one way to think about similarity."}, {"video_title": "Similar triangle basics   Similarity   Geometry   Khan Academy.mp3", "Sentence": "But we can't say it the other way around. If triangle ABC is similar to XYZ, we can't say that it's necessarily congruent. And we see for this particular example, they definitely are not congruent. So this is one way to think about similarity. The other way to think about similarity is that all of the corresponding angles will be equal. So if something is similar, then all of the corresponding angles are going to be congruent. I always have trouble spelling this."}, {"video_title": "Similar triangle basics   Similarity   Geometry   Khan Academy.mp3", "Sentence": "So this is one way to think about similarity. The other way to think about similarity is that all of the corresponding angles will be equal. So if something is similar, then all of the corresponding angles are going to be congruent. I always have trouble spelling this. It is two Rs, one S. Corresponding angles are congruent. So if we say that triangle ABC is similar to triangle XYZ, that is equivalent to saying that angle ABC is congruent, or we can say that their measures are equal to angle XYZ. That angle BAC is going to be congruent to angle YXZ."}, {"video_title": "Similar triangle basics   Similarity   Geometry   Khan Academy.mp3", "Sentence": "I always have trouble spelling this. It is two Rs, one S. Corresponding angles are congruent. So if we say that triangle ABC is similar to triangle XYZ, that is equivalent to saying that angle ABC is congruent, or we can say that their measures are equal to angle XYZ. That angle BAC is going to be congruent to angle YXZ. YXZ to angle YXZ. And then finally, angle ACB is going to be congruent to angle XZY. So if you have two triangles, all of their angles are the same, then you can say that they're similar."}, {"video_title": "Similar triangle basics   Similarity   Geometry   Khan Academy.mp3", "Sentence": "That angle BAC is going to be congruent to angle YXZ. YXZ to angle YXZ. And then finally, angle ACB is going to be congruent to angle XZY. So if you have two triangles, all of their angles are the same, then you can say that they're similar. Or if you find two triangles and you're told that they are similar triangles, then you know that all of their corresponding angles are the same. And the last way to think about it is that the sides are all just scaled up versions of each other. So the sides scaled by the same factor."}, {"video_title": "Similar triangle basics   Similarity   Geometry   Khan Academy.mp3", "Sentence": "So if you have two triangles, all of their angles are the same, then you can say that they're similar. Or if you find two triangles and you're told that they are similar triangles, then you know that all of their corresponding angles are the same. And the last way to think about it is that the sides are all just scaled up versions of each other. So the sides scaled by the same factor. In the example we did here, the scaling factor was 3. It doesn't have to be 3. It just has to be the same scaling factor for every side."}, {"video_title": "Similar triangle basics   Similarity   Geometry   Khan Academy.mp3", "Sentence": "So the sides scaled by the same factor. In the example we did here, the scaling factor was 3. It doesn't have to be 3. It just has to be the same scaling factor for every side. If we scale this side up by 3 and we only scale this side up by 2, then we would not be dealing with a similar triangle. But if we scaled all of these sides up by 7, then that's still a similar, as long as you have all of them scaled up or scaled down by the exact same factor. So one way to think about it is, and I want to keep having, well, I want to still visualize those triangles."}, {"video_title": "Similar triangle basics   Similarity   Geometry   Khan Academy.mp3", "Sentence": "It just has to be the same scaling factor for every side. If we scale this side up by 3 and we only scale this side up by 2, then we would not be dealing with a similar triangle. But if we scaled all of these sides up by 7, then that's still a similar, as long as you have all of them scaled up or scaled down by the exact same factor. So one way to think about it is, and I want to keep having, well, I want to still visualize those triangles. Let me redraw them right over here a little bit simpler, because I'm not talking now in general terms, not even for that specific case. So if we say that this is A, B, and C, and this right over here is X, Y, and Z. I just redrew them so I can refer to them when we write down here. If we're saying that these two things right over here are similar, that means that corresponding sides are scaled up versions of each other."}, {"video_title": "Similar triangle basics   Similarity   Geometry   Khan Academy.mp3", "Sentence": "So one way to think about it is, and I want to keep having, well, I want to still visualize those triangles. Let me redraw them right over here a little bit simpler, because I'm not talking now in general terms, not even for that specific case. So if we say that this is A, B, and C, and this right over here is X, Y, and Z. I just redrew them so I can refer to them when we write down here. If we're saying that these two things right over here are similar, that means that corresponding sides are scaled up versions of each other. So we could say that the length of AB is equal to some scaling factor, and this thing could be less than 1, times the length of XY, the corresponding sides. And I know that AB corresponds to XY because of the order in which I wrote this similarity statement. So some scaling factor times XY."}, {"video_title": "Similar triangle basics   Similarity   Geometry   Khan Academy.mp3", "Sentence": "If we're saying that these two things right over here are similar, that means that corresponding sides are scaled up versions of each other. So we could say that the length of AB is equal to some scaling factor, and this thing could be less than 1, times the length of XY, the corresponding sides. And I know that AB corresponds to XY because of the order in which I wrote this similarity statement. So some scaling factor times XY. We know that BC, the length of BC, needs to be that same scaling factor times the length of YZ. So that same scaling factor. And then we know the length of AC is going to be equal to that same scaling factor times XZ."}, {"video_title": "Similar triangle basics   Similarity   Geometry   Khan Academy.mp3", "Sentence": "So some scaling factor times XY. We know that BC, the length of BC, needs to be that same scaling factor times the length of YZ. So that same scaling factor. And then we know the length of AC is going to be equal to that same scaling factor times XZ. So that's XZ, and this could be a scaling factor. So if ABC is larger than XYZ, then these Ks will be larger than 1. If they're the exact same size, if they're essentially congruent triangles, then these Ks will be 1."}, {"video_title": "Similar triangle basics   Similarity   Geometry   Khan Academy.mp3", "Sentence": "And then we know the length of AC is going to be equal to that same scaling factor times XZ. So that's XZ, and this could be a scaling factor. So if ABC is larger than XYZ, then these Ks will be larger than 1. If they're the exact same size, if they're essentially congruent triangles, then these Ks will be 1. And if XYZ is bigger than ABC, then these scaling factors will be less than 1. But another way to write the same statement, notice all I'm saying is corresponding sides are scaled up versions of each other. This first statement right here, if you divide both sides by XY, you get AB over XY is equal to our scaling factor."}, {"video_title": "Similar triangle basics   Similarity   Geometry   Khan Academy.mp3", "Sentence": "If they're the exact same size, if they're essentially congruent triangles, then these Ks will be 1. And if XYZ is bigger than ABC, then these scaling factors will be less than 1. But another way to write the same statement, notice all I'm saying is corresponding sides are scaled up versions of each other. This first statement right here, if you divide both sides by XY, you get AB over XY is equal to our scaling factor. And then the second statement right over here, if you divide both sides by YZ, you get B. Let me do that same color. You get BC divided by YZ is equal to that scaling factor."}, {"video_title": "Similar triangle basics   Similarity   Geometry   Khan Academy.mp3", "Sentence": "This first statement right here, if you divide both sides by XY, you get AB over XY is equal to our scaling factor. And then the second statement right over here, if you divide both sides by YZ, you get B. Let me do that same color. You get BC divided by YZ is equal to that scaling factor. Remember, in the example we just showed, that scaling factor was 3. But now we're saying in the more general terms, similarity, as long as you have the same scaling factor. And then finally, if you divide both sides here by the length between X and XZ, or segment XZ's length, you get AC over XZ is equal to K as well."}, {"video_title": "Similar triangle basics   Similarity   Geometry   Khan Academy.mp3", "Sentence": "You get BC divided by YZ is equal to that scaling factor. Remember, in the example we just showed, that scaling factor was 3. But now we're saying in the more general terms, similarity, as long as you have the same scaling factor. And then finally, if you divide both sides here by the length between X and XZ, or segment XZ's length, you get AC over XZ is equal to K as well. Or another way to think about it is the ratio between corresponding sides. Notice, this is the ratio between AB and XY. The ratio between BC and YZ."}, {"video_title": "Similar triangle basics   Similarity   Geometry   Khan Academy.mp3", "Sentence": "And then finally, if you divide both sides here by the length between X and XZ, or segment XZ's length, you get AC over XZ is equal to K as well. Or another way to think about it is the ratio between corresponding sides. Notice, this is the ratio between AB and XY. The ratio between BC and YZ. The ratio between AC and XZ. That the ratio between corresponding sides all gives us the same constant. Or you could rewrite this as AB over XY is equal to BC over YZ is equal to AC over XZ, which would be equal to some scaling factor, which is equal to K. So if you have similar triangles, let me draw an arrow right over here."}, {"video_title": "Similar triangle basics   Similarity   Geometry   Khan Academy.mp3", "Sentence": "The ratio between BC and YZ. The ratio between AC and XZ. That the ratio between corresponding sides all gives us the same constant. Or you could rewrite this as AB over XY is equal to BC over YZ is equal to AC over XZ, which would be equal to some scaling factor, which is equal to K. So if you have similar triangles, let me draw an arrow right over here. Similar triangles means that they're scaled up versions, and you can also flip and rotate and do all this stuff with congruency. And you can scale them up or down, which means all of the corresponding angles are congruent, which also means that the ratio between corresponding sides is going to be the same constant for all the corresponding sides. Or the ratio between corresponding sides is constant."}, {"video_title": "Pythagorean theorem 2   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "This is our right triangle. This is the 90 degree angle right here. And we're told that this side's length right here is 14. This side's length right over here is 9. And we're told that this side is a. And we need to find the length of a. So as I mentioned already, this is a right triangle."}, {"video_title": "Pythagorean theorem 2   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "This side's length right over here is 9. And we're told that this side is a. And we need to find the length of a. So as I mentioned already, this is a right triangle. And we know that if we have a right triangle, we can always figure out, if we know two of the sides, we can always figure out a third side using the Pythagorean theorem. And what the Pythagorean theorem tells us is that the sum of the squares of the shorter sides is going to be equal to the square of the longer side, or the square of the hypotenuse. And if you're not sure about that, you're probably thinking, hey, Sal, how do I know that a is shorter than this side over here?"}, {"video_title": "Pythagorean theorem 2   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "So as I mentioned already, this is a right triangle. And we know that if we have a right triangle, we can always figure out, if we know two of the sides, we can always figure out a third side using the Pythagorean theorem. And what the Pythagorean theorem tells us is that the sum of the squares of the shorter sides is going to be equal to the square of the longer side, or the square of the hypotenuse. And if you're not sure about that, you're probably thinking, hey, Sal, how do I know that a is shorter than this side over here? How do I know it's not 15 or 16? And the way to tell is that the longest side in a right triangle, this only applies to a right triangle, is the side opposite the 90-degree angle. And in this case, 14 is opposite the 90 degrees."}, {"video_title": "Pythagorean theorem 2   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "And if you're not sure about that, you're probably thinking, hey, Sal, how do I know that a is shorter than this side over here? How do I know it's not 15 or 16? And the way to tell is that the longest side in a right triangle, this only applies to a right triangle, is the side opposite the 90-degree angle. And in this case, 14 is opposite the 90 degrees. This 90-degree angle kind of opens into this longest side, the side that we call the hypotenuse. So now that we know that that's the longest side, let me color code it. So this is the longest side."}, {"video_title": "Pythagorean theorem 2   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "And in this case, 14 is opposite the 90 degrees. This 90-degree angle kind of opens into this longest side, the side that we call the hypotenuse. So now that we know that that's the longest side, let me color code it. So this is the longest side. This is one of the shorter sides. And this is the other of the shorter sides. The Pythagorean theorem tells us that the sum of the squares of the shorter sides, so a squared plus 9 squared is going to be equal to 14 squared."}, {"video_title": "Pythagorean theorem 2   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "So this is the longest side. This is one of the shorter sides. And this is the other of the shorter sides. The Pythagorean theorem tells us that the sum of the squares of the shorter sides, so a squared plus 9 squared is going to be equal to 14 squared. And it's really important that you realize that it's not 9 squared plus 14 squared is going to be equal to a squared. a squared is one of the shorter sides. The sum of the squares of these two sides are going to be equal to 14 squared, the hypotenuse squared."}, {"video_title": "Pythagorean theorem 2   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "The Pythagorean theorem tells us that the sum of the squares of the shorter sides, so a squared plus 9 squared is going to be equal to 14 squared. And it's really important that you realize that it's not 9 squared plus 14 squared is going to be equal to a squared. a squared is one of the shorter sides. The sum of the squares of these two sides are going to be equal to 14 squared, the hypotenuse squared. And from here, we just have to solve for a. So we get a squared plus 81 is equal to 14 squared. And in case we don't know what that is, let's just multiply it out."}, {"video_title": "Pythagorean theorem 2   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "The sum of the squares of these two sides are going to be equal to 14 squared, the hypotenuse squared. And from here, we just have to solve for a. So we get a squared plus 81 is equal to 14 squared. And in case we don't know what that is, let's just multiply it out. 14 times 14. 4 times 4 is 16. 4 times 1 is 4 plus 1 is 5."}, {"video_title": "Pythagorean theorem 2   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "And in case we don't know what that is, let's just multiply it out. 14 times 14. 4 times 4 is 16. 4 times 1 is 4 plus 1 is 5. Stick a 0 there. 1 times 4 is 4. 1 times 1 is 1."}, {"video_title": "Pythagorean theorem 2   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "4 times 1 is 4 plus 1 is 5. Stick a 0 there. 1 times 4 is 4. 1 times 1 is 1. 6 plus 0 is 6. 5 plus 4 is 9. Bring down the 1."}, {"video_title": "Pythagorean theorem 2   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "1 times 1 is 1. 6 plus 0 is 6. 5 plus 4 is 9. Bring down the 1. It's 196. So a squared plus 81 is equal to 14 squared, which is 196. Then we could subtract 81 from both sides of this equation."}, {"video_title": "Pythagorean theorem 2   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "Bring down the 1. It's 196. So a squared plus 81 is equal to 14 squared, which is 196. Then we could subtract 81 from both sides of this equation. Let's subtract 81 from both sides. On the left-hand side, we're going to be left with just the a squared. These two guys cancel out."}, {"video_title": "Pythagorean theorem 2   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "Then we could subtract 81 from both sides of this equation. Let's subtract 81 from both sides. On the left-hand side, we're going to be left with just the a squared. These two guys cancel out. Whole point of subtracting 81. So we're left with a squared is equal to 196 minus 81. 196 minus 81."}, {"video_title": "Pythagorean theorem 2   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "These two guys cancel out. Whole point of subtracting 81. So we're left with a squared is equal to 196 minus 81. 196 minus 81. What is that? See, 190. So if you just subtract 1, 195."}, {"video_title": "Pythagorean theorem 2   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "196 minus 81. What is that? See, 190. So if you just subtract 1, 195. You subtract 80. It'd be 115 if I'm doing that right. It would be 115."}, {"video_title": "Pythagorean theorem 2   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "So if you just subtract 1, 195. You subtract 80. It'd be 115 if I'm doing that right. It would be 115. And then to solve for a, we just take the square root of both sides. The principal square root, the positive square root of both sides of this equation. So let's do that because we're dealing with distances."}, {"video_title": "Pythagorean theorem 2   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "It would be 115. And then to solve for a, we just take the square root of both sides. The principal square root, the positive square root of both sides of this equation. So let's do that because we're dealing with distances. You can't have a negative square root or a negative distance here. And we get a is equal to the square root of 115. Now let's see if we can break down 115 any further."}, {"video_title": "Pythagorean theorem 2   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "So let's do that because we're dealing with distances. You can't have a negative square root or a negative distance here. And we get a is equal to the square root of 115. Now let's see if we can break down 115 any further. So let's see. It's clearly divisible by 5. If you factor it out, it's 5."}, {"video_title": "Pythagorean theorem 2   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "Now let's see if we can break down 115 any further. So let's see. It's clearly divisible by 5. If you factor it out, it's 5. And then what is it? 5 goes into 115 23 times. So both of these are prime numbers."}, {"video_title": "Pythagorean theorem 2   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "If you factor it out, it's 5. And then what is it? 5 goes into 115 23 times. So both of these are prime numbers. So we're done. So you actually can't factor this anymore. So a is just going to be equal to the square root of 115."}, {"video_title": "Pythagorean theorem 2   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "So both of these are prime numbers. So we're done. So you actually can't factor this anymore. So a is just going to be equal to the square root of 115. Now if you want to get a sense of roughly how large the square root of 115 is, if you think about it, the square root of 100 is equal to 10. And the square root of 121 is equal to 11. So this value right here is going to be someplace in between 10 and 11, which makes sense if you think about it visually."}, {"video_title": "Opposite of a number   Negative numbers   6th grade   Khan Academy.mp3", "Sentence": "So let's draw a number line. And let's put some numbers on this number line. So we can start at zero, and if we go to the right, we have positive numbers. One, two, three, four, five. As we go to the left, we get more and more negative. So negative one, negative two, negative three, negative four, and I could keep going on and on and on. So let's pick one of these numbers."}, {"video_title": "Opposite of a number   Negative numbers   6th grade   Khan Academy.mp3", "Sentence": "One, two, three, four, five. As we go to the left, we get more and more negative. So negative one, negative two, negative three, negative four, and I could keep going on and on and on. So let's pick one of these numbers. Let's say that we pick the number three. What is going to be the opposite of the number three? Well, the opposite of the number is a number that's the same distance from zero, but on the other side."}, {"video_title": "Opposite of a number   Negative numbers   6th grade   Khan Academy.mp3", "Sentence": "So let's pick one of these numbers. Let's say that we pick the number three. What is going to be the opposite of the number three? Well, the opposite of the number is a number that's the same distance from zero, but on the other side. So three is three to the right of zero, one, two, three. So its opposite is going to be three to the left of zero, one, two, three. So the opposite of three is negative three."}, {"video_title": "Opposite of a number   Negative numbers   6th grade   Khan Academy.mp3", "Sentence": "Well, the opposite of the number is a number that's the same distance from zero, but on the other side. So three is three to the right of zero, one, two, three. So its opposite is going to be three to the left of zero, one, two, three. So the opposite of three is negative three. Let me make a little table here. So if we have the number, the number, and then we have its opposite. So we just figured out that if you have the number three, its opposite is going to be negative three."}, {"video_title": "Opposite of a number   Negative numbers   6th grade   Khan Academy.mp3", "Sentence": "So the opposite of three is negative three. Let me make a little table here. So if we have the number, the number, and then we have its opposite. So we just figured out that if you have the number three, its opposite is going to be negative three. Now what if your number is negative? What if your number, let's say the number negative four? What's the opposite of that?"}, {"video_title": "Opposite of a number   Negative numbers   6th grade   Khan Academy.mp3", "Sentence": "So we just figured out that if you have the number three, its opposite is going to be negative three. Now what if your number is negative? What if your number, let's say the number negative four? What's the opposite of that? And I encourage you to pause the video and try to think about it on your own. Well, you say, okay, negative four is right over here. That's negative four."}, {"video_title": "Opposite of a number   Negative numbers   6th grade   Khan Academy.mp3", "Sentence": "What's the opposite of that? And I encourage you to pause the video and try to think about it on your own. Well, you say, okay, negative four is right over here. That's negative four. It is four to the left of zero, one, two, three, four to the left of zero. So its opposite is going to be four to the right of it. So one, two, three, four."}, {"video_title": "Opposite of a number   Negative numbers   6th grade   Khan Academy.mp3", "Sentence": "That's negative four. It is four to the left of zero, one, two, three, four to the left of zero. So its opposite is going to be four to the right of it. So one, two, three, four. It's going to be positive. It's going to be positive four. And so you're probably starting to see a pattern here."}, {"video_title": "Opposite of a number   Negative numbers   6th grade   Khan Academy.mp3", "Sentence": "So one, two, three, four. It's going to be positive. It's going to be positive four. And so you're probably starting to see a pattern here. The opposite of a number is going to be the opposite sign of that number. If you had, if you have a positive three here, its opposite is going to be negative three. If you start with negative four, its opposite is going to be positive four."}, {"video_title": "Opposite of a number   Negative numbers   6th grade   Khan Academy.mp3", "Sentence": "And so you're probably starting to see a pattern here. The opposite of a number is going to be the opposite sign of that number. If you had, if you have a positive three here, its opposite is going to be negative three. If you start with negative four, its opposite is going to be positive four. So one way to think about it, it's going to have the same absolute value, but have a different sign. Or another way to think about it is, however, if this is three to the right of zero, its opposite's going to be three to the left of zero. Or if the number is four to the left of zero, its opposite is going to be four to the right of zero."}, {"video_title": "Opposite of a number   Negative numbers   6th grade   Khan Academy.mp3", "Sentence": "If you start with negative four, its opposite is going to be positive four. So one way to think about it, it's going to have the same absolute value, but have a different sign. Or another way to think about it is, however, if this is three to the right of zero, its opposite's going to be three to the left of zero. Or if the number is four to the left of zero, its opposite is going to be four to the right of zero. So we'll do one last one. What is the opposite of, what's going to be the opposite of one? Well, one is one to the right of zero, so its opposite is going to be one to the left of zero, or negative one."}, {"video_title": "Opposite of a number   Negative numbers   6th grade   Khan Academy.mp3", "Sentence": "Or if the number is four to the left of zero, its opposite is going to be four to the right of zero. So we'll do one last one. What is the opposite of, what's going to be the opposite of one? Well, one is one to the right of zero, so its opposite is going to be one to the left of zero, or negative one. Or another way to think about it, one is positive, so its opposite is, we're going to change the sign. Instead of being positive, it's going to be negative. It's going to be negative one."}, {"video_title": "Algebraic midpoint of a segment exercise   Geometry   8th grade   Khan Academy.mp3", "Sentence": "So that tells us that segment jk is going to be congruent to segment kl, that they're going to be the exact same length. And they tell us that segment jk is equal to 8x minus 8. So this distance right over here is equal to 8x minus 8. And then they tell us that segment kl is equal to 7x minus 6, that its length is equal to 7x minus 6. So this length right over here is 7x minus 6. Because k is the midpoint, we know that this length must be equal to this length. So to find jl, we just need to find the whole length."}, {"video_title": "Algebraic midpoint of a segment exercise   Geometry   8th grade   Khan Academy.mp3", "Sentence": "And then they tell us that segment kl is equal to 7x minus 6, that its length is equal to 7x minus 6. So this length right over here is 7x minus 6. Because k is the midpoint, we know that this length must be equal to this length. So to find jl, we just need to find the whole length. We need to find what x is. If we know what x is, then we're going to know what this length is and what this length is. And we could either double one of them or add them together to find the length of the entire segment."}, {"video_title": "Algebraic midpoint of a segment exercise   Geometry   8th grade   Khan Academy.mp3", "Sentence": "So to find jl, we just need to find the whole length. We need to find what x is. If we know what x is, then we're going to know what this length is and what this length is. And we could either double one of them or add them together to find the length of the entire segment. So first, let's figure out x. And the best way to figure out x is based on the fact that we know that 8x minus 8 is equal to 7x minus 6. And I just want to reemphasize, how do we know that?"}, {"video_title": "Algebraic midpoint of a segment exercise   Geometry   8th grade   Khan Academy.mp3", "Sentence": "And we could either double one of them or add them together to find the length of the entire segment. So first, let's figure out x. And the best way to figure out x is based on the fact that we know that 8x minus 8 is equal to 7x minus 6. And I just want to reemphasize, how do we know that? Well, they told us that k is the midpoint of jl. This is the midpoint, which tells us that this distance is equal to this distance, or 8x minus 8 is equal to 7x minus 6. Now to figure out x, we just have to do a little bit of algebra."}, {"video_title": "Algebraic midpoint of a segment exercise   Geometry   8th grade   Khan Academy.mp3", "Sentence": "And I just want to reemphasize, how do we know that? Well, they told us that k is the midpoint of jl. This is the midpoint, which tells us that this distance is equal to this distance, or 8x minus 8 is equal to 7x minus 6. Now to figure out x, we just have to do a little bit of algebra. So let's see what we can do to simplify things. So if we want to get all the x terms on one side, let's say we're going in on the left side, we could subtract 7x from both sides. So let's do that."}, {"video_title": "Algebraic midpoint of a segment exercise   Geometry   8th grade   Khan Academy.mp3", "Sentence": "Now to figure out x, we just have to do a little bit of algebra. So let's see what we can do to simplify things. So if we want to get all the x terms on one side, let's say we're going in on the left side, we could subtract 7x from both sides. So let's do that. Let's subtract 7x from both sides. And if we want to get all of the constant terms on one side, well, let's add 8 to both sides so that we don't have this negative 8 right over here. So let's add 8 to both sides."}, {"video_title": "Algebraic midpoint of a segment exercise   Geometry   8th grade   Khan Academy.mp3", "Sentence": "So let's do that. Let's subtract 7x from both sides. And if we want to get all of the constant terms on one side, well, let's add 8 to both sides so that we don't have this negative 8 right over here. So let's add 8 to both sides. And let's see what we are left with. On the left-hand side, you don't have an 8 anymore. And 8x minus 7x is just an x."}, {"video_title": "Algebraic midpoint of a segment exercise   Geometry   8th grade   Khan Academy.mp3", "Sentence": "So let's add 8 to both sides. And let's see what we are left with. On the left-hand side, you don't have an 8 anymore. And 8x minus 7x is just an x. And that's going to be equal to, on the right-hand side, you don't have any 7x's anymore. And negative 6 plus 8 is just 2. So we get x is equal to 2."}, {"video_title": "Algebraic midpoint of a segment exercise   Geometry   8th grade   Khan Academy.mp3", "Sentence": "And 8x minus 7x is just an x. And that's going to be equal to, on the right-hand side, you don't have any 7x's anymore. And negative 6 plus 8 is just 2. So we get x is equal to 2. But we're not done yet. They didn't say solve for x. They said find jl."}, {"video_title": "Algebraic midpoint of a segment exercise   Geometry   8th grade   Khan Academy.mp3", "Sentence": "So we get x is equal to 2. But we're not done yet. They didn't say solve for x. They said find jl. So jl is just going to be the sum of jk and kl. Or since k is a midpoint, it would just be double jk or double kl. So let's figure out either way."}, {"video_title": "Algebraic midpoint of a segment exercise   Geometry   8th grade   Khan Academy.mp3", "Sentence": "They said find jl. So jl is just going to be the sum of jk and kl. Or since k is a midpoint, it would just be double jk or double kl. So let's figure out either way. So now we can figure out that jk, the length of segment jk, is equal to 8 times 2 minus 8. We know that x is 2 now. So this is equal to, well, this is 16 minus 8."}, {"video_title": "Algebraic midpoint of a segment exercise   Geometry   8th grade   Khan Academy.mp3", "Sentence": "So let's figure out either way. So now we can figure out that jk, the length of segment jk, is equal to 8 times 2 minus 8. We know that x is 2 now. So this is equal to, well, this is 16 minus 8. This is just equal to, this right over here is just equal to 8. And if we wanted to figure out jl, we know that this is halfway. So that this must be 8 as well."}, {"video_title": "Algebraic midpoint of a segment exercise   Geometry   8th grade   Khan Academy.mp3", "Sentence": "So this is equal to, well, this is 16 minus 8. This is just equal to, this right over here is just equal to 8. And if we wanted to figure out jl, we know that this is halfway. So that this must be 8 as well. And the length of the entire thing, the length of jl, must be 16. And if you wanted to spend extra time to make sure that all of the math is consistent, you could put 2 into this right over here. And 7 times 2 is 14 minus 6 is 8."}, {"video_title": "Why aren't we using the multiplication sign   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "We now hopefully know a little bit about variables. And as we covered in the last video, a variable can be really any symbol, although we typically use letters because we're used to writing and typing letters. But it can be anything from an x to a y, a z, an a, a b. And so oftentimes, we'll start using Greek letters like theta. But you can really use any symbol to say, hey, this is going to vary. It can take on multiple values. But out of all of these, the one that's most typically used in algebra, or really in all of mathematics, is the variable x, although all of these are used to some degree."}, {"video_title": "Why aren't we using the multiplication sign   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And so oftentimes, we'll start using Greek letters like theta. But you can really use any symbol to say, hey, this is going to vary. It can take on multiple values. But out of all of these, the one that's most typically used in algebra, or really in all of mathematics, is the variable x, although all of these are used to some degree. But given that x is used so heavily, it does introduce a slight problem. And the problem is it looks a lot like the multiplication symbol or the one that we use in arithmetic. So in arithmetic, if I want to write 2 times 3, I literally write 2 times 3."}, {"video_title": "Why aren't we using the multiplication sign   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "But out of all of these, the one that's most typically used in algebra, or really in all of mathematics, is the variable x, although all of these are used to some degree. But given that x is used so heavily, it does introduce a slight problem. And the problem is it looks a lot like the multiplication symbol or the one that we use in arithmetic. So in arithmetic, if I want to write 2 times 3, I literally write 2 times 3. But now that we're starting to use variables and if I want to write 2 times x, well, if I use this as the multiplication symbol, it would be 2 times x. And the times symbol and the x look awfully similar. And if I'm not really careful with my penmanship, it can get very confusing."}, {"video_title": "Why aren't we using the multiplication sign   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "So in arithmetic, if I want to write 2 times 3, I literally write 2 times 3. But now that we're starting to use variables and if I want to write 2 times x, well, if I use this as the multiplication symbol, it would be 2 times x. And the times symbol and the x look awfully similar. And if I'm not really careful with my penmanship, it can get very confusing. Is this 2xx? Is this 2 times times something? What exactly is going on here?"}, {"video_title": "Why aren't we using the multiplication sign   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And if I'm not really careful with my penmanship, it can get very confusing. Is this 2xx? Is this 2 times times something? What exactly is going on here? And because this is confusing, this right over here is extremely confusing and it can be misinterpreted, we tend to not use this multiplication symbol when we are doing algebra. Instead of that, to represent multiplication, we have several options. Instead of writing 2 times x this way, we could write 2 dot x."}, {"video_title": "Why aren't we using the multiplication sign   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "What exactly is going on here? And because this is confusing, this right over here is extremely confusing and it can be misinterpreted, we tend to not use this multiplication symbol when we are doing algebra. Instead of that, to represent multiplication, we have several options. Instead of writing 2 times x this way, we could write 2 dot x. And this dot, I want to be very clear, this is not a decimal. This is written a little bit higher. And we write this so we don't get confusion between this and one of these variables right here."}, {"video_title": "Why aren't we using the multiplication sign   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Instead of writing 2 times x this way, we could write 2 dot x. And this dot, I want to be very clear, this is not a decimal. This is written a little bit higher. And we write this so we don't get confusion between this and one of these variables right here. But this really can be interpreted as 2 times x. So for example, if someone says 2 dot x, what is 2 dot x when x is equal to 3? Well, this would be the same thing as 2 times 3 when x is equal to 3."}, {"video_title": "Why aren't we using the multiplication sign   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And we write this so we don't get confusion between this and one of these variables right here. But this really can be interpreted as 2 times x. So for example, if someone says 2 dot x, what is 2 dot x when x is equal to 3? Well, this would be the same thing as 2 times 3 when x is equal to 3. Another way that you could write it is you could write 2 and then you could write the x in parentheses right next to it. This is also interpreted as 2 times x. Once again, so if in this situation x were 7, this would be 2 times 7 or 14."}, {"video_title": "Why aren't we using the multiplication sign   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Well, this would be the same thing as 2 times 3 when x is equal to 3. Another way that you could write it is you could write 2 and then you could write the x in parentheses right next to it. This is also interpreted as 2 times x. Once again, so if in this situation x were 7, this would be 2 times 7 or 14. And then the most traditional way of doing it is to just write the x right after the 2. And sometimes this will be read as 2x. But this literally does mean 2 times x."}, {"video_title": "Why aren't we using the multiplication sign   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Once again, so if in this situation x were 7, this would be 2 times 7 or 14. And then the most traditional way of doing it is to just write the x right after the 2. And sometimes this will be read as 2x. But this literally does mean 2 times x. And you might say, well, how come we didn't always do that? Well, it would be literally confusing if we did it over here. If we just wrote, instead of writing 2 times 3, we just wrote 2, 3, well, that looks like 23."}, {"video_title": "Why aren't we using the multiplication sign   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "But this literally does mean 2 times x. And you might say, well, how come we didn't always do that? Well, it would be literally confusing if we did it over here. If we just wrote, instead of writing 2 times 3, we just wrote 2, 3, well, that looks like 23. This doesn't look like 2 times 3. That's why we never did it. But here, since we're using a letter now, it's clear that this isn't part of that number."}, {"video_title": "Why aren't we using the multiplication sign   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "If we just wrote, instead of writing 2 times 3, we just wrote 2, 3, well, that looks like 23. This doesn't look like 2 times 3. That's why we never did it. But here, since we're using a letter now, it's clear that this isn't part of that number. This isn't 20-something. This is 2 times this variable x. So all of these are really the same expression, 2 times x, 2 times x, and 2 times x."}, {"video_title": "Why aren't we using the multiplication sign   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "But here, since we're using a letter now, it's clear that this isn't part of that number. This isn't 20-something. This is 2 times this variable x. So all of these are really the same expression, 2 times x, 2 times x, and 2 times x. And so with that out of the way, let's try a few worked examples, a few practice problems. And this will hopefully prepare you for the next exercise where you get a lot of chance to practice this. So if I were to say, what is 10 minus 3y?"}, {"video_title": "Why aren't we using the multiplication sign   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "So all of these are really the same expression, 2 times x, 2 times x, and 2 times x. And so with that out of the way, let's try a few worked examples, a few practice problems. And this will hopefully prepare you for the next exercise where you get a lot of chance to practice this. So if I were to say, what is 10 minus 3y? And what does this equal when y is equal to 2? Well, every time you see the y, you'd want that 2 there. So this is when y is equal to 2."}, {"video_title": "Why aren't we using the multiplication sign   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "So if I were to say, what is 10 minus 3y? And what does this equal when y is equal to 2? Well, every time you see the y, you'd want that 2 there. So this is when y is equal to 2. Let's set that y equal to 2. This is the same thing as 10 minus 3 times 2. You do the multiplication first."}, {"video_title": "Why aren't we using the multiplication sign   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "So this is when y is equal to 2. Let's set that y equal to 2. This is the same thing as 10 minus 3 times 2. You do the multiplication first. Multiplication takes precedence in order of operations. So 3 times 2 is 6. 10 minus 6 is equal to 4."}, {"video_title": "Why aren't we using the multiplication sign   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "You do the multiplication first. Multiplication takes precedence in order of operations. So 3 times 2 is 6. 10 minus 6 is equal to 4. Let's do another one. Let's say that we had 7x minus 4. And let me do it in that same green color."}, {"video_title": "Why aren't we using the multiplication sign   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "10 minus 6 is equal to 4. Let's do another one. Let's say that we had 7x minus 4. And let me do it in that same green color. 7x minus 4. And we want to evaluate that when x is equal to 3. Where will we see the x?"}, {"video_title": "Why aren't we using the multiplication sign   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And let me do it in that same green color. 7x minus 4. And we want to evaluate that when x is equal to 3. Where will we see the x? We want to put a 3 there. So this is the same thing as 7 times 3. And I'll actually use this notation, because I can use that with numbers."}, {"video_title": "Why aren't we using the multiplication sign   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Where will we see the x? We want to put a 3 there. So this is the same thing as 7 times 3. And I'll actually use this notation, because I can use that with numbers. 7 times 3 minus 4. And once again, multiplication takes precedence by order of operations over addition or subtraction. So we want to do the multiplying first."}, {"video_title": "Why aren't we using the multiplication sign   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And I'll actually use this notation, because I can use that with numbers. 7 times 3 minus 4. And once again, multiplication takes precedence by order of operations over addition or subtraction. So we want to do the multiplying first. 7 times 3 is 21. 21 minus 4 is equal to 17. So hopefully that gives you a little bit of background."}, {"video_title": "Multiplying positive and negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "we know that if we were to multiply two times three, that would give us positive six. And since we're gonna start thinking about negative numbers in this video, one way to think about it is I have a positive number times another positive number, and that gave me a positive number. So if I have a positive times a positive, that will give me a positive number. Now let's mix it up a little bit. Introduce some negative numbers. So what happens if I had negative two, negative two times three, negative two times three? Well, one way to think about it, and we'll talk more about the intuition in this video and in future videos is, well, you could view this as negative two repeatedly added three times."}, {"video_title": "Multiplying positive and negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "Now let's mix it up a little bit. Introduce some negative numbers. So what happens if I had negative two, negative two times three, negative two times three? Well, one way to think about it, and we'll talk more about the intuition in this video and in future videos is, well, you could view this as negative two repeatedly added three times. So this could be negative two, negative two plus negative two plus negative two, not negative six, plus a negative two, which would be equal to, well, negative two plus negative two is negative four plus another negative two is negative six. So this would be equal to negative negative six. Or another way to think about it is if I had two times three, I would get six, but because one of these two numbers is negative, then my product is going to be negative."}, {"video_title": "Multiplying positive and negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "Well, one way to think about it, and we'll talk more about the intuition in this video and in future videos is, well, you could view this as negative two repeatedly added three times. So this could be negative two, negative two plus negative two plus negative two, not negative six, plus a negative two, which would be equal to, well, negative two plus negative two is negative four plus another negative two is negative six. So this would be equal to negative negative six. Or another way to think about it is if I had two times three, I would get six, but because one of these two numbers is negative, then my product is going to be negative. So if I multiply a negative, a negative times a positive, I am going to get a, I am going to get a negative. Now what if we swap the order in which we multiply? So if we were to multiply three, three times negative two."}, {"video_title": "Multiplying positive and negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "Or another way to think about it is if I had two times three, I would get six, but because one of these two numbers is negative, then my product is going to be negative. So if I multiply a negative, a negative times a positive, I am going to get a, I am going to get a negative. Now what if we swap the order in which we multiply? So if we were to multiply three, three times negative two. Well, it shouldn't matter. The order in which we multiply things don't change, or it shouldn't change the product. Whether we multiply two times three, we'll get six, or if we multiply three times two, we'll get six."}, {"video_title": "Multiplying positive and negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "So if we were to multiply three, three times negative two. Well, it shouldn't matter. The order in which we multiply things don't change, or it shouldn't change the product. Whether we multiply two times three, we'll get six, or if we multiply three times two, we'll get six. And so we should have the same property here. Three times negative two should give us the same result. It's going to be equal to, it's going to be equal to negative six."}, {"video_title": "Multiplying positive and negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "Whether we multiply two times three, we'll get six, or if we multiply three times two, we'll get six. And so we should have the same property here. Three times negative two should give us the same result. It's going to be equal to, it's going to be equal to negative six. And once again, we say three times two would be six. One of these two numbers is negative, and so our product is going to be negative. So we could write a positive times a negative is also going to be a negative."}, {"video_title": "Multiplying positive and negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "It's going to be equal to, it's going to be equal to negative six. And once again, we say three times two would be six. One of these two numbers is negative, and so our product is going to be negative. So we could write a positive times a negative is also going to be a negative. And both of these are just the same thing with the order in which we're multiplying switched around, but this is one of the two numbers are negative, exactly one. So one negative, one negative, one positive, one positive number is being multiplied. Then you will get a negative product."}, {"video_title": "Multiplying positive and negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "So we could write a positive times a negative is also going to be a negative. And both of these are just the same thing with the order in which we're multiplying switched around, but this is one of the two numbers are negative, exactly one. So one negative, one negative, one positive, one positive number is being multiplied. Then you will get a negative product. Now let's think about the third circumstance when both of the numbers are negative. So if I were to multiply, I'll just switch colors for fun here. If I were to multiply negative two times negative three, and this might be the least intuitive for you of all, and here I'm just going to introduce you to the rule, and in future videos we'll explore why this is and why this makes mathematics more all fit together."}, {"video_title": "Multiplying positive and negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "Then you will get a negative product. Now let's think about the third circumstance when both of the numbers are negative. So if I were to multiply, I'll just switch colors for fun here. If I were to multiply negative two times negative three, and this might be the least intuitive for you of all, and here I'm just going to introduce you to the rule, and in future videos we'll explore why this is and why this makes mathematics more all fit together. But this is going to be, you say, well two times three would be six, and I have a negative times a negative, and one way you can think about it is that the negatives cancel out. And so you will actually end up with a positive six. I actually don't have to write a positive here, but I'll write it here just to reemphasize."}, {"video_title": "Multiplying positive and negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "If I were to multiply negative two times negative three, and this might be the least intuitive for you of all, and here I'm just going to introduce you to the rule, and in future videos we'll explore why this is and why this makes mathematics more all fit together. But this is going to be, you say, well two times three would be six, and I have a negative times a negative, and one way you can think about it is that the negatives cancel out. And so you will actually end up with a positive six. I actually don't have to write a positive here, but I'll write it here just to reemphasize. This right over here is a positive six. So we have another rule of thumb here. If I have a negative times a negative, that they're actually going to, the negatives are going to cancel out, and that's going to give me a positive number."}, {"video_title": "Multiplying positive and negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "I actually don't have to write a positive here, but I'll write it here just to reemphasize. This right over here is a positive six. So we have another rule of thumb here. If I have a negative times a negative, that they're actually going to, the negatives are going to cancel out, and that's going to give me a positive number. Now with these out of the way, let's just do a bunch of examples, and I encourage you to try them out before I do them. Pause the video, try them out, and see if you get the same answer. So let's try negative one times negative one."}, {"video_title": "Multiplying positive and negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "If I have a negative times a negative, that they're actually going to, the negatives are going to cancel out, and that's going to give me a positive number. Now with these out of the way, let's just do a bunch of examples, and I encourage you to try them out before I do them. Pause the video, try them out, and see if you get the same answer. So let's try negative one times negative one. Well, one times one would be one, and we have a negative times a negative. They cancel out. Negative times a negative give me a positive, so this is going to be positive one."}, {"video_title": "Multiplying positive and negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "So let's try negative one times negative one. Well, one times one would be one, and we have a negative times a negative. They cancel out. Negative times a negative give me a positive, so this is going to be positive one. I could just write one, or I could literally write a plus sign there to emphasize that this is a positive one. What happens if I did negative one times zero? Now this might say, wait, this doesn't really fit into any of these circumstances."}, {"video_title": "Multiplying positive and negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "Negative times a negative give me a positive, so this is going to be positive one. I could just write one, or I could literally write a plus sign there to emphasize that this is a positive one. What happens if I did negative one times zero? Now this might say, wait, this doesn't really fit into any of these circumstances. Zero is neither positive nor negative, and here you just have to remember anything times zero is going to be zero. So negative one times zero is going to be zero, or I could have said zero times negative 783. That is also going to be zero."}, {"video_title": "Multiplying positive and negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "Now this might say, wait, this doesn't really fit into any of these circumstances. Zero is neither positive nor negative, and here you just have to remember anything times zero is going to be zero. So negative one times zero is going to be zero, or I could have said zero times negative 783. That is also going to be zero. Now what about two, let me do some interesting ones. What about, I'll pick a new color, 12 times negative four? Well, once again, 12 times positive four would be 48, and we're in the circumstance where one of these two numbers right over here is negative."}, {"video_title": "Multiplying positive and negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "That is also going to be zero. Now what about two, let me do some interesting ones. What about, I'll pick a new color, 12 times negative four? Well, once again, 12 times positive four would be 48, and we're in the circumstance where one of these two numbers right over here is negative. This one right over here. If exactly one of the two numbers is negative, then the product is going to be negative. We are in this circumstance."}, {"video_title": "Multiplying positive and negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "Well, once again, 12 times positive four would be 48, and we're in the circumstance where one of these two numbers right over here is negative. This one right over here. If exactly one of the two numbers is negative, then the product is going to be negative. We are in this circumstance. We are in this circumstance right over here. We have one negative, so the product is negative. You can imagine this as repeatedly adding negative four 12 times, and so you would get to negative 48."}, {"video_title": "Multiplying positive and negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "We are in this circumstance. We are in this circumstance right over here. We have one negative, so the product is negative. You can imagine this as repeatedly adding negative four 12 times, and so you would get to negative 48. Let's do another one. What is seven times three? Well, this is a bit of a trick."}, {"video_title": "Multiplying positive and negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "You can imagine this as repeatedly adding negative four 12 times, and so you would get to negative 48. Let's do another one. What is seven times three? Well, this is a bit of a trick. There are no negative numbers here. This is just going to be seven times three, positive seven times positive three. The first circumstance, which you already knew how to do before this video, this would just be equal to 21."}, {"video_title": "Multiplying positive and negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "Well, this is a bit of a trick. There are no negative numbers here. This is just going to be seven times three, positive seven times positive three. The first circumstance, which you already knew how to do before this video, this would just be equal to 21. Let's do one more. So if I were to say negative five times negative 10. Well, once again, negative times a negative, the negatives cancel out."}, {"video_title": "Multiplying positive and negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "The first circumstance, which you already knew how to do before this video, this would just be equal to 21. Let's do one more. So if I were to say negative five times negative 10. Well, once again, negative times a negative, the negatives cancel out. Then you're just left with a positive product, so it's gonna be five times 10. It's gonna be 50. The negative and the negative cancel out."}, {"video_title": "Finding the slope of a line from its graph   Algebra I   Khan Academy.mp3", "Sentence": "Find the slope of the line in the graph. And just as a bit of a review, slope is just telling us how steep a line is. And the best way to view it, slope is equal to change in y, change in y over change in x. Over change in x. And for a line, this will always be constant. And sometimes you might see it written like this. You might see this triangle."}, {"video_title": "Finding the slope of a line from its graph   Algebra I   Khan Academy.mp3", "Sentence": "Over change in x. And for a line, this will always be constant. And sometimes you might see it written like this. You might see this triangle. That's a capital delta. That means change in y over change in x. That's just a fancy way of saying change in y over change in x."}, {"video_title": "Finding the slope of a line from its graph   Algebra I   Khan Academy.mp3", "Sentence": "You might see this triangle. That's a capital delta. That means change in y over change in x. That's just a fancy way of saying change in y over change in x. So let's see what this change in y is for any change in x. So let's start at some point that seems pretty reasonable to read from this table right here, from this graph. So let's see, we're starting here."}, {"video_title": "Finding the slope of a line from its graph   Algebra I   Khan Academy.mp3", "Sentence": "That's just a fancy way of saying change in y over change in x. So let's see what this change in y is for any change in x. So let's start at some point that seems pretty reasonable to read from this table right here, from this graph. So let's see, we're starting here. Let me do it in a more vibrant color. So let's say we start at that point right there. And we want to go to another point that's pretty straightforward to read."}, {"video_title": "Finding the slope of a line from its graph   Algebra I   Khan Academy.mp3", "Sentence": "So let's see, we're starting here. Let me do it in a more vibrant color. So let's say we start at that point right there. And we want to go to another point that's pretty straightforward to read. So we can move to that point right there. We can literally pick any two points on this line. I'm just picking ones that are at nice integer coordinates so it's easy to read."}, {"video_title": "Finding the slope of a line from its graph   Algebra I   Khan Academy.mp3", "Sentence": "And we want to go to another point that's pretty straightforward to read. So we can move to that point right there. We can literally pick any two points on this line. I'm just picking ones that are at nice integer coordinates so it's easy to read. So what is the change in y and what is the change in x? So first let's look at the change in x. So if we go from there to there, what is the change in x?"}, {"video_title": "Finding the slope of a line from its graph   Algebra I   Khan Academy.mp3", "Sentence": "I'm just picking ones that are at nice integer coordinates so it's easy to read. So what is the change in y and what is the change in x? So first let's look at the change in x. So if we go from there to there, what is the change in x? My change in x is equal to what? Well, I could just count it out. I went one step, two steps, three steps."}, {"video_title": "Finding the slope of a line from its graph   Algebra I   Khan Academy.mp3", "Sentence": "So if we go from there to there, what is the change in x? My change in x is equal to what? Well, I could just count it out. I went one step, two steps, three steps. My change in x is 3. And you could even see it from the x values. If I go from negative 3 to 0, I went up by 3."}, {"video_title": "Finding the slope of a line from its graph   Algebra I   Khan Academy.mp3", "Sentence": "I went one step, two steps, three steps. My change in x is 3. And you could even see it from the x values. If I go from negative 3 to 0, I went up by 3. So my change in x is 3. So let me write this. Change in x delta x is equal to 3."}, {"video_title": "Finding the slope of a line from its graph   Algebra I   Khan Academy.mp3", "Sentence": "If I go from negative 3 to 0, I went up by 3. So my change in x is 3. So let me write this. Change in x delta x is equal to 3. And what's my change in y? Well, my change in y, I'm going from negative 3 up to negative 1. Or you could just say 1, 2."}, {"video_title": "Finding the slope of a line from its graph   Algebra I   Khan Academy.mp3", "Sentence": "Change in x delta x is equal to 3. And what's my change in y? Well, my change in y, I'm going from negative 3 up to negative 1. Or you could just say 1, 2. So my change in y is equal to positive 2. So let me write that down. Change in y is equal to 2."}, {"video_title": "Finding the slope of a line from its graph   Algebra I   Khan Academy.mp3", "Sentence": "Or you could just say 1, 2. So my change in y is equal to positive 2. So let me write that down. Change in y is equal to 2. So what is my change in y for a change in x? Well, when my change in x was 3, my change in y is 2. So this is my slope."}, {"video_title": "Finding the slope of a line from its graph   Algebra I   Khan Academy.mp3", "Sentence": "Change in y is equal to 2. So what is my change in y for a change in x? Well, when my change in x was 3, my change in y is 2. So this is my slope. And one thing I want to do, I want to show you that I could have really picked any two points here. Let's say I didn't pick those two points. Let me pick some other points."}, {"video_title": "Finding the slope of a line from its graph   Algebra I   Khan Academy.mp3", "Sentence": "So this is my slope. And one thing I want to do, I want to show you that I could have really picked any two points here. Let's say I didn't pick those two points. Let me pick some other points. And I'll even go in a different direction. And I want to show you that you're going to get the same answer. Let's say I viewed this as my starting point."}, {"video_title": "Finding the slope of a line from its graph   Algebra I   Khan Academy.mp3", "Sentence": "Let me pick some other points. And I'll even go in a different direction. And I want to show you that you're going to get the same answer. Let's say I viewed this as my starting point. And I want to go all the way over there. So what is my, well, let's think about the change in y first. So the change in y, I am going down by how many units?"}, {"video_title": "Finding the slope of a line from its graph   Algebra I   Khan Academy.mp3", "Sentence": "Let's say I viewed this as my starting point. And I want to go all the way over there. So what is my, well, let's think about the change in y first. So the change in y, I am going down by how many units? 1, 2, 3, 4 units. So my change in y in this example is negative 4. I went from 1 to negative 3."}, {"video_title": "Finding the slope of a line from its graph   Algebra I   Khan Academy.mp3", "Sentence": "So the change in y, I am going down by how many units? 1, 2, 3, 4 units. So my change in y in this example is negative 4. I went from 1 to negative 3. That's negative 4. That's my change in y. Change in y is equal to negative 4."}, {"video_title": "Finding the slope of a line from its graph   Algebra I   Khan Academy.mp3", "Sentence": "I went from 1 to negative 3. That's negative 4. That's my change in y. Change in y is equal to negative 4. Now what is my change in x? Well, I'm going from this point, or from this x value, all the way back like this. So I'm going to the left."}, {"video_title": "Finding the slope of a line from its graph   Algebra I   Khan Academy.mp3", "Sentence": "Change in y is equal to negative 4. Now what is my change in x? Well, I'm going from this point, or from this x value, all the way back like this. So I'm going to the left. So it's going to be a negative change in x. And I went 1, 2, 3, 4, 5, 6 units back. So my change in x is equal to negative 6."}, {"video_title": "Finding the slope of a line from its graph   Algebra I   Khan Academy.mp3", "Sentence": "So I'm going to the left. So it's going to be a negative change in x. And I went 1, 2, 3, 4, 5, 6 units back. So my change in x is equal to negative 6. Change in x is equal to negative 6. And you can even see I started at an x is equal to 3. And I went all the way to x is equal to a negative 3."}, {"video_title": "Finding the slope of a line from its graph   Algebra I   Khan Academy.mp3", "Sentence": "So my change in x is equal to negative 6. Change in x is equal to negative 6. And you can even see I started at an x is equal to 3. And I went all the way to x is equal to a negative 3. That's a change of negative 6. I went 6 to the left, or a change of negative 6. So what is my change in y over change in x?"}, {"video_title": "Finding the slope of a line from its graph   Algebra I   Khan Academy.mp3", "Sentence": "And I went all the way to x is equal to a negative 3. That's a change of negative 6. I went 6 to the left, or a change of negative 6. So what is my change in y over change in x? My change in y over change in x is equal to negative 4 over negative 6. The negatives cancel out. And what's 4 over 6?"}, {"video_title": "Finding the slope of a line from its graph   Algebra I   Khan Academy.mp3", "Sentence": "So what is my change in y over change in x? My change in y over change in x is equal to negative 4 over negative 6. The negatives cancel out. And what's 4 over 6? Well, that's just 2 over 3. So it's the same value. You just have to be consistent."}, {"video_title": "Finding the slope of a line from its graph   Algebra I   Khan Academy.mp3", "Sentence": "And what's 4 over 6? Well, that's just 2 over 3. So it's the same value. You just have to be consistent. If this is my start point, I went down 4. And then I went back 6. Negative 4 over negative 6."}, {"video_title": "Finding the slope of a line from its graph   Algebra I   Khan Academy.mp3", "Sentence": "You just have to be consistent. If this is my start point, I went down 4. And then I went back 6. Negative 4 over negative 6. If I viewed this as my starting point, I could say that I went up 4. So it would be a change in y would be 4. And then my change in x would be 6."}, {"video_title": "How to create a function from an equation (example)   Functions   Algebra I   Khan Academy.mp3", "Sentence": "For a given input value b, the function f outputs a value a to satisfy the following equation. Four a plus seven b is equal to negative 52. So for a given input b, the function f, the function f will output an a that satisfies this relationship right over here for the a and the b. Write a formula for f of b in terms of b. So what we wanna do is we just wanna solve for, if we're given a b, what a does that imply that we have to output? Or another way to think about it is, let's just solve this for a. Or we could think about a as being a function of b."}, {"video_title": "How to create a function from an equation (example)   Functions   Algebra I   Khan Academy.mp3", "Sentence": "Write a formula for f of b in terms of b. So what we wanna do is we just wanna solve for, if we're given a b, what a does that imply that we have to output? Or another way to think about it is, let's just solve this for a. Or we could think about a as being a function of b. So let's write this. So we have four a plus seven b is equal to negative 52. So if I can solve for a in terms of b, then any b that I have, let's say the b's are on the right-hand side, I can put it in, I can substitute that value for b, and then I can just solve for a. I can solve for the a that needs to be outputted."}, {"video_title": "How to create a function from an equation (example)   Functions   Algebra I   Khan Academy.mp3", "Sentence": "Or we could think about a as being a function of b. So let's write this. So we have four a plus seven b is equal to negative 52. So if I can solve for a in terms of b, then any b that I have, let's say the b's are on the right-hand side, I can put it in, I can substitute that value for b, and then I can just solve for a. I can solve for the a that needs to be outputted. So let's do that. Let's solve for a. So I wanna get all the a, I wanna just have an a left over on the left-hand side."}, {"video_title": "How to create a function from an equation (example)   Functions   Algebra I   Khan Academy.mp3", "Sentence": "So if I can solve for a in terms of b, then any b that I have, let's say the b's are on the right-hand side, I can put it in, I can substitute that value for b, and then I can just solve for a. I can solve for the a that needs to be outputted. So let's do that. Let's solve for a. So I wanna get all the a, I wanna just have an a left over on the left-hand side. I wanna have everything else on the right-hand side, including the b's. So let's get rid of this b on the left-hand side. And I can do that by subtracting seven b."}, {"video_title": "How to create a function from an equation (example)   Functions   Algebra I   Khan Academy.mp3", "Sentence": "So I wanna get all the a, I wanna just have an a left over on the left-hand side. I wanna have everything else on the right-hand side, including the b's. So let's get rid of this b on the left-hand side. And I can do that by subtracting seven b. Now, of course, I wanna do that to both sides. I can't just do an equation and do an operation only on one side like that. So let's subtract, and we are left with, we are left with the seven b's add up to zero, seven b minus seven b, and we are left with four a is equal to negative 52 minus seven b, minus seven b."}, {"video_title": "How to create a function from an equation (example)   Functions   Algebra I   Khan Academy.mp3", "Sentence": "And I can do that by subtracting seven b. Now, of course, I wanna do that to both sides. I can't just do an equation and do an operation only on one side like that. So let's subtract, and we are left with, we are left with the seven b's add up to zero, seven b minus seven b, and we are left with four a is equal to negative 52 minus seven b, minus seven b. Now, to isolate the a here, to just have an a here instead of a four a, we can divide both sides by four. We can divide both sides by four, so I'm just gonna divide everything by four. And on the left-hand side, we got our goal."}, {"video_title": "How to create a function from an equation (example)   Functions   Algebra I   Khan Academy.mp3", "Sentence": "So let's subtract, and we are left with, we are left with the seven b's add up to zero, seven b minus seven b, and we are left with four a is equal to negative 52 minus seven b, minus seven b. Now, to isolate the a here, to just have an a here instead of a four a, we can divide both sides by four. We can divide both sides by four, so I'm just gonna divide everything by four. And on the left-hand side, we got our goal. We are left with an a is equal to, now what's negative 54 divided by, what's negative 52 divided by four? So let's think about it. 52 is 40 plus 12."}, {"video_title": "How to create a function from an equation (example)   Functions   Algebra I   Khan Academy.mp3", "Sentence": "And on the left-hand side, we got our goal. We are left with an a is equal to, now what's negative 54 divided by, what's negative 52 divided by four? So let's think about it. 52 is 40 plus 12. 40 divided by four is 10, 12 divided by four is three, so it's gonna be 13, negative 13. So it's negative 13 minus seven for its b, minus seven for its b. So given a b, if you give me a b, I can put that value right over here, and then I can calculate what the corresponding a needs to be in order to satisfy this relationship."}, {"video_title": "How to create a function from an equation (example)   Functions   Algebra I   Khan Academy.mp3", "Sentence": "52 is 40 plus 12. 40 divided by four is 10, 12 divided by four is three, so it's gonna be 13, negative 13. So it's negative 13 minus seven for its b, minus seven for its b. So given a b, if you give me a b, I can put that value right over here, and then I can calculate what the corresponding a needs to be in order to satisfy this relationship. So if I want a formula for f of b in terms of b, I could say, look, you give me a b, the output of our function, which is f of a, the output of our function is going to be, is going to be negative 13 minus seven for its b. Because the output of our function needs to be an a that will satisfy, that will satisfy this equation up here. So hopefully that helped."}, {"video_title": "Intervals and interval notation   Functions   Algebra I   Khan Academy.mp3", "Sentence": "Now let's say I wanted to talk about the interval on the number line that goes from, let's say, from negative three to two. So I care about this, let me do this in a different color. Let's say I care about this interval right over here. So I care about all of the numbers from negative three, negative three to two. So in order to be more precise, I have to be clear. Am I including negative three and two, or am I not including negative three and two? Or maybe I'm just including one of them."}, {"video_title": "Intervals and interval notation   Functions   Algebra I   Khan Academy.mp3", "Sentence": "So I care about all of the numbers from negative three, negative three to two. So in order to be more precise, I have to be clear. Am I including negative three and two, or am I not including negative three and two? Or maybe I'm just including one of them. So if I'm including negative three and two, then I would fill them in. So this right over here, I'm filling negative three and two in, which means that negative three and two are part of this interval. And when you include the end points, this is called a closed interval."}, {"video_title": "Intervals and interval notation   Functions   Algebra I   Khan Academy.mp3", "Sentence": "Or maybe I'm just including one of them. So if I'm including negative three and two, then I would fill them in. So this right over here, I'm filling negative three and two in, which means that negative three and two are part of this interval. And when you include the end points, this is called a closed interval. Closed interval. Closed interval. And I just showed you how I can depict it on a number line by actually filling in the end points."}, {"video_title": "Intervals and interval notation   Functions   Algebra I   Khan Academy.mp3", "Sentence": "And when you include the end points, this is called a closed interval. Closed interval. Closed interval. And I just showed you how I can depict it on a number line by actually filling in the end points. And there's multiple ways to talk about this interval mathematically. I could say that this is all of the, let's say that this is, let's say this number line is showing different values for x. I could say these are all of the x's that are between negative three, negative three, and two, and two. And notice, I have negative three is less than or equal to x."}, {"video_title": "Intervals and interval notation   Functions   Algebra I   Khan Academy.mp3", "Sentence": "And I just showed you how I can depict it on a number line by actually filling in the end points. And there's multiple ways to talk about this interval mathematically. I could say that this is all of the, let's say that this is, let's say this number line is showing different values for x. I could say these are all of the x's that are between negative three, negative three, and two, and two. And notice, I have negative three is less than or equal to x. So that's telling us that x could be equal to, that x could be equal to negative three. And then we have x is less than or equal to positive two. So that means that x could be equal to positive two."}, {"video_title": "Intervals and interval notation   Functions   Algebra I   Khan Academy.mp3", "Sentence": "And notice, I have negative three is less than or equal to x. So that's telling us that x could be equal to, that x could be equal to negative three. And then we have x is less than or equal to positive two. So that means that x could be equal to positive two. So it is a closed interval. Now another way that we could depict this closed interval is we could say, okay, we're talking about the interval between, and we could use brackets because it's a closed interval, negative three and two. And once again, I'm using brackets here."}, {"video_title": "Intervals and interval notation   Functions   Algebra I   Khan Academy.mp3", "Sentence": "So that means that x could be equal to positive two. So it is a closed interval. Now another way that we could depict this closed interval is we could say, okay, we're talking about the interval between, and we could use brackets because it's a closed interval, negative three and two. And once again, I'm using brackets here. These brackets tell us that we include. So this bracket on the left say that we include negative three, and this bracket on the right says that we include positive two in our interval. Now sometimes you might see things written a little bit more math."}, {"video_title": "Intervals and interval notation   Functions   Algebra I   Khan Academy.mp3", "Sentence": "And once again, I'm using brackets here. These brackets tell us that we include. So this bracket on the left say that we include negative three, and this bracket on the right says that we include positive two in our interval. Now sometimes you might see things written a little bit more math. You might see something like x is a member of the real numbers such that, so, and I could put these curly brackets around like this. These curly brackets say that we're talking about a set of values, and we're saying the set of all x's that are a member of the real numbers, so this is just fancy math notation to say member of the real numbers. I'm using the Greek letter epsilon right over here to say member of the real numbers such that, this little, this vertical line here means such that negative three is less than x is less than, negative three is less than or equal to x is less than or equal to two."}, {"video_title": "Intervals and interval notation   Functions   Algebra I   Khan Academy.mp3", "Sentence": "Now sometimes you might see things written a little bit more math. You might see something like x is a member of the real numbers such that, so, and I could put these curly brackets around like this. These curly brackets say that we're talking about a set of values, and we're saying the set of all x's that are a member of the real numbers, so this is just fancy math notation to say member of the real numbers. I'm using the Greek letter epsilon right over here to say member of the real numbers such that, this little, this vertical line here means such that negative three is less than x is less than, negative three is less than or equal to x is less than or equal to two. I could also write it this way. I could write x is a member of the real numbers such that x is a member, such that x is a member of this closed set. I'm including the end points here."}, {"video_title": "Intervals and interval notation   Functions   Algebra I   Khan Academy.mp3", "Sentence": "I'm using the Greek letter epsilon right over here to say member of the real numbers such that, this little, this vertical line here means such that negative three is less than x is less than, negative three is less than or equal to x is less than or equal to two. I could also write it this way. I could write x is a member of the real numbers such that x is a member, such that x is a member of this closed set. I'm including the end points here. So these are all different ways of denoting or depicting the same interval. Let's do some more examples here. Let me draw a number line again."}, {"video_title": "Intervals and interval notation   Functions   Algebra I   Khan Academy.mp3", "Sentence": "I'm including the end points here. So these are all different ways of denoting or depicting the same interval. Let's do some more examples here. Let me draw a number line again. So number line. Now let me do, let me just do an open interval, an open interval just so that we see, we clearly can see the difference. Let's say that I want to talk about the values between negative one and four."}, {"video_title": "Intervals and interval notation   Functions   Algebra I   Khan Academy.mp3", "Sentence": "Let me draw a number line again. So number line. Now let me do, let me just do an open interval, an open interval just so that we see, we clearly can see the difference. Let's say that I want to talk about the values between negative one and four. Negative, let me do this in a different color. The values between negative one and four, but I don't want to include negative one and four. So this is going to be an open interval."}, {"video_title": "Intervals and interval notation   Functions   Algebra I   Khan Academy.mp3", "Sentence": "Let's say that I want to talk about the values between negative one and four. Negative, let me do this in a different color. The values between negative one and four, but I don't want to include negative one and four. So this is going to be an open interval. So I'm not going to include four and I'm not going to include negative one. Notice I have open circles here. Over here I had closed circles."}, {"video_title": "Intervals and interval notation   Functions   Algebra I   Khan Academy.mp3", "Sentence": "So this is going to be an open interval. So I'm not going to include four and I'm not going to include negative one. Notice I have open circles here. Over here I had closed circles. The closed circles told me that I included negative three and two. Now I have open circles here. So that says that I'm not, it's all the values in between negative one and four."}, {"video_title": "Intervals and interval notation   Functions   Algebra I   Khan Academy.mp3", "Sentence": "Over here I had closed circles. The closed circles told me that I included negative three and two. Now I have open circles here. So that says that I'm not, it's all the values in between negative one and four. So negative one or negative.99999999 is going to be included, but negative one is not going to be included. And 3.99999999 is going to be included, but four is not going to be included. So how would we, what would be the notation for this?"}, {"video_title": "Intervals and interval notation   Functions   Algebra I   Khan Academy.mp3", "Sentence": "So that says that I'm not, it's all the values in between negative one and four. So negative one or negative.99999999 is going to be included, but negative one is not going to be included. And 3.99999999 is going to be included, but four is not going to be included. So how would we, what would be the notation for this? Well here we could say, x is going to be a member of the real numbers such that negative one, I'm not going to say less than or equal to because x can't be equal to negative one. So negative one is strictly less than x, is strictly less than four. Notice, not less than or equal because I can't be equal to four."}, {"video_title": "Intervals and interval notation   Functions   Algebra I   Khan Academy.mp3", "Sentence": "So how would we, what would be the notation for this? Well here we could say, x is going to be a member of the real numbers such that negative one, I'm not going to say less than or equal to because x can't be equal to negative one. So negative one is strictly less than x, is strictly less than four. Notice, not less than or equal because I can't be equal to four. Four is not included. So that's one way to say it. Or another way, I could write it like this."}, {"video_title": "Intervals and interval notation   Functions   Algebra I   Khan Academy.mp3", "Sentence": "Notice, not less than or equal because I can't be equal to four. Four is not included. So that's one way to say it. Or another way, I could write it like this. x is a member of the real numbers such that x is a member of, now the interval is from negative one to four, but I'm not going to use these brackets. These brackets say, hey, let me include the endpoints, but I'm not going to include them. So I'm going to put the parentheses right over here."}, {"video_title": "Intervals and interval notation   Functions   Algebra I   Khan Academy.mp3", "Sentence": "Or another way, I could write it like this. x is a member of the real numbers such that x is a member of, now the interval is from negative one to four, but I'm not going to use these brackets. These brackets say, hey, let me include the endpoints, but I'm not going to include them. So I'm going to put the parentheses right over here. Parentheses. So this tells us that we're dealing with an open interval. So this right over here, let me make it clear, this is an open, an open interval."}, {"video_title": "Intervals and interval notation   Functions   Algebra I   Khan Academy.mp3", "Sentence": "So I'm going to put the parentheses right over here. Parentheses. So this tells us that we're dealing with an open interval. So this right over here, let me make it clear, this is an open, an open interval. Now you're probably wondering, okay, in this case, both endpoints were included, it's a closed interval. In this case, both endpoints were excluded, it's an open interval. Can you have things that have one endpoint included and one endpoint excluded?"}, {"video_title": "Intervals and interval notation   Functions   Algebra I   Khan Academy.mp3", "Sentence": "So this right over here, let me make it clear, this is an open, an open interval. Now you're probably wondering, okay, in this case, both endpoints were included, it's a closed interval. In this case, both endpoints were excluded, it's an open interval. Can you have things that have one endpoint included and one endpoint excluded? And the answer is absolutely. Let's see an example of that. So I'll get another number line here."}, {"video_title": "Intervals and interval notation   Functions   Algebra I   Khan Academy.mp3", "Sentence": "Can you have things that have one endpoint included and one endpoint excluded? And the answer is absolutely. Let's see an example of that. So I'll get another number line here. Another number line. And let's say that we want to, actually let me do it the other way around. Let me write it first, and then I'll graph it."}, {"video_title": "Intervals and interval notation   Functions   Algebra I   Khan Academy.mp3", "Sentence": "So I'll get another number line here. Another number line. And let's say that we want to, actually let me do it the other way around. Let me write it first, and then I'll graph it. So let's say we're thinking about all of the x's that are a member of the real numbers, such that, let's say negative four, let's say negative four is not included, is less than x, is less than or equal to negative one. So now negative one is included. So we're not going to include negative four."}, {"video_title": "Intervals and interval notation   Functions   Algebra I   Khan Academy.mp3", "Sentence": "Let me write it first, and then I'll graph it. So let's say we're thinking about all of the x's that are a member of the real numbers, such that, let's say negative four, let's say negative four is not included, is less than x, is less than or equal to negative one. So now negative one is included. So we're not going to include negative four. Negative four is strictly less than, not less than or equal to. So x can't be equal to negative four, so open circle there. But x could be equal to negative one."}, {"video_title": "Intervals and interval notation   Functions   Algebra I   Khan Academy.mp3", "Sentence": "So we're not going to include negative four. Negative four is strictly less than, not less than or equal to. So x can't be equal to negative four, so open circle there. But x could be equal to negative one. It has to be less than or equal to negative one. So it could be equal to negative one, so I'm going to fill that in right over there. And then it's everything in between."}, {"video_title": "Intervals and interval notation   Functions   Algebra I   Khan Academy.mp3", "Sentence": "But x could be equal to negative one. It has to be less than or equal to negative one. So it could be equal to negative one, so I'm going to fill that in right over there. And then it's everything in between. If I want to write it with this notation, I could write x is a member of the real numbers, such that x is a member of the interval. So it's going to go between negative four and negative one. But we're not including negative four."}, {"video_title": "Intervals and interval notation   Functions   Algebra I   Khan Academy.mp3", "Sentence": "And then it's everything in between. If I want to write it with this notation, I could write x is a member of the real numbers, such that x is a member of the interval. So it's going to go between negative four and negative one. But we're not including negative four. We have an open circle here, so I'm going to put a parentheses on that side. But we are including negative one. We are including negative one."}, {"video_title": "Intervals and interval notation   Functions   Algebra I   Khan Academy.mp3", "Sentence": "But we're not including negative four. We have an open circle here, so I'm going to put a parentheses on that side. But we are including negative one. We are including negative one. So we put a bracket on that side. And so that would be, that right over there would be the notation. Now there's other things that you could do with interval notation."}, {"video_title": "Intervals and interval notation   Functions   Algebra I   Khan Academy.mp3", "Sentence": "We are including negative one. So we put a bracket on that side. And so that would be, that right over there would be the notation. Now there's other things that you could do with interval notation. You could say, well hey, everything except for some value. So let me give another example. Let's get another example here."}, {"video_title": "Intervals and interval notation   Functions   Algebra I   Khan Academy.mp3", "Sentence": "Now there's other things that you could do with interval notation. You could say, well hey, everything except for some value. So let me give another example. Let's get another example here. Let's say that we want to talk about all the real numbers except for one. So we want to include all of the real numbers except for one. So we're going to exclude one right over here, so open circle, but it can be any other real number."}, {"video_title": "Intervals and interval notation   Functions   Algebra I   Khan Academy.mp3", "Sentence": "Let's get another example here. Let's say that we want to talk about all the real numbers except for one. So we want to include all of the real numbers except for one. So we're going to exclude one right over here, so open circle, but it can be any other real number. So how would we denote this? Well, we could write, x is a member of the real numbers such that x does not equal one. So here I'm saying, look, x can be a member of the real numbers, but x cannot be equal to one."}, {"video_title": "Intervals and interval notation   Functions   Algebra I   Khan Academy.mp3", "Sentence": "So we're going to exclude one right over here, so open circle, but it can be any other real number. So how would we denote this? Well, we could write, x is a member of the real numbers such that x does not equal one. So here I'm saying, look, x can be a member of the real numbers, but x cannot be equal to one. It can be anything else, but it cannot be equal to one. And there's other ways of denoting this exact same interval. You could say, x is a member of the real numbers such that x is less than one, or x is greater than one."}, {"video_title": "Intervals and interval notation   Functions   Algebra I   Khan Academy.mp3", "Sentence": "So here I'm saying, look, x can be a member of the real numbers, but x cannot be equal to one. It can be anything else, but it cannot be equal to one. And there's other ways of denoting this exact same interval. You could say, x is a member of the real numbers such that x is less than one, or x is greater than one. So you could write it just like that, or you could do something interesting. This is the one that I would use. This is the shortest, and it makes it very clear."}, {"video_title": "Intervals and interval notation   Functions   Algebra I   Khan Academy.mp3", "Sentence": "You could say, x is a member of the real numbers such that x is less than one, or x is greater than one. So you could write it just like that, or you could do something interesting. This is the one that I would use. This is the shortest, and it makes it very clear. You're saying, hey, everything except for one. But you could even do something fancy. Like you could say, x is a member of the real numbers such that x is a member of the set going from negative infinity to one, not including one, or x is a member of the set going from, or a member of the interval going from one, not including one, all the way to positive, all the way to positive infinity."}, {"video_title": "Intervals and interval notation   Functions   Algebra I   Khan Academy.mp3", "Sentence": "This is the shortest, and it makes it very clear. You're saying, hey, everything except for one. But you could even do something fancy. Like you could say, x is a member of the real numbers such that x is a member of the set going from negative infinity to one, not including one, or x is a member of the set going from, or a member of the interval going from one, not including one, all the way to positive, all the way to positive infinity. And when you're talking about negative infinity or positive infinity, you always put a parentheses. And the view there is you can never include everything all the way up to infinity. It needs to be at least open at that end point, because infinity just keeps going on and on."}, {"video_title": "Intervals and interval notation   Functions   Algebra I   Khan Academy.mp3", "Sentence": "Like you could say, x is a member of the real numbers such that x is a member of the set going from negative infinity to one, not including one, or x is a member of the set going from, or a member of the interval going from one, not including one, all the way to positive, all the way to positive infinity. And when you're talking about negative infinity or positive infinity, you always put a parentheses. And the view there is you can never include everything all the way up to infinity. It needs to be at least open at that end point, because infinity just keeps going on and on. So you always want to put a parentheses if you're talking about infinity or negative infinity. It's not really an endpoint. It keeps going on and on forever."}, {"video_title": "Intervals and interval notation   Functions   Algebra I   Khan Academy.mp3", "Sentence": "It needs to be at least open at that end point, because infinity just keeps going on and on. So you always want to put a parentheses if you're talking about infinity or negative infinity. It's not really an endpoint. It keeps going on and on forever. So you use the notation for open interval, at least at that end. And notice we're not including one either. So if x is a member of this interval or that interval, it essentially can be anything other than one."}, {"video_title": "Finding dimensions of triangle from area   Quadratic equations   Algebra I   Khan Academy.mp3", "Sentence": "The area of the triangle is 30 inches squared. Find the height and base. Use the formula area equals 1 half base times height for the area of a triangle. Okay, so let's think about it a little bit. We have the base. Let me draw a triangle here. So this is our triangle."}, {"video_title": "Finding dimensions of triangle from area   Quadratic equations   Algebra I   Khan Academy.mp3", "Sentence": "Okay, so let's think about it a little bit. We have the base. Let me draw a triangle here. So this is our triangle. And let's say that the length of this bottom side, that's the base. Let's call that b. And then this is the height."}, {"video_title": "Finding dimensions of triangle from area   Quadratic equations   Algebra I   Khan Academy.mp3", "Sentence": "So this is our triangle. And let's say that the length of this bottom side, that's the base. Let's call that b. And then this is the height. This is the height right over here. And then the area is equal to 1 half base times height. Now in this first sentence, they tell us that the height of a triangle is 4 inches less than the length of the base."}, {"video_title": "Finding dimensions of triangle from area   Quadratic equations   Algebra I   Khan Academy.mp3", "Sentence": "And then this is the height. This is the height right over here. And then the area is equal to 1 half base times height. Now in this first sentence, they tell us that the height of a triangle is 4 inches less than the length of the base. So the height is equal to the base minus 4. That's what that first sentence tells us. The area of the triangle is 30 inches squared."}, {"video_title": "Finding dimensions of triangle from area   Quadratic equations   Algebra I   Khan Academy.mp3", "Sentence": "Now in this first sentence, they tell us that the height of a triangle is 4 inches less than the length of the base. So the height is equal to the base minus 4. That's what that first sentence tells us. The area of the triangle is 30 inches squared. So if we take 1 half the base times the height, we'll get 30 inches squared. Or we could say that 30 inches squared is equal to 1 half times the base times the height. Now instead of putting an h in for height, we know that the height is the same thing as 4 less than the base."}, {"video_title": "Finding dimensions of triangle from area   Quadratic equations   Algebra I   Khan Academy.mp3", "Sentence": "The area of the triangle is 30 inches squared. So if we take 1 half the base times the height, we'll get 30 inches squared. Or we could say that 30 inches squared is equal to 1 half times the base times the height. Now instead of putting an h in for height, we know that the height is the same thing as 4 less than the base. So let's put that in there. 4 less than the base. And then let's see what we get here."}, {"video_title": "Finding dimensions of triangle from area   Quadratic equations   Algebra I   Khan Academy.mp3", "Sentence": "Now instead of putting an h in for height, we know that the height is the same thing as 4 less than the base. So let's put that in there. 4 less than the base. And then let's see what we get here. We get 30 is equal to 1 half times b over 2 times b minus 4. I just multiplied the 1 half times the b. Now let's distribute the b over 2."}, {"video_title": "Finding dimensions of triangle from area   Quadratic equations   Algebra I   Khan Academy.mp3", "Sentence": "And then let's see what we get here. We get 30 is equal to 1 half times b over 2 times b minus 4. I just multiplied the 1 half times the b. Now let's distribute the b over 2. 30 is equal to b squared over 2. Be careful. b over 2 times b is just b squared over 2."}, {"video_title": "Finding dimensions of triangle from area   Quadratic equations   Algebra I   Khan Academy.mp3", "Sentence": "Now let's distribute the b over 2. 30 is equal to b squared over 2. Be careful. b over 2 times b is just b squared over 2. And then b over 2 times negative 4 is negative 2b. Now just to get rid of this fraction here, let's multiply both sides of this equation by 2. So let's multiply that side by 2 and let's multiply that side by 2."}, {"video_title": "Finding dimensions of triangle from area   Quadratic equations   Algebra I   Khan Academy.mp3", "Sentence": "b over 2 times b is just b squared over 2. And then b over 2 times negative 4 is negative 2b. Now just to get rid of this fraction here, let's multiply both sides of this equation by 2. So let's multiply that side by 2 and let's multiply that side by 2. On the left-hand side, you get 60. On the right-hand side, 2 times b squared over 2 is just b squared. Negative 2b times 2 is negative 4b."}, {"video_title": "Finding dimensions of triangle from area   Quadratic equations   Algebra I   Khan Academy.mp3", "Sentence": "So let's multiply that side by 2 and let's multiply that side by 2. On the left-hand side, you get 60. On the right-hand side, 2 times b squared over 2 is just b squared. Negative 2b times 2 is negative 4b. And now we have a quadratic here. And the best way to solve a quadratic, we have a second degree term right here, is to get all of the terms on one side of the equation having them equal 0. So let's subtract 60 from both sides of this equation."}, {"video_title": "Finding dimensions of triangle from area   Quadratic equations   Algebra I   Khan Academy.mp3", "Sentence": "Negative 2b times 2 is negative 4b. And now we have a quadratic here. And the best way to solve a quadratic, we have a second degree term right here, is to get all of the terms on one side of the equation having them equal 0. So let's subtract 60 from both sides of this equation. Let's subtract 60 from both sides. And we get 0 is equal to b squared minus 4b minus 60. And so what we need to do here is just factor this thing right now, or factor it, and then know that if I have the product of some things and that equals 0, that means that either one or both of those things need to be equal to 0."}, {"video_title": "Finding dimensions of triangle from area   Quadratic equations   Algebra I   Khan Academy.mp3", "Sentence": "So let's subtract 60 from both sides of this equation. Let's subtract 60 from both sides. And we get 0 is equal to b squared minus 4b minus 60. And so what we need to do here is just factor this thing right now, or factor it, and then know that if I have the product of some things and that equals 0, that means that either one or both of those things need to be equal to 0. So we need to factor b squared minus 4b minus 60. So what we want to do, we want to find 2 numbers whose sum is negative 4 and whose product is negative 60. So we want to find 2 numbers whose sum is equal to negative 4 and whose product is equal to negative 60."}, {"video_title": "Finding dimensions of triangle from area   Quadratic equations   Algebra I   Khan Academy.mp3", "Sentence": "And so what we need to do here is just factor this thing right now, or factor it, and then know that if I have the product of some things and that equals 0, that means that either one or both of those things need to be equal to 0. So we need to factor b squared minus 4b minus 60. So what we want to do, we want to find 2 numbers whose sum is negative 4 and whose product is negative 60. So we want to find 2 numbers whose sum is equal to negative 4 and whose product is equal to negative 60. Now, given that their product is negative, we know there are different signs, and this tells us that their absolute values are going to be 4 apart, that one is going to be 4 less than the other. So you can look at the factors of 60. 1 and 60 are too far apart."}, {"video_title": "Finding dimensions of triangle from area   Quadratic equations   Algebra I   Khan Academy.mp3", "Sentence": "So we want to find 2 numbers whose sum is equal to negative 4 and whose product is equal to negative 60. Now, given that their product is negative, we know there are different signs, and this tells us that their absolute values are going to be 4 apart, that one is going to be 4 less than the other. So you can look at the factors of 60. 1 and 60 are too far apart. If you made one of them negative, you would either get positive 59 as a sum or negative 59 as a sum. 2 and 30, still too far apart. 3 and 20, sorry, still too far apart."}, {"video_title": "Finding dimensions of triangle from area   Quadratic equations   Algebra I   Khan Academy.mp3", "Sentence": "1 and 60 are too far apart. If you made one of them negative, you would either get positive 59 as a sum or negative 59 as a sum. 2 and 30, still too far apart. 3 and 20, sorry, still too far apart. If you had made one negative, you would either get negative 17 or positive 17. Then you can have 4 and 15, still too far apart. If you made one of them negative, their sum would be either negative 11 or positive 11."}, {"video_title": "Finding dimensions of triangle from area   Quadratic equations   Algebra I   Khan Academy.mp3", "Sentence": "3 and 20, sorry, still too far apart. If you had made one negative, you would either get negative 17 or positive 17. Then you can have 4 and 15, still too far apart. If you made one of them negative, their sum would be either negative 11 or positive 11. Then you have 5 and 12, still seems too far apart from each other. One of them is negative, then you would either have their sum being positive 7 or negative 7. Then you have 6 and 10."}, {"video_title": "Finding dimensions of triangle from area   Quadratic equations   Algebra I   Khan Academy.mp3", "Sentence": "If you made one of them negative, their sum would be either negative 11 or positive 11. Then you have 5 and 12, still seems too far apart from each other. One of them is negative, then you would either have their sum being positive 7 or negative 7. Then you have 6 and 10. Now this looks interesting. They're 4 apart. So if we make, and we want the larger absolute magnitude number to be negative so that their sum is negative."}, {"video_title": "Finding dimensions of triangle from area   Quadratic equations   Algebra I   Khan Academy.mp3", "Sentence": "Then you have 6 and 10. Now this looks interesting. They're 4 apart. So if we make, and we want the larger absolute magnitude number to be negative so that their sum is negative. So if we make it 6 and negative 10, their sum will be negative 4 and their product is negative 60. So that works. So you can literally say that this is equal to B plus 6 times B minus 10."}, {"video_title": "Finding dimensions of triangle from area   Quadratic equations   Algebra I   Khan Academy.mp3", "Sentence": "So if we make, and we want the larger absolute magnitude number to be negative so that their sum is negative. So if we make it 6 and negative 10, their sum will be negative 4 and their product is negative 60. So that works. So you can literally say that this is equal to B plus 6 times B minus 10. B plus the A times B minus, or plus B minus the B. And let me be very careful here. This B over here, I want to make it very clear, is different than the B that we're using in the equation."}, {"video_title": "Finding dimensions of triangle from area   Quadratic equations   Algebra I   Khan Academy.mp3", "Sentence": "So you can literally say that this is equal to B plus 6 times B minus 10. B plus the A times B minus, or plus B minus the B. And let me be very careful here. This B over here, I want to make it very clear, is different than the B that we're using in the equation. I just used this B here to say, look, we're looking for two numbers that add up to this second term right over here. It's a different B. I could have said X plus Y is equal to negative 4 and X times Y is equal to negative 60. In fact, let me do it that way just so we don't get confused."}, {"video_title": "Finding dimensions of triangle from area   Quadratic equations   Algebra I   Khan Academy.mp3", "Sentence": "This B over here, I want to make it very clear, is different than the B that we're using in the equation. I just used this B here to say, look, we're looking for two numbers that add up to this second term right over here. It's a different B. I could have said X plus Y is equal to negative 4 and X times Y is equal to negative 60. In fact, let me do it that way just so we don't get confused. So we could write X plus Y is equal to negative 4 and then we have X times Y is equal to negative 60. So we have B plus 6 times B plus Y. X is 6, Y is negative 10. And that is equal to 0."}, {"video_title": "Finding dimensions of triangle from area   Quadratic equations   Algebra I   Khan Academy.mp3", "Sentence": "In fact, let me do it that way just so we don't get confused. So we could write X plus Y is equal to negative 4 and then we have X times Y is equal to negative 60. So we have B plus 6 times B plus Y. X is 6, Y is negative 10. And that is equal to 0. And you could, well, let's just solve this right here and then we'll go back and show you that you could also factor this by grouping. But just from this, we know that either one of these is equal to 0. Either B plus 6 is equal to 0 or B minus 10 is equal to 0."}, {"video_title": "Finding dimensions of triangle from area   Quadratic equations   Algebra I   Khan Academy.mp3", "Sentence": "And that is equal to 0. And you could, well, let's just solve this right here and then we'll go back and show you that you could also factor this by grouping. But just from this, we know that either one of these is equal to 0. Either B plus 6 is equal to 0 or B minus 10 is equal to 0. If we subtract 6 from both sides of this equation, we get B is equal to negative 6. Or if you add 10 to both sides of this equation, you get B is equal to 10. And those are our two solutions."}, {"video_title": "Finding dimensions of triangle from area   Quadratic equations   Algebra I   Khan Academy.mp3", "Sentence": "Either B plus 6 is equal to 0 or B minus 10 is equal to 0. If we subtract 6 from both sides of this equation, we get B is equal to negative 6. Or if you add 10 to both sides of this equation, you get B is equal to 10. And those are our two solutions. You could put them back in and verify that they satisfy our constraints. Now, the other way that you could solve this, and we're going to get the exact same answer, is you could just break up this negative 4B into its constituents. So you could have broken this up into 0 is equal to B squared."}, {"video_title": "Finding dimensions of triangle from area   Quadratic equations   Algebra I   Khan Academy.mp3", "Sentence": "And those are our two solutions. You could put them back in and verify that they satisfy our constraints. Now, the other way that you could solve this, and we're going to get the exact same answer, is you could just break up this negative 4B into its constituents. So you could have broken this up into 0 is equal to B squared. And then you could have broken it up into plus 6B minus 10B minus 60. And then factor it by grouping. Group these first two terms, group these second two terms."}, {"video_title": "Finding dimensions of triangle from area   Quadratic equations   Algebra I   Khan Academy.mp3", "Sentence": "So you could have broken this up into 0 is equal to B squared. And then you could have broken it up into plus 6B minus 10B minus 60. And then factor it by grouping. Group these first two terms, group these second two terms. And you're going to add them together. The first one you can factor out a B. So you have B times B plus 6."}, {"video_title": "Finding dimensions of triangle from area   Quadratic equations   Algebra I   Khan Academy.mp3", "Sentence": "Group these first two terms, group these second two terms. And you're going to add them together. The first one you can factor out a B. So you have B times B plus 6. The second one you can factor out a negative 10. So minus 10 times B plus 6. All that's equal to 0."}, {"video_title": "Finding dimensions of triangle from area   Quadratic equations   Algebra I   Khan Academy.mp3", "Sentence": "So you have B times B plus 6. The second one you can factor out a negative 10. So minus 10 times B plus 6. All that's equal to 0. And now you can factor out a B plus 6. So if you factor out a B plus 6 here, you get 0 is equal to B minus 10 times B plus 6. We're literally just factoring out this out of the expression."}, {"video_title": "Finding dimensions of triangle from area   Quadratic equations   Algebra I   Khan Academy.mp3", "Sentence": "All that's equal to 0. And now you can factor out a B plus 6. So if you factor out a B plus 6 here, you get 0 is equal to B minus 10 times B plus 6. We're literally just factoring out this out of the expression. You're just left with a B minus 10. It's the same thing that we did in one step over here. Whatever works for you."}, {"video_title": "Finding dimensions of triangle from area   Quadratic equations   Algebra I   Khan Academy.mp3", "Sentence": "We're literally just factoring out this out of the expression. You're just left with a B minus 10. It's the same thing that we did in one step over here. Whatever works for you. But either way, the solutions are either B is equal to negative 6 or B is equal to 10. And we have to be careful here. Remember, this is a word problem."}, {"video_title": "Finding dimensions of triangle from area   Quadratic equations   Algebra I   Khan Academy.mp3", "Sentence": "Whatever works for you. But either way, the solutions are either B is equal to negative 6 or B is equal to 10. And we have to be careful here. Remember, this is a word problem. We can't just state, oh, B could be negative 6 or B could be 10. We have to think about whether this makes sense in the context of the actual problem. We're talking about lengths of triangles or lengths of the sides of triangles."}, {"video_title": "Finding dimensions of triangle from area   Quadratic equations   Algebra I   Khan Academy.mp3", "Sentence": "Remember, this is a word problem. We can't just state, oh, B could be negative 6 or B could be 10. We have to think about whether this makes sense in the context of the actual problem. We're talking about lengths of triangles or lengths of the sides of triangles. We can't have a negative length. So because of that, the base of a triangle can't have length negative 6. So we can cross that out."}, {"video_title": "Finding dimensions of triangle from area   Quadratic equations   Algebra I   Khan Academy.mp3", "Sentence": "We're talking about lengths of triangles or lengths of the sides of triangles. We can't have a negative length. So because of that, the base of a triangle can't have length negative 6. So we can cross that out. So we actually only have one solution here. Almost made a careless mistake. Forgot that we were dealing with a word problem."}, {"video_title": "Finding dimensions of triangle from area   Quadratic equations   Algebra I   Khan Academy.mp3", "Sentence": "So we can cross that out. So we actually only have one solution here. Almost made a careless mistake. Forgot that we were dealing with a word problem. The only possible base is 10. And let's see. They say find the height and the base."}, {"video_title": "Finding dimensions of triangle from area   Quadratic equations   Algebra I   Khan Academy.mp3", "Sentence": "Forgot that we were dealing with a word problem. The only possible base is 10. And let's see. They say find the height and the base. Once again, not done. So the base we're saying is 10. The height is 4 inches less."}, {"video_title": "Finding dimensions of triangle from area   Quadratic equations   Algebra I   Khan Academy.mp3", "Sentence": "They say find the height and the base. Once again, not done. So the base we're saying is 10. The height is 4 inches less. It's B minus 4. So the height is 6. And then you can verify."}, {"video_title": "Recursive formulas for arithmetic sequences   Mathematics I   High School Math   Khan Academy (2).mp3", "Sentence": "When n is two, f of n is five. When n is three, f of n is negative two. When n is four, f of n is negative nine. And so one way to think about it is, this function f is defining a sequence where the first term of this sequence is 12. The second term of this sequence is five. The third term of this sequence is negative two. The fourth term of this sequence is negative nine."}, {"video_title": "Recursive formulas for arithmetic sequences   Mathematics I   High School Math   Khan Academy (2).mp3", "Sentence": "And so one way to think about it is, this function f is defining a sequence where the first term of this sequence is 12. The second term of this sequence is five. The third term of this sequence is negative two. The fourth term of this sequence is negative nine. It goes on and on and on. And you might notice that it's an arithmetic sequence. We start with a 12, and then the next term, what have we done?"}, {"video_title": "Recursive formulas for arithmetic sequences   Mathematics I   High School Math   Khan Academy (2).mp3", "Sentence": "The fourth term of this sequence is negative nine. It goes on and on and on. And you might notice that it's an arithmetic sequence. We start with a 12, and then the next term, what have we done? We've subtracted seven. Now to go from the second to the third term, what do we do? We subtract seven."}, {"video_title": "Recursive formulas for arithmetic sequences   Mathematics I   High School Math   Khan Academy (2).mp3", "Sentence": "We start with a 12, and then the next term, what have we done? We've subtracted seven. Now to go from the second to the third term, what do we do? We subtract seven. So each term is seven less than the term before it. Now with that out of the way, see if you can define this function, this function of n, if you can define it explicitly. So figure out a function definition."}, {"video_title": "Recursive formulas for arithmetic sequences   Mathematics I   High School Math   Khan Academy (2).mp3", "Sentence": "We subtract seven. So each term is seven less than the term before it. Now with that out of the way, see if you can define this function, this function of n, if you can define it explicitly. So figure out a function definition. So I want to figure out f of n is equal. I want you to figure out what this needs to be so that if you input n here, it gives you the appropriate f of n. So let's think about it a little bit. It's going to be, we could think of it as, we're starting at, the first term is going to be 12, but then we are going to subtract, we're going to subtract seven."}, {"video_title": "Recursive formulas for arithmetic sequences   Mathematics I   High School Math   Khan Academy (2).mp3", "Sentence": "So figure out a function definition. So I want to figure out f of n is equal. I want you to figure out what this needs to be so that if you input n here, it gives you the appropriate f of n. So let's think about it a little bit. It's going to be, we could think of it as, we're starting at, the first term is going to be 12, but then we are going to subtract, we're going to subtract seven. And what are we going to subtract seven? How many times are we going to subtract seven? So for the first term, we subtract seven zero times, and so we just get 12."}, {"video_title": "Recursive formulas for arithmetic sequences   Mathematics I   High School Math   Khan Academy (2).mp3", "Sentence": "It's going to be, we could think of it as, we're starting at, the first term is going to be 12, but then we are going to subtract, we're going to subtract seven. And what are we going to subtract seven? How many times are we going to subtract seven? So for the first term, we subtract seven zero times, and so we just get 12. For the second term, we subtract seven once. For the third term, we subtract seven twice, one, two times. For the fourth term, we subtract seven three times."}, {"video_title": "Recursive formulas for arithmetic sequences   Mathematics I   High School Math   Khan Academy (2).mp3", "Sentence": "So for the first term, we subtract seven zero times, and so we just get 12. For the second term, we subtract seven once. For the third term, we subtract seven twice, one, two times. For the fourth term, we subtract seven three times. So it looks like whatever term we're on, we're subtracting seven n minus one, we're subtracting seven, whatever term we're on, that term minus one times. So it's n minus one times. And let's see if this actually works out."}, {"video_title": "Recursive formulas for arithmetic sequences   Mathematics I   High School Math   Khan Academy (2).mp3", "Sentence": "For the fourth term, we subtract seven three times. So it looks like whatever term we're on, we're subtracting seven n minus one, we're subtracting seven, whatever term we're on, that term minus one times. So it's n minus one times. And let's see if this actually works out. So f of one is going to be 12 minus seven times one minus one, that's a zero, so that's all just going to be 12. F of two is going to be 12 minus seven times two minus one. So it's going to be 12 minus seven times one, or we're just going to subtract seven once, which is exactly the case."}, {"video_title": "Recursive formulas for arithmetic sequences   Mathematics I   High School Math   Khan Academy (2).mp3", "Sentence": "And let's see if this actually works out. So f of one is going to be 12 minus seven times one minus one, that's a zero, so that's all just going to be 12. F of two is going to be 12 minus seven times two minus one. So it's going to be 12 minus seven times one, or we're just going to subtract seven once, which is exactly the case. We start at 12, and we subtract seven once. F of three, you can keep testing this, 12 minus, and we should have to subtract seven twice. And we see three minus one is two times, so we're going to subtract seven two times."}, {"video_title": "Recursive formulas for arithmetic sequences   Mathematics I   High School Math   Khan Academy (2).mp3", "Sentence": "So it's going to be 12 minus seven times one, or we're just going to subtract seven once, which is exactly the case. We start at 12, and we subtract seven once. F of three, you can keep testing this, 12 minus, and we should have to subtract seven twice. And we see three minus one is two times, so we're going to subtract seven two times. So this looks right on. We've defined the function explicitly, we've defined f explicitly for this sequence. Let's do another example here."}, {"video_title": "Recursive formulas for arithmetic sequences   Mathematics I   High School Math   Khan Academy (2).mp3", "Sentence": "And we see three minus one is two times, so we're going to subtract seven two times. So this looks right on. We've defined the function explicitly, we've defined f explicitly for this sequence. Let's do another example here. So in this case, we have some function definitions already here. So you have your sequence, it's kind of viewed in this table, or you could view it as the first term is negative 100, next term is negative 50, next term is zero, next term is 50. And it's very clear that this is also an arithmetic sequence."}, {"video_title": "Recursive formulas for arithmetic sequences   Mathematics I   High School Math   Khan Academy (2).mp3", "Sentence": "Let's do another example here. So in this case, we have some function definitions already here. So you have your sequence, it's kind of viewed in this table, or you could view it as the first term is negative 100, next term is negative 50, next term is zero, next term is 50. And it's very clear that this is also an arithmetic sequence. We're starting at negative 100, and then what are we doing here? We're adding 50, and then we're adding 50, and then we are adding 50, so each term is 50 more than the term before it. And what I want you to do is pause the video and think about which of these definitions of the function f are correct."}, {"video_title": "Recursive formulas for arithmetic sequences   Mathematics I   High School Math   Khan Academy (2).mp3", "Sentence": "And it's very clear that this is also an arithmetic sequence. We're starting at negative 100, and then what are we doing here? We're adding 50, and then we're adding 50, and then we are adding 50, so each term is 50 more than the term before it. And what I want you to do is pause the video and think about which of these definitions of the function f are correct. And it might be more than one. All right, so let's think about it. So this definition right over here, one way to think about it is saying, okay, I'm going to start at negative 100, and I'm going to add 50 n minus one times."}, {"video_title": "Recursive formulas for arithmetic sequences   Mathematics I   High School Math   Khan Academy (2).mp3", "Sentence": "And what I want you to do is pause the video and think about which of these definitions of the function f are correct. And it might be more than one. All right, so let's think about it. So this definition right over here, one way to think about it is saying, okay, I'm going to start at negative 100, and I'm going to add 50 n minus one times. Does this make sense? Well, for the first term, we're going to, if we start at negative 100, we don't want to add 50 at all. We want to add 50 zero times, and it works out, because one minus one is going to be zero."}, {"video_title": "Recursive formulas for arithmetic sequences   Mathematics I   High School Math   Khan Academy (2).mp3", "Sentence": "So this definition right over here, one way to think about it is saying, okay, I'm going to start at negative 100, and I'm going to add 50 n minus one times. Does this make sense? Well, for the first term, we're going to, if we start at negative 100, we don't want to add 50 at all. We want to add 50 zero times, and it works out, because one minus one is going to be zero. So it checks out for n equals one. Let's see, for n equals two, you start at negative 100. I want to add 50 once."}, {"video_title": "Recursive formulas for arithmetic sequences   Mathematics I   High School Math   Khan Academy (2).mp3", "Sentence": "We want to add 50 zero times, and it works out, because one minus one is going to be zero. So it checks out for n equals one. Let's see, for n equals two, you start at negative 100. I want to add 50 once. So I want to add 50 once. So this should be a one. Two minus one, yep, it's a one."}, {"video_title": "Recursive formulas for arithmetic sequences   Mathematics I   High School Math   Khan Academy (2).mp3", "Sentence": "I want to add 50 once. So I want to add 50 once. So this should be a one. Two minus one, yep, it's a one. We're adding 50, whatever this number is, whatever n is, we're adding 50 one less that number of times. So for here, we're adding 50 twice. For when n is four, we're adding 50 three times."}, {"video_title": "Recursive formulas for arithmetic sequences   Mathematics I   High School Math   Khan Academy (2).mp3", "Sentence": "Two minus one, yep, it's a one. We're adding 50, whatever this number is, whatever n is, we're adding 50 one less that number of times. So for here, we're adding 50 twice. For when n is four, we're adding 50 three times. And this one checks out. When n is four, we're adding 50, four minus one, three times. Negative 100 plus 50 times three, we're adding 50 three times."}, {"video_title": "Recursive formulas for arithmetic sequences   Mathematics I   High School Math   Khan Academy (2).mp3", "Sentence": "For when n is four, we're adding 50 three times. And this one checks out. When n is four, we're adding 50, four minus one, three times. Negative 100 plus 50 times three, we're adding 50 three times. Adding 50 one, two, three times, well, that gives you 50. So I like this one. Now let's see about this one over here."}, {"video_title": "Recursive formulas for arithmetic sequences   Mathematics I   High School Math   Khan Academy (2).mp3", "Sentence": "Negative 100 plus 50 times three, we're adding 50 three times. Adding 50 one, two, three times, well, that gives you 50. So I like this one. Now let's see about this one over here. Negative 150 plus 50 n. All right, that's one way of saying, so let's see, if n is equal to one, it's going to be negative, actually, let me draw a table for this one. So if we have n and we have f of n, and this is going to be for this character right over here. So if n is one, it's going to be negative 150 plus 50, which is negative 100."}, {"video_title": "Recursive formulas for arithmetic sequences   Mathematics I   High School Math   Khan Academy (2).mp3", "Sentence": "Now let's see about this one over here. Negative 150 plus 50 n. All right, that's one way of saying, so let's see, if n is equal to one, it's going to be negative, actually, let me draw a table for this one. So if we have n and we have f of n, and this is going to be for this character right over here. So if n is one, it's going to be negative 150 plus 50, which is negative 100. Yep, that checks out. When n is two, we get negative 150 plus 50 times two, which is going to be, this is 100, and there's negative 150, this is going to be negative 50. When n is three, and that checks out, of course."}, {"video_title": "Recursive formulas for arithmetic sequences   Mathematics I   High School Math   Khan Academy (2).mp3", "Sentence": "So if n is one, it's going to be negative 150 plus 50, which is negative 100. Yep, that checks out. When n is two, we get negative 150 plus 50 times two, which is going to be, this is 100, and there's negative 150, this is going to be negative 50. When n is three, and that checks out, of course. When n is three, you get negative 150 plus 50 times three, which is equal to zero. This checks out. This one over here is going to work."}, {"video_title": "Recursive formulas for arithmetic sequences   Mathematics I   High School Math   Khan Academy (2).mp3", "Sentence": "When n is three, and that checks out, of course. When n is three, you get negative 150 plus 50 times three, which is equal to zero. This checks out. This one over here is going to work. And you might say, well, hey, these formulas look different. But you can algebraically manipulate them to see that they are the exact same thing. If you were to take this first one, it's negative 100 plus, let's distribute this 50, plus 50n minus 50."}, {"video_title": "Recursive formulas for arithmetic sequences   Mathematics I   High School Math   Khan Academy (2).mp3", "Sentence": "This one over here is going to work. And you might say, well, hey, these formulas look different. But you can algebraically manipulate them to see that they are the exact same thing. If you were to take this first one, it's negative 100 plus, let's distribute this 50, plus 50n minus 50. Well, negative 100 minus 50, that's negative 150, and then you have plus 50n. So these are algebraically the exact same definition for our function. Now what about this one here?"}, {"video_title": "Recursive formulas for arithmetic sequences   Mathematics I   High School Math   Khan Academy (2).mp3", "Sentence": "If you were to take this first one, it's negative 100 plus, let's distribute this 50, plus 50n minus 50. Well, negative 100 minus 50, that's negative 150, and then you have plus 50n. So these are algebraically the exact same definition for our function. Now what about this one here? Negative 100 plus 50n. Does this one work? Well, let's see, when n is equal to one, this would be negative 100 plus 50, which is negative 50."}, {"video_title": "Recursive formulas for arithmetic sequences   Mathematics I   High School Math   Khan Academy (2).mp3", "Sentence": "Now what about this one here? Negative 100 plus 50n. Does this one work? Well, let's see, when n is equal to one, this would be negative 100 plus 50, which is negative 50. Well, no, this doesn't work. We need to get a negative 100 here. So this one is not, not correct."}, {"video_title": "Solving quadratics by factoring leading coefficient \u00c3\u00a2\u00c2  \u00c2\u00a0 1   High School Math   Khan Academy.mp3", "Sentence": "Alright, let's work through this together. So this, the numbers here don't seem like outlandish numbers they seem like something that I might be able to deal with and I might be able to factor, so let's try to do that. So the first thing I like to do is see if I can get a coefficient of one on the second degree term on the x squared term and it looks like actually all of these terms are divisible by six. So if we divide both sides of this equation by six I'm still going to have nice integer coefficients. So let's do that, let's divide both sides by six. So if we divide the left side by six, divide by six, divide by six, divide by six and I divide the right side by six. So if I do that and clearly if I do the same thing to both sides of the equation then the equality still holds."}, {"video_title": "Solving quadratics by factoring leading coefficient \u00c3\u00a2\u00c2  \u00c2\u00a0 1   High School Math   Khan Academy.mp3", "Sentence": "So if we divide both sides of this equation by six I'm still going to have nice integer coefficients. So let's do that, let's divide both sides by six. So if we divide the left side by six, divide by six, divide by six, divide by six and I divide the right side by six. So if I do that and clearly if I do the same thing to both sides of the equation then the equality still holds. On the left hand side I am going to be left with x squared and then negative 120 divided by six, that is, let's see, 120 divided by six is 20 so that's minus 20 x and then 600 divided by six is 100 so plus 100 is equal to zero divided by six, is equal to zero. So let's see if we can factor, if we can express this quadratic as a product of two expressions. And the way we think about this, and we've done it multiple times, if we have something, if we have x plus a times x plus b and this is hopefully a review for you, if you multiply that out that is going to be equal to, that equals to x squared plus a plus b x plus ab and so what we want to do is see if we can factor this into an x plus a and an x plus b and so a plus b needs to be equal to negative 20, that needs to be a plus b and then a times b right over here, that needs to be equal to the constant term, that needs to be a times b right over there."}, {"video_title": "Solving quadratics by factoring leading coefficient \u00c3\u00a2\u00c2  \u00c2\u00a0 1   High School Math   Khan Academy.mp3", "Sentence": "So if I do that and clearly if I do the same thing to both sides of the equation then the equality still holds. On the left hand side I am going to be left with x squared and then negative 120 divided by six, that is, let's see, 120 divided by six is 20 so that's minus 20 x and then 600 divided by six is 100 so plus 100 is equal to zero divided by six, is equal to zero. So let's see if we can factor, if we can express this quadratic as a product of two expressions. And the way we think about this, and we've done it multiple times, if we have something, if we have x plus a times x plus b and this is hopefully a review for you, if you multiply that out that is going to be equal to, that equals to x squared plus a plus b x plus ab and so what we want to do is see if we can factor this into an x plus a and an x plus b and so a plus b needs to be equal to negative 20, that needs to be a plus b and then a times b right over here, that needs to be equal to the constant term, that needs to be a times b right over there. So can we think of two numbers that if we take their product we get positive 100 and if we take their sum we get negative 20? Well since their product is positive we know that they have the same sign, so they're both going to have the same sign, so they're either both going to be positive or they're both going to be negative since we know that we have a positive product and since their sum is negative, well they both must both be negative, although you can't add up two positive numbers and get a negative, so they both must be negative. So let's think about it a little bit."}, {"video_title": "Solving quadratics by factoring leading coefficient \u00c3\u00a2\u00c2  \u00c2\u00a0 1   High School Math   Khan Academy.mp3", "Sentence": "And the way we think about this, and we've done it multiple times, if we have something, if we have x plus a times x plus b and this is hopefully a review for you, if you multiply that out that is going to be equal to, that equals to x squared plus a plus b x plus ab and so what we want to do is see if we can factor this into an x plus a and an x plus b and so a plus b needs to be equal to negative 20, that needs to be a plus b and then a times b right over here, that needs to be equal to the constant term, that needs to be a times b right over there. So can we think of two numbers that if we take their product we get positive 100 and if we take their sum we get negative 20? Well since their product is positive we know that they have the same sign, so they're both going to have the same sign, so they're either both going to be positive or they're both going to be negative since we know that we have a positive product and since their sum is negative, well they both must both be negative, although you can't add up two positive numbers and get a negative, so they both must be negative. So let's think about it a little bit. What negative numbers, when I add them together I get negative 20 and when I multiply I get 100? Well you could try to factor 100, you could say well negative two times negative 50 or negative four times negative 25, but the one that might jump out at you is this is negative 10 times, I'll write it this way, negative 10 times negative 10 and this is negative 10 plus negative 10. So in that case both our A and our B would be negative 10 and so we can rewrite the left side of this equation as, we can rewrite it as X, and I'll write it this way at first, X plus negative 10 times, times X plus negative 10 again, X plus negative 10 and that is going to be equal to zero."}, {"video_title": "Solving quadratics by factoring leading coefficient \u00c3\u00a2\u00c2  \u00c2\u00a0 1   High School Math   Khan Academy.mp3", "Sentence": "So let's think about it a little bit. What negative numbers, when I add them together I get negative 20 and when I multiply I get 100? Well you could try to factor 100, you could say well negative two times negative 50 or negative four times negative 25, but the one that might jump out at you is this is negative 10 times, I'll write it this way, negative 10 times negative 10 and this is negative 10 plus negative 10. So in that case both our A and our B would be negative 10 and so we can rewrite the left side of this equation as, we can rewrite it as X, and I'll write it this way at first, X plus negative 10 times, times X plus negative 10 again, X plus negative 10 and that is going to be equal to zero. So all I've done is I've factored this quadratic. Or another way, these are both the same thing as X minus 10, I could rewrite this as X minus 10 squared is equal to zero and so the only way that the left hand side is going to be equal to zero is if X minus 10 is equal to zero. You could think of this as taking the square root of both sides and it doesn't matter if you're taking the positive or negative square root or both of them, the square root of zero is zero and so we would say that X minus 10 needs to be equal to zero and so X, adding 10 to both sides of this, you have X is equal to 10 is the solution to this quadratic equation up here."}, {"video_title": "Introduction to square roots   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "And now that we know a little bit about exponents, we'll see that the square root symbol, or the root symbol, or the radical, is not so, is not so hard to understand. So let's start with an example. So we know that three to the second power is what? Three squared is what? Well, that's the same thing as three times three. That's going to be equal to nine. But what if we went the other way around?"}, {"video_title": "Introduction to square roots   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "Three squared is what? Well, that's the same thing as three times three. That's going to be equal to nine. But what if we went the other way around? What if we started with a nine, and we said, well, what times itself is equal to nine? We already know that answer is three, but how could we, how could we use a symbol that tells us that? So, well, as you can imagine, that symbol is, this is going to be the radical here."}, {"video_title": "Introduction to square roots   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "But what if we went the other way around? What if we started with a nine, and we said, well, what times itself is equal to nine? We already know that answer is three, but how could we, how could we use a symbol that tells us that? So, well, as you can imagine, that symbol is, this is going to be the radical here. So we could write the square root of nine, and when you look at this way, you say, okay, what squared is equal to nine? And you would say, well, this is going to be equal to, this is going to be equal to three. And I want you to really look at these two, these two, these two equations right over here, because this is the essence of the square root symbol."}, {"video_title": "Introduction to square roots   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "So, well, as you can imagine, that symbol is, this is going to be the radical here. So we could write the square root of nine, and when you look at this way, you say, okay, what squared is equal to nine? And you would say, well, this is going to be equal to, this is going to be equal to three. And I want you to really look at these two, these two, these two equations right over here, because this is the essence of the square root symbol. If you say the square root of nine, you're saying what times itself is equal to nine? And, well, that's going to be three. And three squared is equal to nine."}, {"video_title": "Introduction to square roots   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "And I want you to really look at these two, these two, these two equations right over here, because this is the essence of the square root symbol. If you say the square root of nine, you're saying what times itself is equal to nine? And, well, that's going to be three. And three squared is equal to nine. I can do that again. I can do that many times. I can write four, four squared is equal to 16."}, {"video_title": "Introduction to square roots   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "And three squared is equal to nine. I can do that again. I can do that many times. I can write four, four squared is equal to 16. Well, what's the square root of 16 going to be? Well, it's going to be equal to, it's going to be equal to four. Let me do it again."}, {"video_title": "Introduction to square roots   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "I can write four, four squared is equal to 16. Well, what's the square root of 16 going to be? Well, it's going to be equal to, it's going to be equal to four. Let me do it again. And actually, let me start with the square root. What is the square root of 25 going to be? Well, this is the number that times itself is going to be equal to 25, or the number where if I were to square it, I'd get to 25."}, {"video_title": "Introduction to square roots   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "Let me do it again. And actually, let me start with the square root. What is the square root of 25 going to be? Well, this is the number that times itself is going to be equal to 25, or the number where if I were to square it, I'd get to 25. Well, what number is that? Well, that's going to be equal to five. Why?"}, {"video_title": "Introduction to square roots   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "Well, this is the number that times itself is going to be equal to 25, or the number where if I were to square it, I'd get to 25. Well, what number is that? Well, that's going to be equal to five. Why? Because we know that five squared is equal to, five squared is equal to 25. Now, I know that there's a nagging feeling that some of you might be having, because if I were to take negative three and square it, I would also get positive nine. And the same thing, if I were to take negative four and I were to square the whole thing, I would also get positive 16, or negative five, and if I were to square it, I would also get positive 25."}, {"video_title": "Introduction to square roots   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "Why? Because we know that five squared is equal to, five squared is equal to 25. Now, I know that there's a nagging feeling that some of you might be having, because if I were to take negative three and square it, I would also get positive nine. And the same thing, if I were to take negative four and I were to square the whole thing, I would also get positive 16, or negative five, and if I were to square it, I would also get positive 25. So why couldn't this thing right over here, why can't this square root be positive three or negative three? Well, depending on who you talk to, that's actually a reasonable thing to think about. But when you see a radical symbol like this, people usually call this the principal root."}, {"video_title": "Introduction to square roots   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "And the same thing, if I were to take negative four and I were to square the whole thing, I would also get positive 16, or negative five, and if I were to square it, I would also get positive 25. So why couldn't this thing right over here, why can't this square root be positive three or negative three? Well, depending on who you talk to, that's actually a reasonable thing to think about. But when you see a radical symbol like this, people usually call this the principal root. Principal root. Principal, principal square root. Square root."}, {"video_title": "Introduction to square roots   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "But when you see a radical symbol like this, people usually call this the principal root. Principal root. Principal, principal square root. Square root. And another way to think about it, it's the positive. This is going to be the positive square root. If someone wants the negative square root of nine, they might say something like this."}, {"video_title": "Introduction to square roots   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "Square root. And another way to think about it, it's the positive. This is going to be the positive square root. If someone wants the negative square root of nine, they might say something like this. They might say the negative, let me scroll up a little bit, they might say something like the negative square root of nine, well that's going to be equal to negative three. And what's interesting about this is, well, if you square both sides of this, of this equation, if you were to square both sides of this equation, what do you get? Well, negative, anything negative squared becomes a positive, and then the square root of nine squared, well that's just going to be nine."}, {"video_title": "Introduction to square roots   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "If someone wants the negative square root of nine, they might say something like this. They might say the negative, let me scroll up a little bit, they might say something like the negative square root of nine, well that's going to be equal to negative three. And what's interesting about this is, well, if you square both sides of this, of this equation, if you were to square both sides of this equation, what do you get? Well, negative, anything negative squared becomes a positive, and then the square root of nine squared, well that's just going to be nine. And on the right-hand side, negative three squared, well negative three times negative three is positive nine. So it all works out. Nine is equal, nine is equal to nine."}, {"video_title": "Introduction to square roots   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "Well, negative, anything negative squared becomes a positive, and then the square root of nine squared, well that's just going to be nine. And on the right-hand side, negative three squared, well negative three times negative three is positive nine. So it all works out. Nine is equal, nine is equal to nine. And so this is an interesting thing. Actually, let me write this a little bit more algebraically now. If we were to write, if we were to write the principal root of nine is equal to x, this is, there's only one possible x here that satisfies it because the standard convention, what most mathematicians have agreed to view this radical symbol as is that this is the principal square root, this is the positive square root."}, {"video_title": "Introduction to square roots   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "Nine is equal, nine is equal to nine. And so this is an interesting thing. Actually, let me write this a little bit more algebraically now. If we were to write, if we were to write the principal root of nine is equal to x, this is, there's only one possible x here that satisfies it because the standard convention, what most mathematicians have agreed to view this radical symbol as is that this is the principal square root, this is the positive square root. So there's only one x here. There's only one x that would satisfy this, and that is x is equal to three. Now, if I were to write x squared is equal to nine, now this is slightly different."}, {"video_title": "Introduction to square roots   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "If we were to write, if we were to write the principal root of nine is equal to x, this is, there's only one possible x here that satisfies it because the standard convention, what most mathematicians have agreed to view this radical symbol as is that this is the principal square root, this is the positive square root. So there's only one x here. There's only one x that would satisfy this, and that is x is equal to three. Now, if I were to write x squared is equal to nine, now this is slightly different. x equals three definitely satisfies this. This could be x equals three. But the other thing, the other x that satisfies this could also be, x could also be equal to negative three because negative three squared is also equal to nine."}, {"video_title": "Example 2 Adding polynomials   Algebra I   Khan Academy.mp3", "Sentence": "We're asked to simplify 5x squared plus 8x minus 3 plus 2x squared minus 7x plus 13x. So really all we have to do is we have to combine like terms, terms that have x raised to the same power. And the first thing we can do, we can actually get rid of these parentheses right here, because we have this whole expression and then we're adding it to this whole expression. The parentheses really don't change our order of operations here. So let me just rewrite it once without the parentheses. So we have 5x squared plus 8x minus 3 plus 2x squared. If this was a minus, then we'd have to distribute the negative sign, but it's not."}, {"video_title": "Example 2 Adding polynomials   Algebra I   Khan Academy.mp3", "Sentence": "The parentheses really don't change our order of operations here. So let me just rewrite it once without the parentheses. So we have 5x squared plus 8x minus 3 plus 2x squared. If this was a minus, then we'd have to distribute the negative sign, but it's not. So plus 2x squared minus 7x plus 13x. Now let's just look at the different terms that have different degrees of x. Let's start with the x squared terms."}, {"video_title": "Example 2 Adding polynomials   Algebra I   Khan Academy.mp3", "Sentence": "If this was a minus, then we'd have to distribute the negative sign, but it's not. So plus 2x squared minus 7x plus 13x. Now let's just look at the different terms that have different degrees of x. Let's start with the x squared terms. So you have a 5x squared term here and you have a 2x squared term right there. So 5 of something plus 2 of that same something is going to be 7 of that something. So that's going to be 7x squared."}, {"video_title": "Example 2 Adding polynomials   Algebra I   Khan Academy.mp3", "Sentence": "Let's start with the x squared terms. So you have a 5x squared term here and you have a 2x squared term right there. So 5 of something plus 2 of that same something is going to be 7 of that something. So that's going to be 7x squared. And then let's look at the x terms here. So we have an 8x right there. We have a minus 7x and then we have a plus 13x."}, {"video_title": "Example 2 Adding polynomials   Algebra I   Khan Academy.mp3", "Sentence": "So that's going to be 7x squared. And then let's look at the x terms here. So we have an 8x right there. We have a minus 7x and then we have a plus 13x. So if you have 8 of something minus 7 of something, you're just going to have 1 of that something. And if you add 14 of that something more, you're going to have 15. So this is going to be plus 15x."}, {"video_title": "Example 2 Adding polynomials   Algebra I   Khan Academy.mp3", "Sentence": "We have a minus 7x and then we have a plus 13x. So if you have 8 of something minus 7 of something, you're just going to have 1 of that something. And if you add 14 of that something more, you're going to have 15. So this is going to be plus 15x. 8x minus 7x. Oh, sorry, you're going to have 14x. 8 minus 7 is 1 plus 13 is 14."}, {"video_title": "Example 2 Adding polynomials   Algebra I   Khan Academy.mp3", "Sentence": "So this is going to be plus 15x. 8x minus 7x. Oh, sorry, you're going to have 14x. 8 minus 7 is 1 plus 13 is 14. Plus 14x. That's these three terms. 8x minus 7x plus 13x."}, {"video_title": "Example 2 Adding polynomials   Algebra I   Khan Academy.mp3", "Sentence": "8 minus 7 is 1 plus 13 is 14. Plus 14x. That's these three terms. 8x minus 7x plus 13x. And then finally, you have a negative 3 or minus 3, depending on how you want to view it. And that's the only constant term. You can imagine it's, you could say it's x times x to the 0."}, {"video_title": "Example 2 Adding polynomials   Algebra I   Khan Academy.mp3", "Sentence": "8x minus 7x plus 13x. And then finally, you have a negative 3 or minus 3, depending on how you want to view it. And that's the only constant term. You can imagine it's, you could say it's x times x to the 0. But it's a constant term. It's not being multiplied by x. And that's the only one there."}, {"video_title": "Example 2 Adding polynomials   Algebra I   Khan Academy.mp3", "Sentence": "You can imagine it's, you could say it's x times x to the 0. But it's a constant term. It's not being multiplied by x. And that's the only one there. So minus 3. And we've simplified it as far as we can go. We are done."}, {"video_title": "Adding decimals example 2   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "And when I rewrite it, I want to line up the decimals so that we add the right place to the right place. And so we could write either number first, although I like to write the larger number first. So let's write 5.65. And remember, the important thing is that we line up the decimal points. So if we write 0.822, so we line up the decimal. Let me line up the decimal first. So I'll write the decimal right below the other decimal."}, {"video_title": "Adding decimals example 2   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "And remember, the important thing is that we line up the decimal points. So if we write 0.822, so we line up the decimal. Let me line up the decimal first. So I'll write the decimal right below the other decimal. And it is 0.822. And now we are ready to add. So let's see what's going on here."}, {"video_title": "Adding decimals example 2   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "So I'll write the decimal right below the other decimal. And it is 0.822. And now we are ready to add. So let's see what's going on here. So I like to start in the smallest place. That way the carrying works out well. So you might say, wait, I need to add this 2,000th to something."}, {"video_title": "Adding decimals example 2   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "So let's see what's going on here. So I like to start in the smallest place. That way the carrying works out well. So you might say, wait, I need to add this 2,000th to something. I don't see anything up here. Well, you could say there's just a 0,000th right up here. And then it makes it very clear."}, {"video_title": "Adding decimals example 2   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "So you might say, wait, I need to add this 2,000th to something. I don't see anything up here. Well, you could say there's just a 0,000th right up here. And then it makes it very clear. Well, 0,000th plus 2,000th is going to be 2,000th. 5 hundredths plus 2 hundredths is 7 hundredths. 6 tenths plus 8 tenths is 14 tenths."}, {"video_title": "Adding decimals example 2   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "And then it makes it very clear. Well, 0,000th plus 2,000th is going to be 2,000th. 5 hundredths plus 2 hundredths is 7 hundredths. 6 tenths plus 8 tenths is 14 tenths. Well, 14 tenths is the same thing as 4 tenths and 1 one. Another way of thinking about it is you're carrying the one. But really what you're saying is, look, this is 14 tenths."}, {"video_title": "Adding decimals example 2   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "6 tenths plus 8 tenths is 14 tenths. Well, 14 tenths is the same thing as 4 tenths and 1 one. Another way of thinking about it is you're carrying the one. But really what you're saying is, look, this is 14 tenths. I could write that as 4 tenths and a one. Or a one's place. Or one and a one's place."}, {"video_title": "Adding decimals example 2   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "But really what you're saying is, look, this is 14 tenths. I could write that as 4 tenths and a one. Or a one's place. Or one and a one's place. And then you have 1 plus 5 is 6. And of course, you cannot forget the decimal. The decimal goes right there."}, {"video_title": "Introduction to order of operations   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "And I want you to pay close attention because really everything else that you're going to do in mathematics is going to be based on you having a solid grounding in order of operations. So what do we even mean when we say order of operations? So let me give you an example. The whole point is so that we have one way to interpret a mathematical statement. So let's say I have the mathematical statement 7 plus 3 times 5. Now, if we didn't all agree on order of operations, there would be two ways of interpreting this statement. You could just read it left to right."}, {"video_title": "Introduction to order of operations   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "The whole point is so that we have one way to interpret a mathematical statement. So let's say I have the mathematical statement 7 plus 3 times 5. Now, if we didn't all agree on order of operations, there would be two ways of interpreting this statement. You could just read it left to right. So you could say, well, let me just take 7 plus 3. You could say 7 plus 3 and then multiply that times 5. And 7 plus 3 is 10."}, {"video_title": "Introduction to order of operations   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "You could just read it left to right. So you could say, well, let me just take 7 plus 3. You could say 7 plus 3 and then multiply that times 5. And 7 plus 3 is 10. And then you multiply that by 5. 10 times 5, it would get you 50. So that's one way you would interpret it if we didn't agree on an order of operations."}, {"video_title": "Introduction to order of operations   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "And 7 plus 3 is 10. And then you multiply that by 5. 10 times 5, it would get you 50. So that's one way you would interpret it if we didn't agree on an order of operations. Maybe it's a natural way. You just go left to right. Another way you could interpret it, you say, oh, I like to do multiplication before I do addition."}, {"video_title": "Introduction to order of operations   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "So that's one way you would interpret it if we didn't agree on an order of operations. Maybe it's a natural way. You just go left to right. Another way you could interpret it, you say, oh, I like to do multiplication before I do addition. So you might interpret it as, I'll try to color code it, 7 plus, and you do the 3 times 5 first. 7 plus 3 times 5, which would be 7 plus 3 times 5 is 15. And 7 plus 15 is 22."}, {"video_title": "Introduction to order of operations   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "Another way you could interpret it, you say, oh, I like to do multiplication before I do addition. So you might interpret it as, I'll try to color code it, 7 plus, and you do the 3 times 5 first. 7 plus 3 times 5, which would be 7 plus 3 times 5 is 15. And 7 plus 15 is 22. So notice, we interpreted the statement in two different ways. This was just straight left to right, doing the addition, then the multiplication. This way, we did the multiplication first, then the addition."}, {"video_title": "Introduction to order of operations   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "And 7 plus 15 is 22. So notice, we interpreted the statement in two different ways. This was just straight left to right, doing the addition, then the multiplication. This way, we did the multiplication first, then the addition. We got two different answers. And that's just not cool in mathematics. If this was part of some effort to send something to the moon because two people interpreted it a different way, or one computer interpreted it one way and another computer interpreted it another way, the satellite might go to Mars."}, {"video_title": "Introduction to order of operations   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "This way, we did the multiplication first, then the addition. We got two different answers. And that's just not cool in mathematics. If this was part of some effort to send something to the moon because two people interpreted it a different way, or one computer interpreted it one way and another computer interpreted it another way, the satellite might go to Mars. So this is just completely unacceptable. And that's why we have to have an agreed upon order of operations, an agreed upon way to interpret this statement. So the agreed upon order of operations is to do parentheses first."}, {"video_title": "Introduction to order of operations   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "If this was part of some effort to send something to the moon because two people interpreted it a different way, or one computer interpreted it one way and another computer interpreted it another way, the satellite might go to Mars. So this is just completely unacceptable. And that's why we have to have an agreed upon order of operations, an agreed upon way to interpret this statement. So the agreed upon order of operations is to do parentheses first. Let me write it over here. Then do exponents. If you don't know what exponents are, don't worry about it right now."}, {"video_title": "Introduction to order of operations   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "So the agreed upon order of operations is to do parentheses first. Let me write it over here. Then do exponents. If you don't know what exponents are, don't worry about it right now. And in this video, we're not going to have any exponents in our examples. So you don't really have to worry about them for this video. Then you do multiplication."}, {"video_title": "Introduction to order of operations   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "If you don't know what exponents are, don't worry about it right now. And in this video, we're not going to have any exponents in our examples. So you don't really have to worry about them for this video. Then you do multiplication. I'll just write mult short for multiplication. Then you do multiplication and division next. They kind of have the same level of priority."}, {"video_title": "Introduction to order of operations   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "Then you do multiplication. I'll just write mult short for multiplication. Then you do multiplication and division next. They kind of have the same level of priority. And then finally you do addition and subtraction. So what is this order of operations? Let me label it."}, {"video_title": "Introduction to order of operations   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "They kind of have the same level of priority. And then finally you do addition and subtraction. So what is this order of operations? Let me label it. This right here, that is the agreed upon order of operations. And if we follow these order of operations, we should always get to the same answer for a given statement. So what does this tell us?"}, {"video_title": "Introduction to order of operations   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "Let me label it. This right here, that is the agreed upon order of operations. And if we follow these order of operations, we should always get to the same answer for a given statement. So what does this tell us? What is the best way to interpret this up here? Well, we have no parentheses. Parentheses look like that."}, {"video_title": "Introduction to order of operations   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "So what does this tell us? What is the best way to interpret this up here? Well, we have no parentheses. Parentheses look like that. Those little curly things around numbers. We don't have any parentheses here. I'll do some examples that do have parentheses."}, {"video_title": "Introduction to order of operations   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "Parentheses look like that. Those little curly things around numbers. We don't have any parentheses here. I'll do some examples that do have parentheses. We don't have any exponents here. But we do have some multiplication and division. Or we actually just have some multiplication."}, {"video_title": "Introduction to order of operations   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "I'll do some examples that do have parentheses. We don't have any exponents here. But we do have some multiplication and division. Or we actually just have some multiplication. So the order of operations say, do the multiplication and division first. So it says, do the multiplication first. That's a multiplication."}, {"video_title": "Introduction to order of operations   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "Or we actually just have some multiplication. So the order of operations say, do the multiplication and division first. So it says, do the multiplication first. That's a multiplication. So it says, do this operation first. It gets priority over addition or subtraction. So if we do this first, we get the 3 times 5, which is 15."}, {"video_title": "Introduction to order of operations   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "That's a multiplication. So it says, do this operation first. It gets priority over addition or subtraction. So if we do this first, we get the 3 times 5, which is 15. And then we add the 7. The addition or subtraction, I'll do it here. Addition, we just have addition."}, {"video_title": "Introduction to order of operations   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "So if we do this first, we get the 3 times 5, which is 15. And then we add the 7. The addition or subtraction, I'll do it here. Addition, we just have addition. Just like that. So we do the multiplication first, get 15. Then add the 7."}, {"video_title": "Introduction to order of operations   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "Addition, we just have addition. Just like that. So we do the multiplication first, get 15. Then add the 7. 22. So based upon the agreed order of operations, this right here is the correct answer. The correct way to interpret this statement."}, {"video_title": "Introduction to order of operations   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "Then add the 7. 22. So based upon the agreed order of operations, this right here is the correct answer. The correct way to interpret this statement. Let's do another example. I think it will make things a little bit more clear. And I'll do the example in pink."}, {"video_title": "Introduction to order of operations   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "The correct way to interpret this statement. Let's do another example. I think it will make things a little bit more clear. And I'll do the example in pink. So let's say I have 7 plus 3. Put some parentheses there. Times 4 divided by 2 minus 5 times 6."}, {"video_title": "Introduction to order of operations   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "And I'll do the example in pink. So let's say I have 7 plus 3. Put some parentheses there. Times 4 divided by 2 minus 5 times 6. So there's all sorts of crazy things here. But if you just follow the order of operations, you'll simplify it in a very, I guess, clean way. And hopefully we'll all get the same answer."}, {"video_title": "Introduction to order of operations   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "Times 4 divided by 2 minus 5 times 6. So there's all sorts of crazy things here. But if you just follow the order of operations, you'll simplify it in a very, I guess, clean way. And hopefully we'll all get the same answer. So let's just follow the order of operations. The first thing we have to do is look for parentheses. Are there parentheses here?"}, {"video_title": "Introduction to order of operations   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "And hopefully we'll all get the same answer. So let's just follow the order of operations. The first thing we have to do is look for parentheses. Are there parentheses here? Yes, there are. There's parentheses around the 7 plus 3. So it says, let's do that first."}, {"video_title": "Introduction to order of operations   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "Are there parentheses here? Yes, there are. There's parentheses around the 7 plus 3. So it says, let's do that first. So 7 plus 3 is 10. So this we can simplify, just looking at this order of operations, to 10 times all of that. Let me copy and paste that so I don't have to keep rewriting it."}, {"video_title": "Introduction to order of operations   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "So it says, let's do that first. So 7 plus 3 is 10. So this we can simplify, just looking at this order of operations, to 10 times all of that. Let me copy and paste that so I don't have to keep rewriting it. So let me copy and let me paste it. So that simplifies to 10 times all of that. We did our parentheses first."}, {"video_title": "Introduction to order of operations   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "Let me copy and paste that so I don't have to keep rewriting it. So let me copy and let me paste it. So that simplifies to 10 times all of that. We did our parentheses first. Then what do we do? There are no more parentheses in this expression. Then we should do exponents."}, {"video_title": "Introduction to order of operations   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "We did our parentheses first. Then what do we do? There are no more parentheses in this expression. Then we should do exponents. I don't see any exponents here. And just so, if you're curious what exponents look like, an exponent would look like 7 squared. You'd see these little small numbers up in the top right."}, {"video_title": "Introduction to order of operations   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "Then we should do exponents. I don't see any exponents here. And just so, if you're curious what exponents look like, an exponent would look like 7 squared. You'd see these little small numbers up in the top right. We don't have any exponents here, so we don't have to worry about it. Then it says to do multiplication and division next. So where do we see multiplication and division?"}, {"video_title": "Introduction to order of operations   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "You'd see these little small numbers up in the top right. We don't have any exponents here, so we don't have to worry about it. Then it says to do multiplication and division next. So where do we see multiplication and division? We have a multiplication, a division, a multiplication again. Now, when you have multiple operations at the same level, when our order of operations, multiplication and division, are at the same level, then you do left to right. So in this situation, you're going to multiply by 4 and then divide by 2."}, {"video_title": "Introduction to order of operations   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "So where do we see multiplication and division? We have a multiplication, a division, a multiplication again. Now, when you have multiple operations at the same level, when our order of operations, multiplication and division, are at the same level, then you do left to right. So in this situation, you're going to multiply by 4 and then divide by 2. You won't multiply by 4 divided by 2. Then we'll do the 5 times 6 before we do this subtraction right here. So let's figure out what this is."}, {"video_title": "Introduction to order of operations   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "So in this situation, you're going to multiply by 4 and then divide by 2. You won't multiply by 4 divided by 2. Then we'll do the 5 times 6 before we do this subtraction right here. So let's figure out what this is. So we'll do this multiplication first. We'll do that multiplication first. We could simultaneously do this multiplication because it's not going to change things."}, {"video_title": "Introduction to order of operations   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "So let's figure out what this is. So we'll do this multiplication first. We'll do that multiplication first. We could simultaneously do this multiplication because it's not going to change things. But I'll do things one step at a time. So the next step we're going to do is this 10 times 4. 10 times 4 is 40."}, {"video_title": "Introduction to order of operations   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "We could simultaneously do this multiplication because it's not going to change things. But I'll do things one step at a time. So the next step we're going to do is this 10 times 4. 10 times 4 is 40. Then you have 40 divided by 2. Let me copy and paste all of that again. Then it simplifies to that right there."}, {"video_title": "Introduction to order of operations   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "10 times 4 is 40. Then you have 40 divided by 2. Let me copy and paste all of that again. Then it simplifies to that right there. Remember, multiplication and division, they're at the exact same level, so we're going to do it left to right. You could also express this as multiplying by 1 half, and then it wouldn't matter the order. But for simplicity, multiplication and division, go left to right."}, {"video_title": "Introduction to order of operations   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "Then it simplifies to that right there. Remember, multiplication and division, they're at the exact same level, so we're going to do it left to right. You could also express this as multiplying by 1 half, and then it wouldn't matter the order. But for simplicity, multiplication and division, go left to right. Then you have 40 divided by 2 minus 5 times 6. So you just have one division here. You want to do that."}, {"video_title": "Introduction to order of operations   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "But for simplicity, multiplication and division, go left to right. Then you have 40 divided by 2 minus 5 times 6. So you just have one division here. You want to do that. You have this division, then you have this multiplication. They're not together, so you can actually do them simultaneously. To make it clear that you do this before you do the subtraction, because multiplication and division take priority over addition and subtraction, we could put parentheses around them to say, we're going to do that and that first before I do that subtraction, because multiplication and division have priority."}, {"video_title": "Introduction to order of operations   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "You want to do that. You have this division, then you have this multiplication. They're not together, so you can actually do them simultaneously. To make it clear that you do this before you do the subtraction, because multiplication and division take priority over addition and subtraction, we could put parentheses around them to say, we're going to do that and that first before I do that subtraction, because multiplication and division have priority. So 40 divided by 2 is 20. We're going to have that minus sign. Minus 5 times 6 is 30."}, {"video_title": "Introduction to order of operations   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "To make it clear that you do this before you do the subtraction, because multiplication and division take priority over addition and subtraction, we could put parentheses around them to say, we're going to do that and that first before I do that subtraction, because multiplication and division have priority. So 40 divided by 2 is 20. We're going to have that minus sign. Minus 5 times 6 is 30. 20 minus 30 is equal to negative 10. That is the correct interpretation of that. I want to make something very, very, very clear."}, {"video_title": "Introduction to order of operations   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "Minus 5 times 6 is 30. 20 minus 30 is equal to negative 10. That is the correct interpretation of that. I want to make something very, very, very clear. If you have things at the same level, so if you have 1 plus 2 minus 3 plus 4 minus 1, so addition and subtraction are all at the same level in order of operations. You should go left to right. You should interpret this as 1 plus 2 is 3."}, {"video_title": "Introduction to order of operations   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "I want to make something very, very, very clear. If you have things at the same level, so if you have 1 plus 2 minus 3 plus 4 minus 1, so addition and subtraction are all at the same level in order of operations. You should go left to right. You should interpret this as 1 plus 2 is 3. So this is the same thing as 3 minus 3 plus 4 minus 1. Then you do 3 minus 3 is 0 plus 4 minus 1. Or this is the same thing as 4 minus 1, which is the same thing as 3."}, {"video_title": "Introduction to order of operations   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "You should interpret this as 1 plus 2 is 3. So this is the same thing as 3 minus 3 plus 4 minus 1. Then you do 3 minus 3 is 0 plus 4 minus 1. Or this is the same thing as 4 minus 1, which is the same thing as 3. You just go left to right. Same thing if you have multiplication and division. They're at the same level."}, {"video_title": "Introduction to order of operations   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "Or this is the same thing as 4 minus 1, which is the same thing as 3. You just go left to right. Same thing if you have multiplication and division. They're at the same level. So if you have 4 times 2 divided by 3 times 2, you do 4 times 2 is 8 divided by 3 times 2. You say 8 divided by 3 is, well, we've got a fraction there. It would be 8 thirds."}, {"video_title": "Introduction to order of operations   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "They're at the same level. So if you have 4 times 2 divided by 3 times 2, you do 4 times 2 is 8 divided by 3 times 2. You say 8 divided by 3 is, well, we've got a fraction there. It would be 8 thirds. So this would be 8 thirds times 2. And then 8 thirds times 2 is equal to 16 over 3. That's how you interpret it."}, {"video_title": "Introduction to order of operations   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "It would be 8 thirds. So this would be 8 thirds times 2. And then 8 thirds times 2 is equal to 16 over 3. That's how you interpret it. You don't do this multiplication first and divide the 2 by that and all of that. Now, the one time where you can be loosey-goosey with order of operations, if you have all addition or all multiplication. So if you have 1 plus 5 plus 7 plus 3 plus 2 does not matter what order you do it in."}, {"video_title": "Introduction to order of operations   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "That's how you interpret it. You don't do this multiplication first and divide the 2 by that and all of that. Now, the one time where you can be loosey-goosey with order of operations, if you have all addition or all multiplication. So if you have 1 plus 5 plus 7 plus 3 plus 2 does not matter what order you do it in. You can do the 2 plus 3. You can go from the right to the left. You can go from the left to the right."}, {"video_title": "Introduction to order of operations   Arithmetic properties   Pre-Algebra   Khan Academy.mp3", "Sentence": "So if you have 1 plus 5 plus 7 plus 3 plus 2 does not matter what order you do it in. You can do the 2 plus 3. You can go from the right to the left. You can go from the left to the right. You can start someplace in between. If it's only all addition, and the same thing is true if you have all multiplication. If it's 1 times 5 times 7 times 3 times 2, does not matter what order you're doing it."}, {"video_title": "Solving systems of equations graphically   Algebra II   Khan Academy.mp3", "Sentence": "Just in case we encounter any more trolls who want us to figure out what types of money they have in their pockets, we have devised an exercise for you to practice with. And this is to solve systems of equations visually. So they say right over here, graph this system of equations and solve. And they give us two equations, this first one in blue. Y is equal to 7 5ths x minus five, and then this one in green. Y is equal to 3 5ths x minus one. So let's graph each of these, and we'll do it in the corresponding color."}, {"video_title": "Solving systems of equations graphically   Algebra II   Khan Academy.mp3", "Sentence": "And they give us two equations, this first one in blue. Y is equal to 7 5ths x minus five, and then this one in green. Y is equal to 3 5ths x minus one. So let's graph each of these, and we'll do it in the corresponding color. So first let's graph this first equation. So the first thing I see is that it's y-intercept is negative five. Or another way to think about it, when x is equal to zero, y is going to be negative five."}, {"video_title": "Solving systems of equations graphically   Algebra II   Khan Academy.mp3", "Sentence": "So let's graph each of these, and we'll do it in the corresponding color. So first let's graph this first equation. So the first thing I see is that it's y-intercept is negative five. Or another way to think about it, when x is equal to zero, y is going to be negative five. So let's try this out. So when x is equal to zero, y is going to be equal to negative five. So that makes sense."}, {"video_title": "Solving systems of equations graphically   Algebra II   Khan Academy.mp3", "Sentence": "Or another way to think about it, when x is equal to zero, y is going to be negative five. So let's try this out. So when x is equal to zero, y is going to be equal to negative five. So that makes sense. And then we see its slope is 7 5ths. This was conveniently placed in slope intercept form for us. So it's rise over run."}, {"video_title": "Solving systems of equations graphically   Algebra II   Khan Academy.mp3", "Sentence": "So that makes sense. And then we see its slope is 7 5ths. This was conveniently placed in slope intercept form for us. So it's rise over run. So for every time it moves five to the right, it's going to move seven up. So if it moves one, two, three, four, five to the right, it's going to move seven up. One, two, three, four, five, six, seven."}, {"video_title": "Solving systems of equations graphically   Algebra II   Khan Academy.mp3", "Sentence": "So it's rise over run. So for every time it moves five to the right, it's going to move seven up. So if it moves one, two, three, four, five to the right, it's going to move seven up. One, two, three, four, five, six, seven. So it'll get right over there. Another way you could have done it is you could have just tested out some values. You could have said, oh, when x is equal to zero, y is equal to negative five."}, {"video_title": "Solving systems of equations graphically   Algebra II   Khan Academy.mp3", "Sentence": "One, two, three, four, five, six, seven. So it'll get right over there. Another way you could have done it is you could have just tested out some values. You could have said, oh, when x is equal to zero, y is equal to negative five. When x is equal to five, 7 5ths times five is seven, minus five is two. So I think we've properly graphed this top one. Let's try this bottom one right over here."}, {"video_title": "Solving systems of equations graphically   Algebra II   Khan Academy.mp3", "Sentence": "You could have said, oh, when x is equal to zero, y is equal to negative five. When x is equal to five, 7 5ths times five is seven, minus five is two. So I think we've properly graphed this top one. Let's try this bottom one right over here. So we have negative, when x is equal to zero, y is equal to negative one. So when x is equal to zero, y is equal to negative one. And the slope is 3 5ths."}, {"video_title": "Solving systems of equations graphically   Algebra II   Khan Academy.mp3", "Sentence": "Let's try this bottom one right over here. So we have negative, when x is equal to zero, y is equal to negative one. So when x is equal to zero, y is equal to negative one. And the slope is 3 5ths. So if we move over five to the right, if we move over five to the right, we will move up three. So we will go right over there. And it looks like they intersect right at that point, right at the point x is equal to five, y is equal to two."}, {"video_title": "Solving systems of equations graphically   Algebra II   Khan Academy.mp3", "Sentence": "And the slope is 3 5ths. So if we move over five to the right, if we move over five to the right, we will move up three. So we will go right over there. And it looks like they intersect right at that point, right at the point x is equal to five, y is equal to two. So we'll type in x is equal to five, y is equal to two. And you could even verify by substituting those values into both equations to show that it does satisfy both constraints. So let's check our answer."}, {"video_title": "Finding intercepts from an equation   Algebra I   Khan Academy.mp3", "Sentence": "And we're told to find the intercepts of this equation. So we have to find the intercepts. And then use the intercepts to graph this line on the coordinate plane. So then graph the line. So whenever someone talks about intercepts, they're talking about where you're intersecting the x and the y axes. So let me label my axes here. So this is the x-axis, and that is the y-axis there."}, {"video_title": "Finding intercepts from an equation   Algebra I   Khan Academy.mp3", "Sentence": "So then graph the line. So whenever someone talks about intercepts, they're talking about where you're intersecting the x and the y axes. So let me label my axes here. So this is the x-axis, and that is the y-axis there. And when I intersect the x-axis, what's going on? What is my y value when I'm at the x-axis? Well, my y value is 0."}, {"video_title": "Finding intercepts from an equation   Algebra I   Khan Academy.mp3", "Sentence": "So this is the x-axis, and that is the y-axis there. And when I intersect the x-axis, what's going on? What is my y value when I'm at the x-axis? Well, my y value is 0. I'm not above or below the x-axis. So the x-intercept is when y is equal to 0. And then by that same argument, what's the y-intercept?"}, {"video_title": "Finding intercepts from an equation   Algebra I   Khan Academy.mp3", "Sentence": "Well, my y value is 0. I'm not above or below the x-axis. So the x-intercept is when y is equal to 0. And then by that same argument, what's the y-intercept? Well, if I'm somewhere along the y-axis, what's my x value? Well, I'm not to the right or the left, so my x value has to be 0. So the y-intercept occurs when x is equal to 0."}, {"video_title": "Finding intercepts from an equation   Algebra I   Khan Academy.mp3", "Sentence": "And then by that same argument, what's the y-intercept? Well, if I'm somewhere along the y-axis, what's my x value? Well, I'm not to the right or the left, so my x value has to be 0. So the y-intercept occurs when x is equal to 0. So to figure out the intercepts, let's set y equal to 0 in this equation and solve for x. And then let's set x is equal to 0 and then solve for y. So when y is equal to 0, what does this equation become?"}, {"video_title": "Finding intercepts from an equation   Algebra I   Khan Academy.mp3", "Sentence": "So the y-intercept occurs when x is equal to 0. So to figure out the intercepts, let's set y equal to 0 in this equation and solve for x. And then let's set x is equal to 0 and then solve for y. So when y is equal to 0, what does this equation become? I'll do it in orange. You get negative 5x plus 4y. Well, we're saying y is 0, so 4 times 0 is equal to 20."}, {"video_title": "Finding intercepts from an equation   Algebra I   Khan Academy.mp3", "Sentence": "So when y is equal to 0, what does this equation become? I'll do it in orange. You get negative 5x plus 4y. Well, we're saying y is 0, so 4 times 0 is equal to 20. 4 times 0 is just 0, so we can just not write that. So let me just rewrite it. So we have negative 5x is equal to 20."}, {"video_title": "Finding intercepts from an equation   Algebra I   Khan Academy.mp3", "Sentence": "Well, we're saying y is 0, so 4 times 0 is equal to 20. 4 times 0 is just 0, so we can just not write that. So let me just rewrite it. So we have negative 5x is equal to 20. We can divide both sides of this equation by negative 5. The negative 5's cancel out. That was the whole point behind dividing by negative 5."}, {"video_title": "Finding intercepts from an equation   Algebra I   Khan Academy.mp3", "Sentence": "So we have negative 5x is equal to 20. We can divide both sides of this equation by negative 5. The negative 5's cancel out. That was the whole point behind dividing by negative 5. And we get x is equal to 20 divided by negative 5 is negative 4. So when y is equal to 0, we saw that right there, x is equal to negative 4. Or if we wanted to plot that point, we always put the x-coordinate for it first."}, {"video_title": "Finding intercepts from an equation   Algebra I   Khan Academy.mp3", "Sentence": "That was the whole point behind dividing by negative 5. And we get x is equal to 20 divided by negative 5 is negative 4. So when y is equal to 0, we saw that right there, x is equal to negative 4. Or if we wanted to plot that point, we always put the x-coordinate for it first. So that would be the point negative 4 comma 0. So let me graph that. So if we go 1, 2, 3, 4, that's negative 4."}, {"video_title": "Finding intercepts from an equation   Algebra I   Khan Academy.mp3", "Sentence": "Or if we wanted to plot that point, we always put the x-coordinate for it first. So that would be the point negative 4 comma 0. So let me graph that. So if we go 1, 2, 3, 4, that's negative 4. And then the y value is just 0, so that point is right over there. That is the x-intercept. y is 0, x is negative 4."}, {"video_title": "Finding intercepts from an equation   Algebra I   Khan Academy.mp3", "Sentence": "So if we go 1, 2, 3, 4, that's negative 4. And then the y value is just 0, so that point is right over there. That is the x-intercept. y is 0, x is negative 4. Notice we're intersecting the x-axis. Now let's do the exact same thing for the y-intercept. Let's set x equal to 0."}, {"video_title": "Finding intercepts from an equation   Algebra I   Khan Academy.mp3", "Sentence": "y is 0, x is negative 4. Notice we're intersecting the x-axis. Now let's do the exact same thing for the y-intercept. Let's set x equal to 0. So if we set x is equal to 0, we have negative 5 times 0 plus 4y is equal to 20. Well, anything times 0 is 0, so we can just put that out of the way. And remember, this was setting x is equal to 0."}, {"video_title": "Finding intercepts from an equation   Algebra I   Khan Academy.mp3", "Sentence": "Let's set x equal to 0. So if we set x is equal to 0, we have negative 5 times 0 plus 4y is equal to 20. Well, anything times 0 is 0, so we can just put that out of the way. And remember, this was setting x is equal to 0. We're doing the y-intercept now. So this just simplifies to 4y is equal to 20. We can divide both sides of this equation by 4 to get rid of this 4 right there."}, {"video_title": "Finding intercepts from an equation   Algebra I   Khan Academy.mp3", "Sentence": "And remember, this was setting x is equal to 0. We're doing the y-intercept now. So this just simplifies to 4y is equal to 20. We can divide both sides of this equation by 4 to get rid of this 4 right there. And you get y is equal to 20 over 4, which is 5. So when x is equal to 0, y is equal to 5. So the point 0, 5 is on the graph for this line."}, {"video_title": "Finding intercepts from an equation   Algebra I   Khan Academy.mp3", "Sentence": "We can divide both sides of this equation by 4 to get rid of this 4 right there. And you get y is equal to 20 over 4, which is 5. So when x is equal to 0, y is equal to 5. So the point 0, 5 is on the graph for this line. So 0, 5, 0, x is 0, and y is 1, 2, 3, 4, 5 right over there. And notice, when x is 0, we're right on the y-axis. This is our y-intercept right over there."}, {"video_title": "Finding intercepts from an equation   Algebra I   Khan Academy.mp3", "Sentence": "So the point 0, 5 is on the graph for this line. So 0, 5, 0, x is 0, and y is 1, 2, 3, 4, 5 right over there. And notice, when x is 0, we're right on the y-axis. This is our y-intercept right over there. And if we graph the line, all you need is two points to graph any line. So we just have to connect the dots. And that is our line."}, {"video_title": "Finding intercepts from an equation   Algebra I   Khan Academy.mp3", "Sentence": "This is our y-intercept right over there. And if we graph the line, all you need is two points to graph any line. So we just have to connect the dots. And that is our line. So let me connect the dots, try my best to draw as straight of a line as I can. Well, I could do a better job than that. To draw as straight of a line as I can."}, {"video_title": "Finding the x-intercept of a line   Algebra I   Khan Academy.mp3", "Sentence": "So the x-intercept is the x value when y is equal to zero. Or it's the x value where our graph actually intersects the x-axis. Notice right over here our y value is exactly zero. We're sitting on the x-axis. So let's think about what this x value must be. Well just looking at it from, just trying to eyeball a little bit, it looks like it's a little over two. It's between two and three."}, {"video_title": "Finding the x-intercept of a line   Algebra I   Khan Academy.mp3", "Sentence": "We're sitting on the x-axis. So let's think about what this x value must be. Well just looking at it from, just trying to eyeball a little bit, it looks like it's a little over two. It's between two and three. It looks like it's less than two and a half. But we don't know the exact value. So let's go turn to the equation to figure out the exact value."}, {"video_title": "Finding the x-intercept of a line   Algebra I   Khan Academy.mp3", "Sentence": "It's between two and three. It looks like it's less than two and a half. But we don't know the exact value. So let's go turn to the equation to figure out the exact value. So we essentially have to figure out what x value when y is equal to zero will have this equation be true. So we could just say two times zero plus 3x is equal to seven. Well two times zero is just gonna be zero."}, {"video_title": "Finding the x-intercept of a line   Algebra I   Khan Academy.mp3", "Sentence": "So let's go turn to the equation to figure out the exact value. So we essentially have to figure out what x value when y is equal to zero will have this equation be true. So we could just say two times zero plus 3x is equal to seven. Well two times zero is just gonna be zero. So we have 3x is equal to seven. And then we can divide both sides by three to solve for x. And we get x is equal to seven over three."}, {"video_title": "Finding the x-intercept of a line   Algebra I   Khan Academy.mp3", "Sentence": "Well two times zero is just gonna be zero. So we have 3x is equal to seven. And then we can divide both sides by three to solve for x. And we get x is equal to seven over three. Now does that look like 7 3rds? Well we just have to remind ourselves that seven over three is the same thing as six over three plus one over three. And six over three is two."}, {"video_title": "Finding the x-intercept of a line   Algebra I   Khan Academy.mp3", "Sentence": "And we get x is equal to seven over three. Now does that look like 7 3rds? Well we just have to remind ourselves that seven over three is the same thing as six over three plus one over three. And six over three is two. So this is the same thing as two and 1 3rd. Another way you could think about it is three goes into seven two times. And then you have a remainder of one."}, {"video_title": "Finding the x-intercept of a line   Algebra I   Khan Academy.mp3", "Sentence": "And six over three is two. So this is the same thing as two and 1 3rd. Another way you could think about it is three goes into seven two times. And then you have a remainder of one. So you still gotta divide that one by three. So it's two full times and then a 1 3rd. So this looks like two and 1 3rd."}, {"video_title": "Finding the x-intercept of a line   Algebra I   Khan Academy.mp3", "Sentence": "And then you have a remainder of one. So you still gotta divide that one by three. So it's two full times and then a 1 3rd. So this looks like two and 1 3rd. And so that's its x intercept. Seven 7 3rds. If I was doing this on the exercise on Khan Academy, it's always a little easier to type in the improper fraction."}, {"video_title": "Introduction to the coordinate plane   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And he was somewhat of a contemporary of Galileo, he was 32 years younger, although he died shortly after Galileo died, this guy died at a much younger age. Galileo was well into his 70s, Descartes died at, what, this is only 54 years old. And he's probably most known in popular culture for this quote right over here, a very philosophical quote, I think, therefore I am. But I also wanted to throw in, and this isn't that related to algebra, but I just thought it was a really neat quote, probably his least famous quote, this one right over here. And I like it just because it's very practical, it makes you realize that these great minds, these pillars of philosophy and mathematics, at the end of the day, they really were just human beings. And he said, you just keep pushing, you just keep pushing. I made every mistake that could be made, but I just kept pushing, which I think is very, very good life advice."}, {"video_title": "Introduction to the coordinate plane   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "But I also wanted to throw in, and this isn't that related to algebra, but I just thought it was a really neat quote, probably his least famous quote, this one right over here. And I like it just because it's very practical, it makes you realize that these great minds, these pillars of philosophy and mathematics, at the end of the day, they really were just human beings. And he said, you just keep pushing, you just keep pushing. I made every mistake that could be made, but I just kept pushing, which I think is very, very good life advice. Now, he did many things in philosophy and mathematics, but the reason why I'm including him here as we build our foundations of algebra is that he is the individual most responsible for a very strong connection between algebra and geometry. So on the left over here, you have the world of algebra, and we've discussed it a little bit. You have equations that deal with symbols, and these symbols are essentially, they can take on values."}, {"video_title": "Introduction to the coordinate plane   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "I made every mistake that could be made, but I just kept pushing, which I think is very, very good life advice. Now, he did many things in philosophy and mathematics, but the reason why I'm including him here as we build our foundations of algebra is that he is the individual most responsible for a very strong connection between algebra and geometry. So on the left over here, you have the world of algebra, and we've discussed it a little bit. You have equations that deal with symbols, and these symbols are essentially, they can take on values. So you could have something like y is equal to 2x, y is equal to 2x minus 1. This gives us a relationship between whatever x is and whatever y is. And we can even set up a table here and pick values for x and see what the values of y would be."}, {"video_title": "Introduction to the coordinate plane   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "You have equations that deal with symbols, and these symbols are essentially, they can take on values. So you could have something like y is equal to 2x, y is equal to 2x minus 1. This gives us a relationship between whatever x is and whatever y is. And we can even set up a table here and pick values for x and see what the values of y would be. And I could just pick random values for x and then figure out what y is, but I'll pick relatively straightforward values just so that the math doesn't get too complicated. So for example, if x is negative 2, then y is going to be 2 times negative 2 minus 1, which is negative 4 minus 1, which is negative 5. If x is negative 1, then y is going to be 2 times negative 1 minus 1, which is equal to, this is negative 2 minus 1, which is negative 3."}, {"video_title": "Introduction to the coordinate plane   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And we can even set up a table here and pick values for x and see what the values of y would be. And I could just pick random values for x and then figure out what y is, but I'll pick relatively straightforward values just so that the math doesn't get too complicated. So for example, if x is negative 2, then y is going to be 2 times negative 2 minus 1, which is negative 4 minus 1, which is negative 5. If x is negative 1, then y is going to be 2 times negative 1 minus 1, which is equal to, this is negative 2 minus 1, which is negative 3. If x is equal to 0, then y is going to be 2 times 0 minus 1. 2 times 0 is 0 minus 1, is just negative 1. I'll do a couple more."}, {"video_title": "Introduction to the coordinate plane   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "If x is negative 1, then y is going to be 2 times negative 1 minus 1, which is equal to, this is negative 2 minus 1, which is negative 3. If x is equal to 0, then y is going to be 2 times 0 minus 1. 2 times 0 is 0 minus 1, is just negative 1. I'll do a couple more. If x is 1, and I could have picked any values here. I could have said, well, what happens if x is the negative square root of 2? Or what happens if x is negative 5 halves or positive 6 sevenths?"}, {"video_title": "Introduction to the coordinate plane   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "I'll do a couple more. If x is 1, and I could have picked any values here. I could have said, well, what happens if x is the negative square root of 2? Or what happens if x is negative 5 halves or positive 6 sevenths? But I'm just picking these numbers because it makes the math a lot easier when I try to figure out what y is going to be. But when x is 1, y is going to be 2 times 1 minus 1. 2 times 1 is 2 minus 1 is 1."}, {"video_title": "Introduction to the coordinate plane   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Or what happens if x is negative 5 halves or positive 6 sevenths? But I'm just picking these numbers because it makes the math a lot easier when I try to figure out what y is going to be. But when x is 1, y is going to be 2 times 1 minus 1. 2 times 1 is 2 minus 1 is 1. And I'll do one more. I'll do one more in a color that I have not used yet. Let's see, this purple."}, {"video_title": "Introduction to the coordinate plane   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "2 times 1 is 2 minus 1 is 1. And I'll do one more. I'll do one more in a color that I have not used yet. Let's see, this purple. If x is 2, then y is going to be 2 times 2, now our x is 2, minus 1. So that is 4 minus 1 is equal to 3. So fair enough."}, {"video_title": "Introduction to the coordinate plane   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Let's see, this purple. If x is 2, then y is going to be 2 times 2, now our x is 2, minus 1. So that is 4 minus 1 is equal to 3. So fair enough. I just kind of sampled this relationship. I said, OK, this describes a general relationship between a variable y and a variable x. And then I just made it a little bit more concrete."}, {"video_title": "Introduction to the coordinate plane   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "So fair enough. I just kind of sampled this relationship. I said, OK, this describes a general relationship between a variable y and a variable x. And then I just made it a little bit more concrete. I said, OK, well, then if x is one of these variables, for each of these values of x, what would be the corresponding value of y? And what Descartes realized is that you could visualize this. One, you could visualize these individual points, but that could also help you in general to visualize this relationship."}, {"video_title": "Introduction to the coordinate plane   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And then I just made it a little bit more concrete. I said, OK, well, then if x is one of these variables, for each of these values of x, what would be the corresponding value of y? And what Descartes realized is that you could visualize this. One, you could visualize these individual points, but that could also help you in general to visualize this relationship. And so what he essentially did is he bridged the worlds of this kind of very abstract symbolic algebra and that and geometry, which was concerned with shapes and sizes and angles. So over here, you have the world of geometry. And obviously, there are people in history, maybe many people who history may have forgotten who might have dabbled in this."}, {"video_title": "Introduction to the coordinate plane   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "One, you could visualize these individual points, but that could also help you in general to visualize this relationship. And so what he essentially did is he bridged the worlds of this kind of very abstract symbolic algebra and that and geometry, which was concerned with shapes and sizes and angles. So over here, you have the world of geometry. And obviously, there are people in history, maybe many people who history may have forgotten who might have dabbled in this. But before Descartes, it's generally viewed that geometry was Euclidean geometry. And that's essentially the geometry that you study in a geometry class in eighth or ninth grade or 10th grade in a traditional high school curriculum. And that's the geometry of studying the relationships between triangles and their angles and the relationships between circles."}, {"video_title": "Introduction to the coordinate plane   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And obviously, there are people in history, maybe many people who history may have forgotten who might have dabbled in this. But before Descartes, it's generally viewed that geometry was Euclidean geometry. And that's essentially the geometry that you study in a geometry class in eighth or ninth grade or 10th grade in a traditional high school curriculum. And that's the geometry of studying the relationships between triangles and their angles and the relationships between circles. And you have radii, and then you have triangles inscribed in circles and all the rest. And we go into some depth in that in the geometry playlist. But Descartes says, well, I think I can represent this visually the same way that Euclid was studying these triangles and these circles."}, {"video_title": "Introduction to the coordinate plane   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And that's the geometry of studying the relationships between triangles and their angles and the relationships between circles. And you have radii, and then you have triangles inscribed in circles and all the rest. And we go into some depth in that in the geometry playlist. But Descartes says, well, I think I can represent this visually the same way that Euclid was studying these triangles and these circles. He said, well, why don't I, if we view a piece of paper, if we think about a two-dimensional plane, you could view a piece of paper as kind of a section of a two-dimensional plane. We call it two dimensions because there's two directions that you could go in. There's the up-down direction."}, {"video_title": "Introduction to the coordinate plane   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "But Descartes says, well, I think I can represent this visually the same way that Euclid was studying these triangles and these circles. He said, well, why don't I, if we view a piece of paper, if we think about a two-dimensional plane, you could view a piece of paper as kind of a section of a two-dimensional plane. We call it two dimensions because there's two directions that you could go in. There's the up-down direction. That's one direction. So let me draw that. I'll do it in blue because we're trying to visualize things."}, {"video_title": "Introduction to the coordinate plane   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "There's the up-down direction. That's one direction. So let me draw that. I'll do it in blue because we're trying to visualize things. So I'll do it in the geometry color. So you have the up-down direction. And you have the left-right direction."}, {"video_title": "Introduction to the coordinate plane   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "I'll do it in blue because we're trying to visualize things. So I'll do it in the geometry color. So you have the up-down direction. And you have the left-right direction. That's why it's called a two-dimensional plane. If we were dealing in three dimensions, you would have an in-out dimension. And it's very easy to do two dimensions on the screen because the screen is two-dimensional."}, {"video_title": "Introduction to the coordinate plane   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And you have the left-right direction. That's why it's called a two-dimensional plane. If we were dealing in three dimensions, you would have an in-out dimension. And it's very easy to do two dimensions on the screen because the screen is two-dimensional. And he says, well, there are two variables here, and they have this relationship. So why don't I associate each of these variables with one of these dimensions over here? And by convention, let's make the y variable, which is really the dependent of variable."}, {"video_title": "Introduction to the coordinate plane   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And it's very easy to do two dimensions on the screen because the screen is two-dimensional. And he says, well, there are two variables here, and they have this relationship. So why don't I associate each of these variables with one of these dimensions over here? And by convention, let's make the y variable, which is really the dependent of variable. The way we did it, it depends on what x is. Let's put that on the vertical axis. And let's put our independent variable, the one where I just randomly picked values for it to see what y would become."}, {"video_title": "Introduction to the coordinate plane   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And by convention, let's make the y variable, which is really the dependent of variable. The way we did it, it depends on what x is. Let's put that on the vertical axis. And let's put our independent variable, the one where I just randomly picked values for it to see what y would become. Let's put that on the horizontal axis. And it actually was Descartes who came up with the convention of using x's and y's, and we'll see later z's in algebra so extensively as unknown variables or the variables that you're manipulating. But he says, well, if we think about it this way, if we number these dimensions, so let's say that in the x direction, let's make this right over here is negative 3."}, {"video_title": "Introduction to the coordinate plane   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And let's put our independent variable, the one where I just randomly picked values for it to see what y would become. Let's put that on the horizontal axis. And it actually was Descartes who came up with the convention of using x's and y's, and we'll see later z's in algebra so extensively as unknown variables or the variables that you're manipulating. But he says, well, if we think about it this way, if we number these dimensions, so let's say that in the x direction, let's make this right over here is negative 3. Let's make this negative 2. This is negative 1. This is 0."}, {"video_title": "Introduction to the coordinate plane   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "But he says, well, if we think about it this way, if we number these dimensions, so let's say that in the x direction, let's make this right over here is negative 3. Let's make this negative 2. This is negative 1. This is 0. Now I'm just numbering the x direction, the left-right direction. Now this is positive 1. This is positive 2."}, {"video_title": "Introduction to the coordinate plane   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "This is 0. Now I'm just numbering the x direction, the left-right direction. Now this is positive 1. This is positive 2. This is positive 3. And we can do the same in the y direction. So let's see."}, {"video_title": "Introduction to the coordinate plane   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "This is positive 2. This is positive 3. And we can do the same in the y direction. So let's see. So this could be, let's say this is negative 5, negative 4, negative 3, negative, actually, let me do it a little bit neater than that. Let me clean this up a little bit. So let me erase this and extend this down a little bit so I can go all the way down to negative 5 without making it look too messy."}, {"video_title": "Introduction to the coordinate plane   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "So let's see. So this could be, let's say this is negative 5, negative 4, negative 3, negative, actually, let me do it a little bit neater than that. Let me clean this up a little bit. So let me erase this and extend this down a little bit so I can go all the way down to negative 5 without making it look too messy. So let's go all the way down here. And so we can number it. This is 1."}, {"video_title": "Introduction to the coordinate plane   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "So let me erase this and extend this down a little bit so I can go all the way down to negative 5 without making it look too messy. So let's go all the way down here. And so we can number it. This is 1. This is 2. This is 3. And then this could be negative 1, negative 2."}, {"video_title": "Introduction to the coordinate plane   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "This is 1. This is 2. This is 3. And then this could be negative 1, negative 2. And these are all just conventions. It could have been labeled the other way. We could have decided to put the x there and the y there and make this the positive direction, make this the negative direction."}, {"video_title": "Introduction to the coordinate plane   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And then this could be negative 1, negative 2. And these are all just conventions. It could have been labeled the other way. We could have decided to put the x there and the y there and make this the positive direction, make this the negative direction. But this is just a convention that people adopted starting with Descartes. Negative 2, negative 3, negative 4, and negative 5. And he says, well, I think I can associate each of these pairs of values with a point in two dimensions."}, {"video_title": "Introduction to the coordinate plane   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "We could have decided to put the x there and the y there and make this the positive direction, make this the negative direction. But this is just a convention that people adopted starting with Descartes. Negative 2, negative 3, negative 4, and negative 5. And he says, well, I think I can associate each of these pairs of values with a point in two dimensions. I can take the x-coordinate. I can take the x value right over here. And I say, OK, that's negative 2."}, {"video_title": "Introduction to the coordinate plane   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And he says, well, I think I can associate each of these pairs of values with a point in two dimensions. I can take the x-coordinate. I can take the x value right over here. And I say, OK, that's negative 2. That would be right over there along the left-right direction. I'm going to the left because it's negative. And that's associated with negative 5 in the vertical direction."}, {"video_title": "Introduction to the coordinate plane   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And I say, OK, that's negative 2. That would be right over there along the left-right direction. I'm going to the left because it's negative. And that's associated with negative 5 in the vertical direction. So I say the y value is negative 5. And so if I go 2 to the left and 5 down, I get to this point right over there. So he says, these two values, negative 2, negative 5, I can associate it with this point in this plane right over here, in this two-dimensional plane."}, {"video_title": "Introduction to the coordinate plane   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And that's associated with negative 5 in the vertical direction. So I say the y value is negative 5. And so if I go 2 to the left and 5 down, I get to this point right over there. So he says, these two values, negative 2, negative 5, I can associate it with this point in this plane right over here, in this two-dimensional plane. So I'll say that point has the coordinates. Tells me where to find that point, negative 2, negative 5. And these coordinates, they're called Cartesian coordinates, named for Rene Descartes."}, {"video_title": "Introduction to the coordinate plane   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "So he says, these two values, negative 2, negative 5, I can associate it with this point in this plane right over here, in this two-dimensional plane. So I'll say that point has the coordinates. Tells me where to find that point, negative 2, negative 5. And these coordinates, they're called Cartesian coordinates, named for Rene Descartes. He's the guy that came up with these. He's associating all of a sudden these relationships with points on a coordinate plane. And then he says, well, OK, let's do another one."}, {"video_title": "Introduction to the coordinate plane   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And these coordinates, they're called Cartesian coordinates, named for Rene Descartes. He's the guy that came up with these. He's associating all of a sudden these relationships with points on a coordinate plane. And then he says, well, OK, let's do another one. There's this other relationship where I have when x is equal to negative 1, y is equal to negative 3. So x is negative 1, y is negative 3. That's that point right over there."}, {"video_title": "Introduction to the coordinate plane   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And then he says, well, OK, let's do another one. There's this other relationship where I have when x is equal to negative 1, y is equal to negative 3. So x is negative 1, y is negative 3. That's that point right over there. And the convention is, once again, when you list the coordinates, you list the x-coordinate, then the y-coordinate. And that's just what people decided to do. Negative 1, negative 3, that would be that point right over there."}, {"video_title": "Introduction to the coordinate plane   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "That's that point right over there. And the convention is, once again, when you list the coordinates, you list the x-coordinate, then the y-coordinate. And that's just what people decided to do. Negative 1, negative 3, that would be that point right over there. And then you have the point when x is 0, y is negative 1. When x is 0, right over here, which means I don't go to the left or the right, y is negative 1, which means I go 1 down. So that's that point right over there, 0, negative 1, right over there."}, {"video_title": "Introduction to the coordinate plane   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Negative 1, negative 3, that would be that point right over there. And then you have the point when x is 0, y is negative 1. When x is 0, right over here, which means I don't go to the left or the right, y is negative 1, which means I go 1 down. So that's that point right over there, 0, negative 1, right over there. And I could keep doing this. When x is 1, y is 1. When x is 2, y is 3."}, {"video_title": "Introduction to the coordinate plane   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "So that's that point right over there, 0, negative 1, right over there. And I could keep doing this. When x is 1, y is 1. When x is 2, y is 3. Actually, let me do that in that same purple color. When x is 2, y is 3. 2, 3."}, {"video_title": "Introduction to the coordinate plane   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "When x is 2, y is 3. Actually, let me do that in that same purple color. When x is 2, y is 3. 2, 3. And then this one right over here in orange was 1, 1. And this is neat by itself. I essentially just sampled possible x's."}, {"video_title": "Introduction to the coordinate plane   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "2, 3. And then this one right over here in orange was 1, 1. And this is neat by itself. I essentially just sampled possible x's. But what he realizes, not only do you sample these possible x's, but if you just kept sampling x's, if you tried sampling all the x's in between, you'd actually end up plotting out a line. So if you were to do every possible x, you would end up getting a line that looks something like that right over there. And if you pick any x and find any y, it really represents a point on this line."}, {"video_title": "Introduction to the coordinate plane   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "I essentially just sampled possible x's. But what he realizes, not only do you sample these possible x's, but if you just kept sampling x's, if you tried sampling all the x's in between, you'd actually end up plotting out a line. So if you were to do every possible x, you would end up getting a line that looks something like that right over there. And if you pick any x and find any y, it really represents a point on this line. Or another way to think about it, any point on this line represents a solution to this equation right over here. So if you have this point right over here, which looks like it's about x is 1 and 1 half, y is 2. So let me write that."}, {"video_title": "Introduction to the coordinate plane   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And if you pick any x and find any y, it really represents a point on this line. Or another way to think about it, any point on this line represents a solution to this equation right over here. So if you have this point right over here, which looks like it's about x is 1 and 1 half, y is 2. So let me write that. 1.5, 1.5, 2. That is a solution to this equation. When x is 1.5, 2 times 1.5 is 3, minus 1 is 2."}, {"video_title": "Introduction to the coordinate plane   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "So let me write that. 1.5, 1.5, 2. That is a solution to this equation. When x is 1.5, 2 times 1.5 is 3, minus 1 is 2. That is right over there. So all of a sudden, he was able to bridge this gap or this relationship between algebra and geometry. We can now visualize all of the x and y pairs that satisfy this equation right over here."}, {"video_title": "Introduction to the coordinate plane   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "When x is 1.5, 2 times 1.5 is 3, minus 1 is 2. That is right over there. So all of a sudden, he was able to bridge this gap or this relationship between algebra and geometry. We can now visualize all of the x and y pairs that satisfy this equation right over here. And so he is responsible for making this bridge. And that's why the coordinates that we use to specify these points are called Cartesian coordinates. And as we'll see, the first type of equations we will study are equations of this form over here."}, {"video_title": "Introduction to the coordinate plane   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "We can now visualize all of the x and y pairs that satisfy this equation right over here. And so he is responsible for making this bridge. And that's why the coordinates that we use to specify these points are called Cartesian coordinates. And as we'll see, the first type of equations we will study are equations of this form over here. And in a traditional algebra curriculum, they're called linear equations. Linear equations. And you might be saying, well, OK, this is an equation."}, {"video_title": "Introduction to the coordinate plane   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And as we'll see, the first type of equations we will study are equations of this form over here. And in a traditional algebra curriculum, they're called linear equations. Linear equations. And you might be saying, well, OK, this is an equation. I see that this is equal to that. But what's so linear about them? What makes them look like a line?"}, {"video_title": "Introduction to the coordinate plane   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And you might be saying, well, OK, this is an equation. I see that this is equal to that. But what's so linear about them? What makes them look like a line? And to realize why they are linear, you have to make this jump that Rene Descartes made. Because if you were to plot this using Cartesian coordinates on a Euclidean plane, you will get a line. And in the future, we'll see that there's other types of equations where you won't get a line."}, {"video_title": "Analyzing model in vertex form.mp3", "Sentence": "Its height in meters, x seconds after the launch, is modeled by h of x is equal to negative five times x minus four squared plus 180. So normally when they talk about seconds or time, they usually would use the variable t, but we can roll with x being that. So let's think about what's going to happen here. Let me just visualize it. So let me draw an h axis for our height, and let me draw an x axis. So an x axis. So at time x is equal to zero, we're on a platform, so we're already gonna have some height."}, {"video_title": "Analyzing model in vertex form.mp3", "Sentence": "Let me just visualize it. So let me draw an h axis for our height, and let me draw an x axis. So an x axis. So at time x is equal to zero, we're on a platform, so we're already gonna have some height. At time x is equal to zero, I'm used to saying time t equals zero, but at time x is equal to zero, we're already gonna have some height because we're on some platform, and then we're going to launch this projectile. And it's gonna go in the shape of a parabola, and it's gonna be a downward opening parabola, and you might say, well, how do you know it's gonna be a downward opening parabola? It's gonna look something like that."}, {"video_title": "Analyzing model in vertex form.mp3", "Sentence": "So at time x is equal to zero, we're on a platform, so we're already gonna have some height. At time x is equal to zero, I'm used to saying time t equals zero, but at time x is equal to zero, we're already gonna have some height because we're on some platform, and then we're going to launch this projectile. And it's gonna go in the shape of a parabola, and it's gonna be a downward opening parabola, and you might say, well, how do you know it's gonna be a downward opening parabola? It's gonna look something like that. I didn't draw it exactly perfectly, but you get hopefully the point. And the reason why I knew it was a parabola, in particular a downward opening parabola, is when you look at what's going on here. This is written in vertex form, but it's a quadratic."}, {"video_title": "Analyzing model in vertex form.mp3", "Sentence": "It's gonna look something like that. I didn't draw it exactly perfectly, but you get hopefully the point. And the reason why I knew it was a parabola, in particular a downward opening parabola, is when you look at what's going on here. This is written in vertex form, but it's a quadratic. And in vertex form, you have an expression with x squared, and then you're multiplying by negative five right over here. This tells us that it's going to be downward opening. If you were to multiply this out, if x minus four squared is gonna be x squared plus something else plus something else, and then you're gonna have to multiply all those terms by negative five, your leading term is gonna be negative five x squared."}, {"video_title": "Analyzing model in vertex form.mp3", "Sentence": "This is written in vertex form, but it's a quadratic. And in vertex form, you have an expression with x squared, and then you're multiplying by negative five right over here. This tells us that it's going to be downward opening. If you were to multiply this out, if x minus four squared is gonna be x squared plus something else plus something else, and then you're gonna have to multiply all those terms by negative five, your leading term is gonna be negative five x squared. So once again, it's gonna be a downward opening parabola that looks something like that. So given this visual intuition that we have, let's see if we can answer some questions about it. The first one I'd like to answer is how high is the platform?"}, {"video_title": "Analyzing model in vertex form.mp3", "Sentence": "If you were to multiply this out, if x minus four squared is gonna be x squared plus something else plus something else, and then you're gonna have to multiply all those terms by negative five, your leading term is gonna be negative five x squared. So once again, it's gonna be a downward opening parabola that looks something like that. So given this visual intuition that we have, let's see if we can answer some questions about it. The first one I'd like to answer is how high is the platform? How high is the platform? How high is the platform? And I encourage you to pause the video and try to figure that out."}, {"video_title": "Analyzing model in vertex form.mp3", "Sentence": "The first one I'd like to answer is how high is the platform? How high is the platform? How high is the platform? And I encourage you to pause the video and try to figure that out. What is that value right over there? Well, as you can see, we are at that value at time x equals zero. So to figure out how high is the platform, we essentially just have to evaluate h of zero."}, {"video_title": "Analyzing model in vertex form.mp3", "Sentence": "And I encourage you to pause the video and try to figure that out. What is that value right over there? Well, as you can see, we are at that value at time x equals zero. So to figure out how high is the platform, we essentially just have to evaluate h of zero. So that's going to be negative five times negative four squared plus 180. I just substituted x with zero. See, negative four squared is 16."}, {"video_title": "Analyzing model in vertex form.mp3", "Sentence": "So to figure out how high is the platform, we essentially just have to evaluate h of zero. So that's going to be negative five times negative four squared plus 180. I just substituted x with zero. See, negative four squared is 16. Negative five times 16 is negative 80 plus 180, so this is going to be equal to 100. So the platform is 100 meters tall. Remember, everything is given in, or the height is given in meters."}, {"video_title": "Analyzing model in vertex form.mp3", "Sentence": "See, negative four squared is 16. Negative five times 16 is negative 80 plus 180, so this is going to be equal to 100. So the platform is 100 meters tall. Remember, everything is given in, or the height is given in meters. Now the next question I have is how many seconds after launch do we hit our maximum height? So our maximum height, if we're talking about a downward opening parabola, it's going to be our vertex. It's going to be our maximum height."}, {"video_title": "Analyzing model in vertex form.mp3", "Sentence": "Remember, everything is given in, or the height is given in meters. Now the next question I have is how many seconds after launch do we hit our maximum height? So our maximum height, if we're talking about a downward opening parabola, it's going to be our vertex. It's going to be our maximum height. And so the x value of that would tell us how long after takeoff, how long after launch, do we hit the maximum height? I'm trying to do this in a color you can see. What is this x value right over here?"}, {"video_title": "Analyzing model in vertex form.mp3", "Sentence": "It's going to be our maximum height. And so the x value of that would tell us how long after takeoff, how long after launch, do we hit the maximum height? I'm trying to do this in a color you can see. What is this x value right over here? And once again, pause the video and see if you can figure it out. All right, so we're trying to answer how long after launch is the maximum height? Well, it's going to be the x coordinate of our vertex."}, {"video_title": "Analyzing model in vertex form.mp3", "Sentence": "What is this x value right over here? And once again, pause the video and see if you can figure it out. All right, so we're trying to answer how long after launch is the maximum height? Well, it's going to be the x coordinate of our vertex. Well, how do we figure that out? Well, this quadratic has actually been written already in vertex form, which makes it sound like it should be relatively easy to figure out the vertex over here. And to appreciate that, we just have to appreciate, we have to see the structure in the expression is one way to think about it."}, {"video_title": "Analyzing model in vertex form.mp3", "Sentence": "Well, it's going to be the x coordinate of our vertex. Well, how do we figure that out? Well, this quadratic has actually been written already in vertex form, which makes it sound like it should be relatively easy to figure out the vertex over here. And to appreciate that, we just have to appreciate, we have to see the structure in the expression is one way to think about it. Let's think about what's going on. You have this 180, and then you have this other term right over here. Anything squared is going to be non-negative."}, {"video_title": "Analyzing model in vertex form.mp3", "Sentence": "And to appreciate that, we just have to appreciate, we have to see the structure in the expression is one way to think about it. Let's think about what's going on. You have this 180, and then you have this other term right over here. Anything squared is going to be non-negative. So x minus four squared is always going to be non-negative. But then you always multiply that times a negative five, so it's going to be non, this whole thing is going to be non-positive. So it will never add to the 180."}, {"video_title": "Analyzing model in vertex form.mp3", "Sentence": "Anything squared is going to be non-negative. So x minus four squared is always going to be non-negative. But then you always multiply that times a negative five, so it's going to be non, this whole thing is going to be non-positive. So it will never add to the 180. So your maximum value is when this term right over here is going to be equal to zero. And when is this term going to be equal to zero? Well, in order to make this term equal to zero, then x minus four needs to be equal to zero."}, {"video_title": "Analyzing model in vertex form.mp3", "Sentence": "So it will never add to the 180. So your maximum value is when this term right over here is going to be equal to zero. And when is this term going to be equal to zero? Well, in order to make this term equal to zero, then x minus four needs to be equal to zero. And the only way to get x minus four to be equal to zero is if x is equal to four. So just by looking at this, you say, hey, what makes this zero? Well, four, x equals four will make this zero."}, {"video_title": "Analyzing model in vertex form.mp3", "Sentence": "Well, in order to make this term equal to zero, then x minus four needs to be equal to zero. And the only way to get x minus four to be equal to zero is if x is equal to four. So just by looking at this, you say, hey, what makes this zero? Well, four, x equals four will make this zero. So this is right over there. If I were to write h of four, this is going to be, this term is going to go to zero, and you're going to be left with the 180. So there you go, this right over here."}, {"video_title": "Analyzing model in vertex form.mp3", "Sentence": "Well, four, x equals four will make this zero. So this is right over there. If I were to write h of four, this is going to be, this term is going to go to zero, and you're going to be left with the 180. So there you go, this right over here. The maximum height is 180. It happens four seconds after launch. Now, the last question I'll ask you is how long after launch do we get to a height of zero?"}, {"video_title": "Analyzing model in vertex form.mp3", "Sentence": "So there you go, this right over here. The maximum height is 180. It happens four seconds after launch. Now, the last question I'll ask you is how long after launch do we get to a height of zero? So for what x makes our height zero? Well, to do that, we have to solve h of x is equal to zero. Or we can write h of x as negative five times x minus four squared plus 180 is equal to zero."}, {"video_title": "Analyzing model in vertex form.mp3", "Sentence": "Now, the last question I'll ask you is how long after launch do we get to a height of zero? So for what x makes our height zero? Well, to do that, we have to solve h of x is equal to zero. Or we can write h of x as negative five times x minus four squared plus 180 is equal to zero. And once again, pause the video and see if you can solve this. All right, we could subtract 180 from both sides. You get negative five times x minus four squared is equal to negative 180."}, {"video_title": "Analyzing model in vertex form.mp3", "Sentence": "Or we can write h of x as negative five times x minus four squared plus 180 is equal to zero. And once again, pause the video and see if you can solve this. All right, we could subtract 180 from both sides. You get negative five times x minus four squared is equal to negative 180. We can divide both sides by negative five. We get x minus four squared is equal to 36. Let me scroll down a little bit."}, {"video_title": "Analyzing model in vertex form.mp3", "Sentence": "You get negative five times x minus four squared is equal to negative 180. We can divide both sides by negative five. We get x minus four squared is equal to 36. Let me scroll down a little bit. And then we can take the, well, we could take the plus and minus square root, I guess you could say. And so that will give us x minus four could be equal to six, or x minus four is equal to negative six. So in this first situation, add four to both sides, you get x is equal to 10."}, {"video_title": "Analyzing model in vertex form.mp3", "Sentence": "Let me scroll down a little bit. And then we can take the, well, we could take the plus and minus square root, I guess you could say. And so that will give us x minus four could be equal to six, or x minus four is equal to negative six. So in this first situation, add four to both sides, you get x is equal to 10. Or you add four to both sides here, you get x is equal to negative two. Now, we're dealing with time here. So negative two would have been in the past if it wasn't sitting on the platform and if it was just continuing its trajectory, I guess you could say backwards in time."}, {"video_title": "Analyzing model in vertex form.mp3", "Sentence": "So in this first situation, add four to both sides, you get x is equal to 10. Or you add four to both sides here, you get x is equal to negative two. Now, we're dealing with time here. So negative two would have been in the past if it wasn't sitting on the platform and if it was just continuing its trajectory, I guess you could say backwards in time. But that's not the x that we want to consider, take into consideration. We want the positive time value, and that's right over here. That is when x is equal to 10."}, {"video_title": "Factoring difference of squares analyzing factorization   High School Math   Khan Academy.mp3", "Sentence": "Musa and Fatu were each asked to factor the quadratic expression 16x squared minus 64. Their responses are shown below. So Musa factored it this way, Fatu factored it this way. Which student wrote an expression that is equivalent to 16x squared minus 64? So I encourage you to pause the video and figure that out. Which student wrote an expression that is equivalent to our original one, 16x squared minus 64? Well let's work through it together."}, {"video_title": "Factoring difference of squares analyzing factorization   High School Math   Khan Academy.mp3", "Sentence": "Which student wrote an expression that is equivalent to 16x squared minus 64? So I encourage you to pause the video and figure that out. Which student wrote an expression that is equivalent to our original one, 16x squared minus 64? Well let's work through it together. So let's see if first we can factor this out somehow to get what Musa got. And it looks like Musa first factored out a 16, and then he was left with a difference of squares. So let's see if we can do that."}, {"video_title": "Factoring difference of squares analyzing factorization   High School Math   Khan Academy.mp3", "Sentence": "Well let's work through it together. So let's see if first we can factor this out somehow to get what Musa got. And it looks like Musa first factored out a 16, and then he was left with a difference of squares. So let's see if we can do that. So we can write our original expression, 16x squared minus 64, we can write that as 16 times x squared minus 16 times four. And when you write it like that, it's very clear that you can factor out a 16. So this is going to be equal to a 16 times what you have left over is x squared minus four."}, {"video_title": "Factoring difference of squares analyzing factorization   High School Math   Khan Academy.mp3", "Sentence": "So let's see if we can do that. So we can write our original expression, 16x squared minus 64, we can write that as 16 times x squared minus 16 times four. And when you write it like that, it's very clear that you can factor out a 16. So this is going to be equal to a 16 times what you have left over is x squared minus four. And then x squared minus four, that's the difference of squares right over there. So we can, that part we can factor as, so we have our original 16. And then this part right over here, we can write as x plus two times x minus two."}, {"video_title": "Factoring difference of squares analyzing factorization   High School Math   Khan Academy.mp3", "Sentence": "So this is going to be equal to a 16 times what you have left over is x squared minus four. And then x squared minus four, that's the difference of squares right over there. So we can, that part we can factor as, so we have our original 16. And then this part right over here, we can write as x plus two times x minus two. x plus two times x minus two. What I just did in this last step, going from x squared minus four to x plus two times x minus two doesn't make any sense. I encourage you to watch some of the introductory videos on factoring difference of squares."}, {"video_title": "Factoring difference of squares analyzing factorization   High School Math   Khan Academy.mp3", "Sentence": "And then this part right over here, we can write as x plus two times x minus two. x plus two times x minus two. What I just did in this last step, going from x squared minus four to x plus two times x minus two doesn't make any sense. I encourage you to watch some of the introductory videos on factoring difference of squares. But the basic idea, I have a form here of a squared minus b squared. So it's going to have the form of a plus b times a minus b. In this case, it's x squared minus two squared."}, {"video_title": "Factoring difference of squares analyzing factorization   High School Math   Khan Academy.mp3", "Sentence": "I encourage you to watch some of the introductory videos on factoring difference of squares. But the basic idea, I have a form here of a squared minus b squared. So it's going to have the form of a plus b times a minus b. In this case, it's x squared minus two squared. So it's going to be x plus two times x minus two. So that's exactly what Moussa got. So this one, so Moussa did get an expression that is equivalent to 16 x squared minus 64."}, {"video_title": "Factoring difference of squares analyzing factorization   High School Math   Khan Academy.mp3", "Sentence": "In this case, it's x squared minus two squared. So it's going to be x plus two times x minus two. So that's exactly what Moussa got. So this one, so Moussa did get an expression that is equivalent to 16 x squared minus 64. Now let's think about Fatou. So Fatou didn't factor out a 16 from the get-go. It looks like he just immediately recognized that what we, our original expression is itself a difference of squares, even if we don't factor out a 16."}, {"video_title": "Factoring difference of squares analyzing factorization   High School Math   Khan Academy.mp3", "Sentence": "So this one, so Moussa did get an expression that is equivalent to 16 x squared minus 64. Now let's think about Fatou. So Fatou didn't factor out a 16 from the get-go. It looks like he just immediately recognized that what we, our original expression is itself a difference of squares, even if we don't factor out a 16. And so let's rewrite it. So our original expression, we could write as, so we could write it, instead of writing, well, let me just write it like this. This is our original expression."}, {"video_title": "Factoring difference of squares analyzing factorization   High School Math   Khan Academy.mp3", "Sentence": "It looks like he just immediately recognized that what we, our original expression is itself a difference of squares, even if we don't factor out a 16. And so let's rewrite it. So our original expression, we could write as, so we could write it, instead of writing, well, let me just write it like this. This is our original expression. 16 x squared minus 64, that's the same thing as, 16 x squared is the same thing as four x, the whole thing squared, and then minus eight squared. So when you write it like this, it's clear that this is a difference of squares. So this is going to be four x plus eight times four x minus eight."}, {"video_title": "Factoring difference of squares analyzing factorization   High School Math   Khan Academy.mp3", "Sentence": "This is our original expression. 16 x squared minus 64, that's the same thing as, 16 x squared is the same thing as four x, the whole thing squared, and then minus eight squared. So when you write it like this, it's clear that this is a difference of squares. So this is going to be four x plus eight times four x minus eight. Four x plus eight times four x minus eight. Once again, if this last step that I did doesn't make a lot of sense, I encourage you to watch the video on factoring difference of squares. We'll go a lot more into the intuition of it."}, {"video_title": "How to write basic expressions with variables   6th grade   Khan Academy (2).mp3", "Sentence": "Let's do some examples of the writing expressions with variables exercise. So it says write an expression to represent 11 more than A. Well, you could just have A, but if you want 11 more than A, you would want to add 11. So you could write that as A plus 11. You could also write that as 11 plus A. Both of them would be 11 more than A. So let's check our answer here."}, {"video_title": "How to write basic expressions with variables   6th grade   Khan Academy (2).mp3", "Sentence": "So you could write that as A plus 11. You could also write that as 11 plus A. Both of them would be 11 more than A. So let's check our answer here. We got it right. Let's do a few more of these. Write an expression to represent the sum of D and nine."}, {"video_title": "How to write basic expressions with variables   6th grade   Khan Academy (2).mp3", "Sentence": "So let's check our answer here. We got it right. Let's do a few more of these. Write an expression to represent the sum of D and nine. So the sum of D and nine. That means you're gonna add D and nine. So I could write that as D plus nine, or I could write that as nine plus nine plus D. Check our answer."}, {"video_title": "How to write basic expressions with variables   6th grade   Khan Academy (2).mp3", "Sentence": "Write an expression to represent the sum of D and nine. So the sum of D and nine. That means you're gonna add D and nine. So I could write that as D plus nine, or I could write that as nine plus nine plus D. Check our answer. Got that right. Let's do a few more of these. Write an expression to represent J minus 15."}, {"video_title": "How to write basic expressions with variables   6th grade   Khan Academy (2).mp3", "Sentence": "So I could write that as D plus nine, or I could write that as nine plus nine plus D. Check our answer. Got that right. Let's do a few more of these. Write an expression to represent J minus 15. Well, I could just write it with math symbols instead of writing the word minus. Instead of writing M-I-N-U-S, I could write J minus 15. And then I check my answer."}, {"video_title": "How to write basic expressions with variables   6th grade   Khan Academy (2).mp3", "Sentence": "Write an expression to represent J minus 15. Well, I could just write it with math symbols instead of writing the word minus. Instead of writing M-I-N-U-S, I could write J minus 15. And then I check my answer. Got it right. Let's do a few more of these. This is a lot of fun."}, {"video_title": "How to write basic expressions with variables   6th grade   Khan Academy (2).mp3", "Sentence": "And then I check my answer. Got it right. Let's do a few more of these. This is a lot of fun. Write an expression to represent seven times R. Well, seven, there's a couple ways I could do it. I could use this little dot right over here. I could do seven times R like that."}, {"video_title": "How to write basic expressions with variables   6th grade   Khan Academy (2).mp3", "Sentence": "This is a lot of fun. Write an expression to represent seven times R. Well, seven, there's a couple ways I could do it. I could use this little dot right over here. I could do seven times R like that. That would be correct. I could literally just write seven R. If I just wrote seven R, that would also count. Let me check my answer."}, {"video_title": "How to write basic expressions with variables   6th grade   Khan Academy (2).mp3", "Sentence": "I could do seven times R like that. That would be correct. I could literally just write seven R. If I just wrote seven R, that would also count. Let me check my answer. That's right. Let me do a couple of other of these just so you can see that I could have just done 10. And this is not a decimal."}, {"video_title": "How to write basic expressions with variables   6th grade   Khan Academy (2).mp3", "Sentence": "Let me check my answer. That's right. Let me do a couple of other of these just so you can see that I could have just done 10. And this is not a decimal. This is, you know, it's a little bit higher than a decimal. It's multiplication. And the reason why once you start doing algebra, you use this symbol instead of that kind of cross for multiplication is that that X-looking thing gets confused with X when you're using X as a variable."}, {"video_title": "How to write basic expressions with variables   6th grade   Khan Academy (2).mp3", "Sentence": "And this is not a decimal. This is, you know, it's a little bit higher than a decimal. It's multiplication. And the reason why once you start doing algebra, you use this symbol instead of that kind of cross for multiplication is that that X-looking thing gets confused with X when you're using X as a variable. So that's why this is a lot more useful. So we want to write 10 times U. 10 times U."}, {"video_title": "How to write basic expressions with variables   6th grade   Khan Academy (2).mp3", "Sentence": "And the reason why once you start doing algebra, you use this symbol instead of that kind of cross for multiplication is that that X-looking thing gets confused with X when you're using X as a variable. So that's why this is a lot more useful. So we want to write 10 times U. 10 times U. Let's check our answer. Got it right. Let's do one more."}, {"video_title": "How to write basic expressions with variables   6th grade   Khan Academy (2).mp3", "Sentence": "10 times U. Let's check our answer. Got it right. Let's do one more. Write an expression to represent eight divided by D. So you could write it as eight, and then I could write a slash like that. Eight divided by D. And there you go. This is eight divided by D. Let me check the answer."}, {"video_title": "How to write basic expressions with variables   6th grade   Khan Academy (2).mp3", "Sentence": "Let's do one more. Write an expression to represent eight divided by D. So you could write it as eight, and then I could write a slash like that. Eight divided by D. And there you go. This is eight divided by D. Let me check the answer. I'll do one more of these. Well, six divided by B. All right, same thing."}, {"video_title": "How to write basic expressions with variables   6th grade   Khan Academy (2).mp3", "Sentence": "This is eight divided by D. Let me check the answer. I'll do one more of these. Well, six divided by B. All right, same thing. So six, I could use this tool right over here. It does the same thing as if I were to press the backslash. So six divided by B."}, {"video_title": "Interpreting absolute value   Negative numbers   6th grade   Khan Academy.mp3", "Sentence": "So we have three folks, Jenna, Sarah, and Bill, and they're keeping track of their weight. And in particular, they're keeping track of their weight over a month. And right over here we have their weight change over the month. So if you take their ending weight, and from that you subtract their starting weight, so let me just write that, that's your end weight minus your starting weight, that's going to be your change in weight over that month. And so we have the change in weights here. So Jenna's change in weight is negative three pounds, which means that she lost weight. Sarah's change in weight is positive two and a half pounds, which means she gained two and a half pounds."}, {"video_title": "Interpreting absolute value   Negative numbers   6th grade   Khan Academy.mp3", "Sentence": "So if you take their ending weight, and from that you subtract their starting weight, so let me just write that, that's your end weight minus your starting weight, that's going to be your change in weight over that month. And so we have the change in weights here. So Jenna's change in weight is negative three pounds, which means that she lost weight. Sarah's change in weight is positive two and a half pounds, which means she gained two and a half pounds. And then Bill's weight, or change in weight, I should say, is positive four pounds, which means he gained four pounds over the month. So given that, let's answer some questions. Who lost the most weight?"}, {"video_title": "Interpreting absolute value   Negative numbers   6th grade   Khan Academy.mp3", "Sentence": "Sarah's change in weight is positive two and a half pounds, which means she gained two and a half pounds. And then Bill's weight, or change in weight, I should say, is positive four pounds, which means he gained four pounds over the month. So given that, let's answer some questions. Who lost the most weight? Who had the largest change in weight? And who had the smallest change in weight? And I encourage you to pause the video and try to work through it on your own."}, {"video_title": "Interpreting absolute value   Negative numbers   6th grade   Khan Academy.mp3", "Sentence": "Who lost the most weight? Who had the largest change in weight? And who had the smallest change in weight? And I encourage you to pause the video and try to work through it on your own. Well, let's just plot all of these on a number line here to help us think through it, although some of you might already have an answer here. So Jenna's weight change is negative three pounds. So that's Jenna right over there."}, {"video_title": "Interpreting absolute value   Negative numbers   6th grade   Khan Academy.mp3", "Sentence": "And I encourage you to pause the video and try to work through it on your own. Well, let's just plot all of these on a number line here to help us think through it, although some of you might already have an answer here. So Jenna's weight change is negative three pounds. So that's Jenna right over there. I'll say J for Jenna. Sarah's weight change, and she had a weight gain, is two and a half pounds. So that's right over there, positive two and a half pounds for Sarah."}, {"video_title": "Interpreting absolute value   Negative numbers   6th grade   Khan Academy.mp3", "Sentence": "So that's Jenna right over there. I'll say J for Jenna. Sarah's weight change, and she had a weight gain, is two and a half pounds. So that's right over there, positive two and a half pounds for Sarah. And then finally Bill gained four pounds. So that is Bill right over there. Now let's answer the question, who lost the most?"}, {"video_title": "Interpreting absolute value   Negative numbers   6th grade   Khan Academy.mp3", "Sentence": "So that's right over there, positive two and a half pounds for Sarah. And then finally Bill gained four pounds. So that is Bill right over there. Now let's answer the question, who lost the most? So when we talk about lost, where we care about directions, we said, hey, who moved most to the left of zero? Well, Sarah and Bill didn't move to the left of zero at all. They gained weight."}, {"video_title": "Interpreting absolute value   Negative numbers   6th grade   Khan Academy.mp3", "Sentence": "Now let's answer the question, who lost the most? So when we talk about lost, where we care about directions, we said, hey, who moved most to the left of zero? Well, Sarah and Bill didn't move to the left of zero at all. They gained weight. That's why they're to the right of zero. Jenna's the only person who lost anything. So the person who lost the most is Jenna."}, {"video_title": "Interpreting absolute value   Negative numbers   6th grade   Khan Academy.mp3", "Sentence": "They gained weight. That's why they're to the right of zero. Jenna's the only person who lost anything. So the person who lost the most is Jenna. Now who had the largest change? And if we're thinking about the largest change, one way to think about it is, who is the furthest away from zero? So one way to think about it is, which of these is the largest absolute value?"}, {"video_title": "Interpreting absolute value   Negative numbers   6th grade   Khan Academy.mp3", "Sentence": "So the person who lost the most is Jenna. Now who had the largest change? And if we're thinking about the largest change, one way to think about it is, who is the furthest away from zero? So one way to think about it is, which of these is the largest absolute value? Well, the absolute value here, the absolute value of negative three, is, and let me be, let's see, we can just call this the absolute change, absolute change is just going to be the absolute value. Absolute value of negative three is positive three. Absolute value of 2.5 is 2.5."}, {"video_title": "Interpreting absolute value   Negative numbers   6th grade   Khan Academy.mp3", "Sentence": "So one way to think about it is, which of these is the largest absolute value? Well, the absolute value here, the absolute value of negative three, is, and let me be, let's see, we can just call this the absolute change, absolute change is just going to be the absolute value. Absolute value of negative three is positive three. Absolute value of 2.5 is 2.5. Absolute value, let me make this a little bit clearer, absolute value of four was just equal to four. So who had the largest absolute change? Well, Bill had the largest absolute change, and then second place was actually Jenna."}, {"video_title": "Interpreting absolute value   Negative numbers   6th grade   Khan Academy.mp3", "Sentence": "Absolute value of 2.5 is 2.5. Absolute value, let me make this a little bit clearer, absolute value of four was just equal to four. So who had the largest absolute change? Well, Bill had the largest absolute change, and then second place was actually Jenna. She lost weight, but she had a pretty big change. She had a three pound change, although it was a three pound loss. Bill's was a four pound gain."}, {"video_title": "Interpreting absolute value   Negative numbers   6th grade   Khan Academy.mp3", "Sentence": "Well, Bill had the largest absolute change, and then second place was actually Jenna. She lost weight, but she had a pretty big change. She had a three pound change, although it was a three pound loss. Bill's was a four pound gain. But the largest change is definitely Bill. Now who had the least change? Well, another way to think about it, this is the least absolute change."}, {"video_title": "Interpreting absolute value   Negative numbers   6th grade   Khan Academy.mp3", "Sentence": "Bill's was a four pound gain. But the largest change is definitely Bill. Now who had the least change? Well, another way to think about it, this is the least absolute change. Who changed the least from zero? Well, if you look at here, Sarah had the smallest absolute change. Jenna lost weight, so the weight change of Jenna, she has negative three versus Sarah's positive 2.5."}, {"video_title": "Interpreting absolute value   Negative numbers   6th grade   Khan Academy.mp3", "Sentence": "Well, another way to think about it, this is the least absolute change. Who changed the least from zero? Well, if you look at here, Sarah had the smallest absolute change. Jenna lost weight, so the weight change of Jenna, she has negative three versus Sarah's positive 2.5. But if we talk about who had the least change from zero, if your weight change was zero, that means you didn't gain or lose weight from the beginning of the month to the end of the month. So who had the least change? Well, that's the least absolute change."}, {"video_title": "Interpreting absolute value   Negative numbers   6th grade   Khan Academy.mp3", "Sentence": "Jenna lost weight, so the weight change of Jenna, she has negative three versus Sarah's positive 2.5. But if we talk about who had the least change from zero, if your weight change was zero, that means you didn't gain or lose weight from the beginning of the month to the end of the month. So who had the least change? Well, that's the least absolute change. Well, we see that right over here. 2.5 is less than three or four. So the least change is Sarah, and we're done."}, {"video_title": "Example 1 Factor a linear binomial by taking a common factor   Algebra I   Khan Academy.mp3", "Sentence": "And the key here is to figure out, are there any common factors to both 4x and 18? And we can factor that common factor out. We're essentially going to be reversing the distributive property. So for example, what is the largest number, or I could really say the largest expression, that is divisible into both 4x and 18? Well, 4x is divisible by 2, because we know that 4 is divisible by 2, and 18 is also divisible by 2. So we can rewrite 4x as being 2 times 2x. If you multiply that, it's obviously going to be 4x."}, {"video_title": "Example 1 Factor a linear binomial by taking a common factor   Algebra I   Khan Academy.mp3", "Sentence": "So for example, what is the largest number, or I could really say the largest expression, that is divisible into both 4x and 18? Well, 4x is divisible by 2, because we know that 4 is divisible by 2, and 18 is also divisible by 2. So we can rewrite 4x as being 2 times 2x. If you multiply that, it's obviously going to be 4x. And then we can write 18 as the same thing as 2 times 9. And now it might be clear that when you apply the distributive property, you usually end up with a step that looks something like this. Now we're just going to undistribute the 2 right over here."}, {"video_title": "Example 1 Factor a linear binomial by taking a common factor   Algebra I   Khan Academy.mp3", "Sentence": "If you multiply that, it's obviously going to be 4x. And then we can write 18 as the same thing as 2 times 9. And now it might be clear that when you apply the distributive property, you usually end up with a step that looks something like this. Now we're just going to undistribute the 2 right over here. We're going to factor the 2 out. Let me actually just draw that. So we're going to factor the 2 out."}, {"video_title": "Example 1 Factor a linear binomial by taking a common factor   Algebra I   Khan Academy.mp3", "Sentence": "Now we're just going to undistribute the 2 right over here. We're going to factor the 2 out. Let me actually just draw that. So we're going to factor the 2 out. And so this is going to be 2 times 2x plus 9. And if you were to want to multiply this out, it would be 2 times 2x plus 2 times 9. It would be exactly this, which you would simplify as this right up here."}, {"video_title": "Example 1 Factor a linear binomial by taking a common factor   Algebra I   Khan Academy.mp3", "Sentence": "So we're going to factor the 2 out. And so this is going to be 2 times 2x plus 9. And if you were to want to multiply this out, it would be 2 times 2x plus 2 times 9. It would be exactly this, which you would simplify as this right up here. So there we have it. We have written this as a product of two expressions, 2 times 2x plus 9. Let's do this again."}, {"video_title": "Example 1 Factor a linear binomial by taking a common factor   Algebra I   Khan Academy.mp3", "Sentence": "It would be exactly this, which you would simplify as this right up here. So there we have it. We have written this as a product of two expressions, 2 times 2x plus 9. Let's do this again. So let's say that I have 12 plus 32x. Actually, just to get a little bit of variety here, let's put a y here. 12 plus 32y."}, {"video_title": "Example 1 Factor a linear binomial by taking a common factor   Algebra I   Khan Academy.mp3", "Sentence": "Let's do this again. So let's say that I have 12 plus 32x. Actually, just to get a little bit of variety here, let's put a y here. 12 plus 32y. Well, what's the largest number that's divisible into both 12 and 32? 2 is clearly divisible into both, but so is 4. And let's see."}, {"video_title": "Example 1 Factor a linear binomial by taking a common factor   Algebra I   Khan Academy.mp3", "Sentence": "12 plus 32y. Well, what's the largest number that's divisible into both 12 and 32? 2 is clearly divisible into both, but so is 4. And let's see. It doesn't look like anything larger than 4 is divisible into both 12 and 32. The greatest common factor of 12 and 32 is 4. And y is only divisible into the second term, not into this first term right over here."}, {"video_title": "Example 1 Factor a linear binomial by taking a common factor   Algebra I   Khan Academy.mp3", "Sentence": "And let's see. It doesn't look like anything larger than 4 is divisible into both 12 and 32. The greatest common factor of 12 and 32 is 4. And y is only divisible into the second term, not into this first term right over here. So it looks like 4 is the greatest common factor. So we can rewrite each of these as a product of 4 and something else. So for example, 12 we can rewrite as 4 times 3."}, {"video_title": "Example 1 Factor a linear binomial by taking a common factor   Algebra I   Khan Academy.mp3", "Sentence": "And y is only divisible into the second term, not into this first term right over here. So it looks like 4 is the greatest common factor. So we can rewrite each of these as a product of 4 and something else. So for example, 12 we can rewrite as 4 times 3. And 32 we can rewrite. So this is going to be plus 4 times, well, if you divide 32y by 4, it's going to be 8y. And now, once again, we can factor out the 4."}, {"video_title": "Example 1 Factor a linear binomial by taking a common factor   Algebra I   Khan Academy.mp3", "Sentence": "So for example, 12 we can rewrite as 4 times 3. And 32 we can rewrite. So this is going to be plus 4 times, well, if you divide 32y by 4, it's going to be 8y. And now, once again, we can factor out the 4. So this is going to be 4 times 3 plus 8y. And once you do more and more examples of this, you're going to find that you can just kind of do this step all at once. You can say, hey, what's the largest number that's divisible into both of these?"}, {"video_title": "Example 1 Factor a linear binomial by taking a common factor   Algebra I   Khan Academy.mp3", "Sentence": "And now, once again, we can factor out the 4. So this is going to be 4 times 3 plus 8y. And once you do more and more examples of this, you're going to find that you can just kind of do this step all at once. You can say, hey, what's the largest number that's divisible into both of these? Well, it's 4. So let me factor a 4 out. 12 divided by 4 is 3."}, {"video_title": "Multiplying challenging decimals   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "And I encourage you to pause this video and try it on your own. So let's just think about a very similar problem, but one where essentially we don't write the decimals. Let's just think about multiplying 121 times 43, which we know how to do. So let's just think about this problem first as kind of a simplification, and then we'll think about how to get from this product to this product. So we can start with, so we're gonna say three times one is three, three times two is six, three times one is three, three times 121 is 363. And now we're gonna go to the tens place, so this is a 40 right over here. So since we're in the tens place, let's put a zero there."}, {"video_title": "Multiplying challenging decimals   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "So let's just think about this problem first as kind of a simplification, and then we'll think about how to get from this product to this product. So we can start with, so we're gonna say three times one is three, three times two is six, three times one is three, three times 121 is 363. And now we're gonna go to the tens place, so this is a 40 right over here. So since we're in the tens place, let's put a zero there. 40 times one is 40. 40 times 20 is 800. 40 times 100 is 4,000."}, {"video_title": "Multiplying challenging decimals   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "So since we're in the tens place, let's put a zero there. 40 times one is 40. 40 times 20 is 800. 40 times 100 is 4,000. And we've already known how to do this in the past. And now we can just add all of this together, and we get, and we get, let me do a new color here, three plus zero is three, six plus four is 10, one plus three plus eight is 12, one plus four is five. So 121 times 43 is 5,203."}, {"video_title": "Multiplying challenging decimals   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "40 times 100 is 4,000. And we've already known how to do this in the past. And now we can just add all of this together, and we get, and we get, let me do a new color here, three plus zero is three, six plus four is 10, one plus three plus eight is 12, one plus four is five. So 121 times 43 is 5,203. Now how is this useful for figuring out this product? Well, to go from 120, to go from 1.21 to 121, we're essentially multiplying by 100, right? We're moving the decimal two places over to the right."}, {"video_title": "Multiplying challenging decimals   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "So 121 times 43 is 5,203. Now how is this useful for figuring out this product? Well, to go from 120, to go from 1.21 to 121, we're essentially multiplying by 100, right? We're moving the decimal two places over to the right. And to go from 0.043 to 43, what are we doing? We're removing the decimal, see we're multiplying by 10, 100,000. We're multiplying by 1,000."}, {"video_title": "Multiplying challenging decimals   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "We're moving the decimal two places over to the right. And to go from 0.043 to 43, what are we doing? We're removing the decimal, see we're multiplying by 10, 100,000. We're multiplying by 1,000. So to go from this product to this product, or to this product, we essentially multiplied by 100, and we multiplied by 1,000. So then to go back to this product, to go back to this product, we have to divide. We should divide by 100, we should divide by 100, and then divide by 1,000, which is equivalent to dividing by 100,000."}, {"video_title": "Multiplying challenging decimals   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "We're multiplying by 1,000. So to go from this product to this product, or to this product, we essentially multiplied by 100, and we multiplied by 1,000. So then to go back to this product, to go back to this product, we have to divide. We should divide by 100, we should divide by 100, and then divide by 1,000, which is equivalent to dividing by 100,000. But let's do that. So let's rewrite this number here. So 5,203, actually let me write it like this, just so it's a little bit more aligned."}, {"video_title": "Multiplying challenging decimals   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "We should divide by 100, we should divide by 100, and then divide by 1,000, which is equivalent to dividing by 100,000. But let's do that. So let's rewrite this number here. So 5,203, actually let me write it like this, just so it's a little bit more aligned. 5,203. And we can imagine a decimal point right over here. If we divide by 100, see divide by 10, divide by 100, and then we wanna divide by another 1,000."}, {"video_title": "Multiplying challenging decimals   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "So 5,203, actually let me write it like this, just so it's a little bit more aligned. 5,203. And we can imagine a decimal point right over here. If we divide by 100, see divide by 10, divide by 100, and then we wanna divide by another 1,000. So divide by 10, divide by 100, divide by 1,000. So our decimal point is going to go right over there, and we're done. 1.21 times 0.043 is 0.05203."}, {"video_title": "Multiplying challenging decimals   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "If we divide by 100, see divide by 10, divide by 100, and then we wanna divide by another 1,000. So divide by 10, divide by 100, divide by 1,000. So our decimal point is going to go right over there, and we're done. 1.21 times 0.043 is 0.05203. So one way you could think about it is just multiply these two numbers as if there were no decimals there. Then you could count how many digits are to the right of the decimal, and you see that there are one, two, three, four, five digits to the right of the decimal. And so in your product, you should have one, two, three, four, five digits, five digits to the right of the decimal."}, {"video_title": "Multiplying challenging decimals   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "1.21 times 0.043 is 0.05203. So one way you could think about it is just multiply these two numbers as if there were no decimals there. Then you could count how many digits are to the right of the decimal, and you see that there are one, two, three, four, five digits to the right of the decimal. And so in your product, you should have one, two, three, four, five digits, five digits to the right of the decimal. Why is that the case? Well, when you ignored the decimals, when you just pretended that this was 121 times 43, you essentially multiply this times 100,000 by 100,000. And so to get from the product you get without the decimals to the one that you need with the decimals, you have to then divide by 100,000 again."}, {"video_title": "Multiplying challenging decimals   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "And so in your product, you should have one, two, three, four, five digits, five digits to the right of the decimal. Why is that the case? Well, when you ignored the decimals, when you just pretended that this was 121 times 43, you essentially multiply this times 100,000 by 100,000. And so to get from the product you get without the decimals to the one that you need with the decimals, you have to then divide by 100,000 again. Multiply by 100,000 is essentially equivalent to moving the decimal place five places to the right, and then dividing by 100,000 is equivalent to moving the decimal five digits to the left. So divide by 10, divide by 100, divide by 1,000, divide by 10,000, divide by 100,000. Either way, we are done."}, {"video_title": "Factoring perfect squares common factor   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "Now what I'd like you to do is pause the video and see if you can factor this polynomial completely. Now let's work through it together. So the first thing that you might notice is that all of the terms are divisible by x. So we can actually factor out an x. So let's do that. Actually, if we look at these coefficients, it looks like, let's see, it looks like they don't have any common factors other than one. So it looks like the largest monomial that we can factor out is just going to be an x."}, {"video_title": "Factoring perfect squares common factor   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "So we can actually factor out an x. So let's do that. Actually, if we look at these coefficients, it looks like, let's see, it looks like they don't have any common factors other than one. So it looks like the largest monomial that we can factor out is just going to be an x. So let's do that. Let's factor out an x. So then this is going to be x times."}, {"video_title": "Factoring perfect squares common factor   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "So it looks like the largest monomial that we can factor out is just going to be an x. So let's do that. Let's factor out an x. So then this is going to be x times. When you factor out an x from 16x to the third, you're gonna be left with 16x squared and then plus 24x and then plus 9. Now this is starting to look interesting. So let me just rewrite it."}, {"video_title": "Factoring perfect squares common factor   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "So then this is going to be x times. When you factor out an x from 16x to the third, you're gonna be left with 16x squared and then plus 24x and then plus 9. Now this is starting to look interesting. So let me just rewrite it. This is gonna be x times. This part over here looks interesting because when I see the 16x squared, this looks like a perfect square. Let me write it out."}, {"video_title": "Factoring perfect squares common factor   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "So let me just rewrite it. This is gonna be x times. This part over here looks interesting because when I see the 16x squared, this looks like a perfect square. Let me write it out. 16x squared, that's the same thing as 4x, 4x squared. And then we have a nine over there, which is clearly a perfect square. That is three squared, three squared."}, {"video_title": "Factoring perfect squares common factor   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "Let me write it out. 16x squared, that's the same thing as 4x, 4x squared. And then we have a nine over there, which is clearly a perfect square. That is three squared, three squared. And when we look at this 24x, we see that it is four times three times two. And so we can write it as, let me write it this way. So this is going to be plus two times four times three x."}, {"video_title": "Factoring perfect squares common factor   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "That is three squared, three squared. And when we look at this 24x, we see that it is four times three times two. And so we can write it as, let me write it this way. So this is going to be plus two times four times three x. So let me make it, so two times four times three times 3x. Now why did I take the trouble? Why did I take the trouble of writing everything like this?"}, {"video_title": "Factoring perfect squares common factor   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "So this is going to be plus two times four times three x. So let me make it, so two times four times three times 3x. Now why did I take the trouble? Why did I take the trouble of writing everything like this? Because we see that it fits the pattern for a perfect square. What do I mean by that? Well in previous videos, we saw that if you have something of the form ax plus b and you were to square it, you're going to get ax squared plus two abx plus b squared."}, {"video_title": "Factoring perfect squares common factor   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "Why did I take the trouble of writing everything like this? Because we see that it fits the pattern for a perfect square. What do I mean by that? Well in previous videos, we saw that if you have something of the form ax plus b and you were to square it, you're going to get ax squared plus two abx plus b squared. And we have that form right over here. This is the ax squared. Let me do the same color."}, {"video_title": "Factoring perfect squares common factor   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "Well in previous videos, we saw that if you have something of the form ax plus b and you were to square it, you're going to get ax squared plus two abx plus b squared. And we have that form right over here. This is the ax squared. Let me do the same color. The ax squared, ax squared. We have the b squared. You have the b squared."}, {"video_title": "Factoring perfect squares common factor   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "Let me do the same color. The ax squared, ax squared. We have the b squared. You have the b squared. And then you have the two abx, two abx right over there. So this section, this entire section, we can rewrite as being, we know what a and b are. A is four and b is three."}, {"video_title": "Factoring perfect squares common factor   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "You have the b squared. And then you have the two abx, two abx right over there. So this section, this entire section, we can rewrite as being, we know what a and b are. A is four and b is three. So this is going to be ax. So 4x plus b, which we know to be three, that whole thing squared. And now we can't forget this x out front."}, {"video_title": "Factoring perfect squares common factor   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "A is four and b is three. So this is going to be ax. So 4x plus b, which we know to be three, that whole thing squared. And now we can't forget this x out front. So we have that x out front and we're done. We have just factored this. We have just factored this completely."}, {"video_title": "Factoring perfect squares common factor   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "And now we can't forget this x out front. So we have that x out front and we're done. We have just factored this. We have just factored this completely. We could write it as x times 4x plus three and then write out and then say times 4x plus three or we could just write x times the quantity 4x plus three squared. And so we've just factored it completely. And the key realizations here is, well one, you know, what can I factor out from all of these terms?"}, {"video_title": "Factoring perfect squares missing values   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "The quadratic expression x squared plus five x plus c is a perfect square, and it can be factored as x plus d squared. Both c and d are positive rational numbers. So what I want to figure out in this video is what is c, given the information that we have right over here, what is c going to be equal to and what is d going to be equal to? And like always, pause the video and see if you can figure it out. Well, let's work through this together. We're saying that x squared plus five x plus c can be rewritten as x plus d squared. So let me write that down."}, {"video_title": "Factoring perfect squares missing values   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "And like always, pause the video and see if you can figure it out. Well, let's work through this together. We're saying that x squared plus five x plus c can be rewritten as x plus d squared. So let me write that down. So this part, this part, x squared plus five x plus c, we're saying that that could be written as x plus d squared. This is equal to x plus d squared. Now we can rewrite x plus d squared is going to be equal to x squared plus two x, two, let me write, two dx, two dx plus d squared."}, {"video_title": "Factoring perfect squares missing values   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "So let me write that down. So this part, this part, x squared plus five x plus c, we're saying that that could be written as x plus d squared. This is equal to x plus d squared. Now we can rewrite x plus d squared is going to be equal to x squared plus two x, two, let me write, two dx, two dx plus d squared. If this step right over here you find strange, I encourage you to watch the videos on squaring binomials or on perfect square polynomials, either one, so you can see the pattern that this is going to be x squared plus two times the product of both of these terms plus d squared. And so when you look at it like this, you can start to pattern match a little bit. You can say, all right, well, five x right over here, that is going to have to be equal to two d, and then you can also say that c is going to have to be equal to d squared."}, {"video_title": "Factoring perfect squares missing values   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "Now we can rewrite x plus d squared is going to be equal to x squared plus two x, two, let me write, two dx, two dx plus d squared. If this step right over here you find strange, I encourage you to watch the videos on squaring binomials or on perfect square polynomials, either one, so you can see the pattern that this is going to be x squared plus two times the product of both of these terms plus d squared. And so when you look at it like this, you can start to pattern match a little bit. You can say, all right, well, five x right over here, that is going to have to be equal to two d, and then you can also say that c is going to have to be equal to d squared. So once again, we could say two d is equal to five, two d is equal to five, or that d is equal to five halves. So we've figured out what d is equal to, and now we can figure out what c is, because we know that c needs to be equal to d squared, c, let me do that orange color actually. So we know that c is equal to d squared, which is the same thing as five halves squared."}, {"video_title": "Factoring perfect squares missing values   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "You can say, all right, well, five x right over here, that is going to have to be equal to two d, and then you can also say that c is going to have to be equal to d squared. So once again, we could say two d is equal to five, two d is equal to five, or that d is equal to five halves. So we've figured out what d is equal to, and now we can figure out what c is, because we know that c needs to be equal to d squared, c, let me do that orange color actually. So we know that c is equal to d squared, which is the same thing as five halves squared. We just figured out what d is equal to. It's going to be five halves squared, which is going to be 25 over four. So c is equal to 25 over four, d is equal to five halves, and we're done."}, {"video_title": "Introduction to intercepts    Algebra I   Khan Academy.mp3", "Sentence": "So let's set up a little table here, x comma y. And then we could just try out a couple of x values here and then figure out what the corresponding y values are. And I'm gonna pick x values where it's gonna be fairly easy to calculate the y values. So let's say when x is equal to zero, then you're gonna have 1 half times zero minus three, well then y is going to be negative three. When x is, let's see, let me try x is equal to two because then 1 half times two is just gonna be one. So when x is equal to two, you're gonna have 1 half times two is one minus three is negative two. When x is equal to, let's try four."}, {"video_title": "Introduction to intercepts    Algebra I   Khan Academy.mp3", "Sentence": "So let's say when x is equal to zero, then you're gonna have 1 half times zero minus three, well then y is going to be negative three. When x is, let's see, let me try x is equal to two because then 1 half times two is just gonna be one. So when x is equal to two, you're gonna have 1 half times two is one minus three is negative two. When x is equal to, let's try four. So 1 half times four is two and then minus three is negative one. And we could keep going, but actually all we do need, all we need is two points for a line. So we're ready to plot this line if we like."}, {"video_title": "Introduction to intercepts    Algebra I   Khan Academy.mp3", "Sentence": "When x is equal to, let's try four. So 1 half times four is two and then minus three is negative one. And we could keep going, but actually all we do need, all we need is two points for a line. So we're ready to plot this line if we like. The point zero comma negative three is on this line. Zero comma negative three. Actually let me do this in a slightly darker color so we can see it on this white background."}, {"video_title": "Introduction to intercepts    Algebra I   Khan Academy.mp3", "Sentence": "So we're ready to plot this line if we like. The point zero comma negative three is on this line. Zero comma negative three. Actually let me do this in a slightly darker color so we can see it on this white background. Zero comma negative three is on the line. Two comma negative two is on the line. So two comma negative two."}, {"video_title": "Introduction to intercepts    Algebra I   Khan Academy.mp3", "Sentence": "Actually let me do this in a slightly darker color so we can see it on this white background. Zero comma negative three is on the line. Two comma negative two is on the line. So two comma negative two. And then we have four comma negative one. So when x is four, y is negative one. And I could draw a line that connects all of these."}, {"video_title": "Introduction to intercepts    Algebra I   Khan Academy.mp3", "Sentence": "So two comma negative two. And then we have four comma negative one. So when x is four, y is negative one. And I could draw a line that connects all of these. So it would look something like, let's see if I can do this. It would look something like, it would look something like, like that. So this right over here, this is literally, this is the graph of y is equal to 1 half x minus three."}, {"video_title": "Introduction to intercepts    Algebra I   Khan Academy.mp3", "Sentence": "And I could draw a line that connects all of these. So it would look something like, let's see if I can do this. It would look something like, it would look something like, like that. So this right over here, this is literally, this is the graph of y is equal to 1 half x minus three. Now when we look at a graph like this, an interesting thing that we might wanna ask ourselves is where does the graph intersect our axes? So first we could say, well where does it intersect our x-axis? And when you look at this, it looks like it happens at this point right over here."}, {"video_title": "Introduction to intercepts    Algebra I   Khan Academy.mp3", "Sentence": "So this right over here, this is literally, this is the graph of y is equal to 1 half x minus three. Now when we look at a graph like this, an interesting thing that we might wanna ask ourselves is where does the graph intersect our axes? So first we could say, well where does it intersect our x-axis? And when you look at this, it looks like it happens at this point right over here. And this point where a graph intersects an axis, this is called an intercept. And this one in particular is called the x-intercept. Why is it called the x-intercept?"}, {"video_title": "Introduction to intercepts    Algebra I   Khan Academy.mp3", "Sentence": "And when you look at this, it looks like it happens at this point right over here. And this point where a graph intersects an axis, this is called an intercept. And this one in particular is called the x-intercept. Why is it called the x-intercept? Because that's where the graph is intersecting the x-axis. And the x-intercept, it looks like, this is at the point six comma zero. Now it's very interesting."}, {"video_title": "Introduction to intercepts    Algebra I   Khan Academy.mp3", "Sentence": "Why is it called the x-intercept? Because that's where the graph is intersecting the x-axis. And the x-intercept, it looks like, this is at the point six comma zero. Now it's very interesting. The x-intercept happens when y is equal to zero. Remember, you're on the x-axis when you haven't moved up or down from that axis, which means y is equal to zero. So your x-intercept happens at x equals six, y equals zero, it's this coordinate."}, {"video_title": "Introduction to intercepts    Algebra I   Khan Academy.mp3", "Sentence": "Now it's very interesting. The x-intercept happens when y is equal to zero. Remember, you're on the x-axis when you haven't moved up or down from that axis, which means y is equal to zero. So your x-intercept happens at x equals six, y equals zero, it's this coordinate. Now what about the y-intercept? Well the y-intercept is this point right over here. This is where you intersect, or I guess you could say, intercept the y-axis."}, {"video_title": "Introduction to intercepts    Algebra I   Khan Academy.mp3", "Sentence": "So your x-intercept happens at x equals six, y equals zero, it's this coordinate. Now what about the y-intercept? Well the y-intercept is this point right over here. This is where you intersect, or I guess you could say, intercept the y-axis. So this right over here, that over there is the y-intercept. And the y-intercept is at the coordinate that has a zero for the x-coordinate. X is zero here and y is negative three."}, {"video_title": "Introduction to intercepts    Algebra I   Khan Academy.mp3", "Sentence": "This is where you intersect, or I guess you could say, intercept the y-axis. So this right over here, that over there is the y-intercept. And the y-intercept is at the coordinate that has a zero for the x-coordinate. X is zero here and y is negative three. X is zero and y is negative three. This was actually one of the points, or one of the pairs that we first tried out. And you can validate that six comma zero satisfies this equation right over here."}, {"video_title": "Introduction to intercepts    Algebra I   Khan Academy.mp3", "Sentence": "X is zero here and y is negative three. X is zero and y is negative three. This was actually one of the points, or one of the pairs that we first tried out. And you can validate that six comma zero satisfies this equation right over here. If x is six, one half times six is three, minus three is indeed equal to zero. So now that we know what an x-intercept is, it's the point where a graph intersects the x-axis, or intercepts the x-axis, and the y-intercept is the point where a graph intercepts the y-axis, or intersects the y-axis, let's try to see if we can find the x and y-intercepts for a few other linear equations. So let's say that I have the linear equation."}, {"video_title": "Introduction to intercepts    Algebra I   Khan Academy.mp3", "Sentence": "And you can validate that six comma zero satisfies this equation right over here. If x is six, one half times six is three, minus three is indeed equal to zero. So now that we know what an x-intercept is, it's the point where a graph intersects the x-axis, or intercepts the x-axis, and the y-intercept is the point where a graph intercepts the y-axis, or intersects the y-axis, let's try to see if we can find the x and y-intercepts for a few other linear equations. So let's say that I have the linear equation. Let's say that I have five x plus six y is equal to 30. I encourage you to pause this video and figure out what are the x and y-intercepts for the graph that represents the solutions, all the xy pairs that satisfy this equation. Well, the easiest thing to do here, let's see what the y value is when x is equal to zero, and what the x value is when y is equal to zero."}, {"video_title": "Introduction to intercepts    Algebra I   Khan Academy.mp3", "Sentence": "So let's say that I have the linear equation. Let's say that I have five x plus six y is equal to 30. I encourage you to pause this video and figure out what are the x and y-intercepts for the graph that represents the solutions, all the xy pairs that satisfy this equation. Well, the easiest thing to do here, let's see what the y value is when x is equal to zero, and what the x value is when y is equal to zero. When x is equal to zero, this becomes six y equal 30, and so six times what is 30? Well, y would be equal to five here. So when x is zero, y is five."}, {"video_title": "Introduction to intercepts    Algebra I   Khan Academy.mp3", "Sentence": "Well, the easiest thing to do here, let's see what the y value is when x is equal to zero, and what the x value is when y is equal to zero. When x is equal to zero, this becomes six y equal 30, and so six times what is 30? Well, y would be equal to five here. So when x is zero, y is five. And what about when y is zero? Well, when y is zero, that's gonna be zero, and you have five x is equal to 30. Well, then x would be equal to six."}, {"video_title": "Introduction to intercepts    Algebra I   Khan Academy.mp3", "Sentence": "So when x is zero, y is five. And what about when y is zero? Well, when y is zero, that's gonna be zero, and you have five x is equal to 30. Well, then x would be equal to six. Then x would be equal to six. So we could plot those points, zero comma five. When x is zero, y is five."}, {"video_title": "Introduction to intercepts    Algebra I   Khan Academy.mp3", "Sentence": "Well, then x would be equal to six. Then x would be equal to six. So we could plot those points, zero comma five. When x is zero, y is five. And when x is six, y is zero. So those are both points on this graph. And then the actual graph is going to, or the actual line that represents the x and y pairs that satisfy this equation, is going to look like, it's going to look like this."}, {"video_title": "Introduction to intercepts    Algebra I   Khan Academy.mp3", "Sentence": "When x is zero, y is five. And when x is six, y is zero. So those are both points on this graph. And then the actual graph is going to, or the actual line that represents the x and y pairs that satisfy this equation, is going to look like, it's going to look like this. I'll just try, so I can make it go. It's going to look like it's gonna go through those two points. And so it's gonna, I can make it go the other way too."}, {"video_title": "Introduction to intercepts    Algebra I   Khan Academy.mp3", "Sentence": "And then the actual graph is going to, or the actual line that represents the x and y pairs that satisfy this equation, is going to look like, it's going to look like this. I'll just try, so I can make it go. It's going to look like it's gonna go through those two points. And so it's gonna, I can make it go the other way too. Let me see. It's gonna go through those two points. So it's gonna look something like that."}, {"video_title": "Introduction to intercepts    Algebra I   Khan Academy.mp3", "Sentence": "And so it's gonna, I can make it go the other way too. Let me see. It's gonna go through those two points. So it's gonna look something like that. Now what are its x and y intercepts? Well, we already kind of figured it out, but the intercepts themselves, these are the points on the graph where they intersect the axes. So this right over here, this is the y-intercept."}, {"video_title": "Introduction to intercepts    Algebra I   Khan Academy.mp3", "Sentence": "So it's gonna look something like that. Now what are its x and y intercepts? Well, we already kind of figured it out, but the intercepts themselves, these are the points on the graph where they intersect the axes. So this right over here, this is the y-intercept. That point is the y-intercept, and it happens, it's always going to happen when x is equal to zero, and when x is equal to zero, we know that y is equal to five. It's that point, the point zero comma five. And what is the y, so what is the x-intercept?"}, {"video_title": "Introduction to intercepts    Algebra I   Khan Academy.mp3", "Sentence": "So this right over here, this is the y-intercept. That point is the y-intercept, and it happens, it's always going to happen when x is equal to zero, and when x is equal to zero, we know that y is equal to five. It's that point, the point zero comma five. And what is the y, so what is the x-intercept? The x-intercept is the point, it's actually the same x-intercept for that other, for this equation right over here. It's the point six comma zero. That point right over there."}, {"video_title": "Similar triangle example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "They both share this angle right over there, so that gives us one angle. We need two to get to angle-angle, which gives us similarity. And we know that these two lines are parallel. We know if two lines are parallel and we have a transversal, that corresponding angles are going to be congruent. So that angle is going to correspond to that angle right over there. And we're done. We have one angle in triangle AEC that is congruent to another angle in BDC."}, {"video_title": "Similar triangle example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "We know if two lines are parallel and we have a transversal, that corresponding angles are going to be congruent. So that angle is going to correspond to that angle right over there. And we're done. We have one angle in triangle AEC that is congruent to another angle in BDC. And then we have this angle that's obviously congruent to itself that's in both triangles. So both triangles have a pair of corresponding angles that are congruent, so they must be similar. So we can write triangle ACE is going to be similar to triangle, and we want to get the letters in the right order."}, {"video_title": "Similar triangle example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "We have one angle in triangle AEC that is congruent to another angle in BDC. And then we have this angle that's obviously congruent to itself that's in both triangles. So both triangles have a pair of corresponding angles that are congruent, so they must be similar. So we can write triangle ACE is going to be similar to triangle, and we want to get the letters in the right order. So where the blue angle is here, the blue angle there is vertex B. Then we go to the white angle, C. And then we go to the unlabeled angle right over there, BCD. So we did that first one."}, {"video_title": "Similar triangle example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "So we can write triangle ACE is going to be similar to triangle, and we want to get the letters in the right order. So where the blue angle is here, the blue angle there is vertex B. Then we go to the white angle, C. And then we go to the unlabeled angle right over there, BCD. So we did that first one. Now let's do this one right over here. This is kind of similar, but at least it looks just superficially looking at it, that YZ is definitely not parallel to ST. So we won't be able to do this corresponding angle argument, especially because they didn't even label it as parallel."}, {"video_title": "Similar triangle example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "So we did that first one. Now let's do this one right over here. This is kind of similar, but at least it looks just superficially looking at it, that YZ is definitely not parallel to ST. So we won't be able to do this corresponding angle argument, especially because they didn't even label it as parallel. And so you don't want to look at things just by the way they look. You definitely want to say, what am I given and what am I not given? If these weren't labeled parallel, we wouldn't be able to make the statement, even if they looked parallel."}, {"video_title": "Similar triangle example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "So we won't be able to do this corresponding angle argument, especially because they didn't even label it as parallel. And so you don't want to look at things just by the way they look. You definitely want to say, what am I given and what am I not given? If these weren't labeled parallel, we wouldn't be able to make the statement, even if they looked parallel. One thing we do have is that we have this angle right here that's common to the inner triangle and to the outer triangle. And they've given us a bunch of sides. So maybe we can use SAS for similarity, meaning if we can show the ratio of the sides on either side of this angle, if they have the same ratio from the smaller triangle to the larger triangle, then we can show similarity."}, {"video_title": "Similar triangle example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "If these weren't labeled parallel, we wouldn't be able to make the statement, even if they looked parallel. One thing we do have is that we have this angle right here that's common to the inner triangle and to the outer triangle. And they've given us a bunch of sides. So maybe we can use SAS for similarity, meaning if we can show the ratio of the sides on either side of this angle, if they have the same ratio from the smaller triangle to the larger triangle, then we can show similarity. So let's go, and we have to go on either side of this angle right over here. Let's look at the shorter side on either side of this angle. So the shorter side is 2."}, {"video_title": "Similar triangle example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "So maybe we can use SAS for similarity, meaning if we can show the ratio of the sides on either side of this angle, if they have the same ratio from the smaller triangle to the larger triangle, then we can show similarity. So let's go, and we have to go on either side of this angle right over here. Let's look at the shorter side on either side of this angle. So the shorter side is 2. And let's look at the shorter side on either side of the angle for the larger triangle. Well, then the shorter side is on the right-hand side, and that's going to be xt. So what we want to compare is the ratio between, let me write it this way, we want to see is xy over xt, is that equal to the ratio of the longer side, or if we're looking relative to this angle, the longer of the two, not necessarily the longest of the triangle, although it looks like that as well, is that equal to the ratio of xz over the longer of the two sides when you're looking at this angle right here on either side of that angle for the larger triangle over xs."}, {"video_title": "Similar triangle example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "So the shorter side is 2. And let's look at the shorter side on either side of the angle for the larger triangle. Well, then the shorter side is on the right-hand side, and that's going to be xt. So what we want to compare is the ratio between, let me write it this way, we want to see is xy over xt, is that equal to the ratio of the longer side, or if we're looking relative to this angle, the longer of the two, not necessarily the longest of the triangle, although it looks like that as well, is that equal to the ratio of xz over the longer of the two sides when you're looking at this angle right here on either side of that angle for the larger triangle over xs. And it's a little confusing because we've kind of flipped which side, but I'm just thinking about the shorter side on either side of this angle in between and then the longer side on either side of this angle. So these are the shorter sides for the smaller triangle and the larger triangle. These are the longer sides for the smaller triangle and the larger triangle."}, {"video_title": "Similar triangle example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "So what we want to compare is the ratio between, let me write it this way, we want to see is xy over xt, is that equal to the ratio of the longer side, or if we're looking relative to this angle, the longer of the two, not necessarily the longest of the triangle, although it looks like that as well, is that equal to the ratio of xz over the longer of the two sides when you're looking at this angle right here on either side of that angle for the larger triangle over xs. And it's a little confusing because we've kind of flipped which side, but I'm just thinking about the shorter side on either side of this angle in between and then the longer side on either side of this angle. So these are the shorter sides for the smaller triangle and the larger triangle. These are the longer sides for the smaller triangle and the larger triangle. And we see xy, this is 2, xt is 3 plus 1 is 4, xz is 3, and xs is 6. So you have 2 over 4, which is 1 half, which is the same thing as 3, 6. So the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle for both triangles, the ratio is the same."}, {"video_title": "Similar triangle example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "These are the longer sides for the smaller triangle and the larger triangle. And we see xy, this is 2, xt is 3 plus 1 is 4, xz is 3, and xs is 6. So you have 2 over 4, which is 1 half, which is the same thing as 3, 6. So the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle for both triangles, the ratio is the same. So by SAS we know that the two triangles are congruent. But we have to be careful on how we state the triangles. We want to make sure we get the corresponding sides."}, {"video_title": "Similar triangle example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "So the ratio between the shorter sides on either side of the angle and the longer sides on either side of the angle for both triangles, the ratio is the same. So by SAS we know that the two triangles are congruent. But we have to be careful on how we state the triangles. We want to make sure we get the corresponding sides. So we could say that triangle, and I'm running out of space here, let me write it right above here. We can write that triangle xyz is similar to triangle. So we started up at x, which is the vertex at the angle, and we went to the shorter side first."}, {"video_title": "Similar triangle example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "We want to make sure we get the corresponding sides. So we could say that triangle, and I'm running out of space here, let me write it right above here. We can write that triangle xyz is similar to triangle. So we started up at x, which is the vertex at the angle, and we went to the shorter side first. So now we want to start at x and go to the shorter side of the large triangle. So you go to xts. xyz is similar to xts."}, {"video_title": "Similar triangle example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "So we started up at x, which is the vertex at the angle, and we went to the shorter side first. So now we want to start at x and go to the shorter side of the large triangle. So you go to xts. xyz is similar to xts. Now let's look at this right over here. So in our larger triangle we have a right angle here, but we really know nothing about what's going on with any of these smaller triangles in terms of their actual angles. Even though this looks like a right angle, we cannot assume it."}, {"video_title": "Similar triangle example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "xyz is similar to xts. Now let's look at this right over here. So in our larger triangle we have a right angle here, but we really know nothing about what's going on with any of these smaller triangles in terms of their actual angles. Even though this looks like a right angle, we cannot assume it. And it shares, if we look at this smaller triangle right over here, it shares one side with the larger triangle, but that's not enough to do anything. And then this triangle over here also shares another side, but that also doesn't do anything. So we really can't make any statement here about any kind of similarity."}, {"video_title": "Similar triangle example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "Even though this looks like a right angle, we cannot assume it. And it shares, if we look at this smaller triangle right over here, it shares one side with the larger triangle, but that's not enough to do anything. And then this triangle over here also shares another side, but that also doesn't do anything. So we really can't make any statement here about any kind of similarity. So there's no similarity going on here. If they gave us, well, there are some shared angles. This guy, they both share that angle."}, {"video_title": "Similar triangle example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "So we really can't make any statement here about any kind of similarity. So there's no similarity going on here. If they gave us, well, there are some shared angles. This guy, they both share that angle. The larger triangle and the smaller triangle. So there could be a statement of similarity we could make if we knew that this definitely was a right angle. Then we could make some interesting statements about similarity."}, {"video_title": "Similar triangle example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "This guy, they both share that angle. The larger triangle and the smaller triangle. So there could be a statement of similarity we could make if we knew that this definitely was a right angle. Then we could make some interesting statements about similarity. But right now we can't really do anything as is. Let's try this one out, or this pair right over here. So these are the first ones that we've actually separated out the triangles."}, {"video_title": "Similar triangle example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "Then we could make some interesting statements about similarity. But right now we can't really do anything as is. Let's try this one out, or this pair right over here. So these are the first ones that we've actually separated out the triangles. So they've given us the three sides of both triangles. So let's just figure out if the ratios between corresponding sides are a constant. So let's start with the short side."}, {"video_title": "Similar triangle example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "So these are the first ones that we've actually separated out the triangles. So they've given us the three sides of both triangles. So let's just figure out if the ratios between corresponding sides are a constant. So let's start with the short side. So the short side here is 3. The shortest side here is 9 square roots of 3. So we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here is 3 square roots of 3, is equal to 3 square roots of 3, over the next longest side over here, which is 27."}, {"video_title": "Similar triangle example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "So let's start with the short side. So the short side here is 3. The shortest side here is 9 square roots of 3. So we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here is 3 square roots of 3, is equal to 3 square roots of 3, over the next longest side over here, which is 27. And then see if that's going to be equal to the ratio of the longest side. So the longest side here is 6. And then the longest side over here is 18 square roots of 3."}, {"video_title": "Similar triangle example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "So we want to see whether the ratio of 3 to 9 square roots of 3 is equal to the next longest side over here is 3 square roots of 3, is equal to 3 square roots of 3, over the next longest side over here, which is 27. And then see if that's going to be equal to the ratio of the longest side. So the longest side here is 6. And then the longest side over here is 18 square roots of 3. So this is going to give us, let's see, this is 3. Let me do this in a neutral color. So this becomes 1 over 3 square roots of 3."}, {"video_title": "Similar triangle example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "And then the longest side over here is 18 square roots of 3. So this is going to give us, let's see, this is 3. Let me do this in a neutral color. So this becomes 1 over 3 square roots of 3. This becomes 1 over square roots of 3, root 3 over 9, which seems like a different number, but we want to be careful here. And then this right over here, this becomes, this is a, if you divide the numerator and denominator by 6, this becomes a 1 and this becomes 3 square roots of 3. So you get 1 square roots of 3, 1 over 3 root 3 needs to be equal to 1, needs to be equal to square root of 3 over 9, which needs to be equal to 1 over 3 square roots of 3."}, {"video_title": "Similar triangle example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "So this becomes 1 over 3 square roots of 3. This becomes 1 over square roots of 3, root 3 over 9, which seems like a different number, but we want to be careful here. And then this right over here, this becomes, this is a, if you divide the numerator and denominator by 6, this becomes a 1 and this becomes 3 square roots of 3. So you get 1 square roots of 3, 1 over 3 root 3 needs to be equal to 1, needs to be equal to square root of 3 over 9, which needs to be equal to 1 over 3 square roots of 3. At first they don't look equal, but we can actually rationalize this denominator right over here. We can show that 1 over 3 square roots of 3, if you multiply it by square root of 3 over square root of 3, this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 times 3 is 9. So these actually are all the same."}, {"video_title": "Similar triangle example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "So you get 1 square roots of 3, 1 over 3 root 3 needs to be equal to 1, needs to be equal to square root of 3 over 9, which needs to be equal to 1 over 3 square roots of 3. At first they don't look equal, but we can actually rationalize this denominator right over here. We can show that 1 over 3 square roots of 3, if you multiply it by square root of 3 over square root of 3, this actually gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3 times 3 is 9. So these actually are all the same. This is actually saying, this is 1 over 3 root 3, which is the same thing as square root of 3 over 9, which is this right over here, which is the same thing as 1 over 3 root 3. So actually these are similar triangles. So we can actually say it, and I'll make sure I get the order right."}, {"video_title": "Similar triangle example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "So these actually are all the same. This is actually saying, this is 1 over 3 root 3, which is the same thing as square root of 3 over 9, which is this right over here, which is the same thing as 1 over 3 root 3. So actually these are similar triangles. So we can actually say it, and I'll make sure I get the order right. So I'll start with E, which is between the blue and the magenta side. So that's between the blue and the magenta side, that is H right over here. So triangle E, I'll do it like this."}, {"video_title": "Similar triangle example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "So we can actually say it, and I'll make sure I get the order right. So I'll start with E, which is between the blue and the magenta side. So that's between the blue and the magenta side, that is H right over here. So triangle E, I'll do it like this. Triangle E, and then I'll go along the blue side, F. Then I'll go along the blue side over here. Oh, sorry, let me do it this way. Actually, let me just write it this way."}, {"video_title": "Similar triangle example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "So triangle E, I'll do it like this. Triangle E, and then I'll go along the blue side, F. Then I'll go along the blue side over here. Oh, sorry, let me do it this way. Actually, let me just write it this way. E, triangle E, F, G, we know is similar to triangle. So E is between the blue and the magenta side. Blue and magenta side, that is H. And then we go along the blue side to F, go along the blue side to I."}, {"video_title": "Similar triangle example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "Actually, let me just write it this way. E, triangle E, F, G, we know is similar to triangle. So E is between the blue and the magenta side. Blue and magenta side, that is H. And then we go along the blue side to F, go along the blue side to I. And then you went along the orange side to G. And then you go along the orange side to J. So triangle E, F, G is similar to triangle H, I, J by side, side, side similarity. They're not congruent sides."}, {"video_title": "Similar triangle example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "Blue and magenta side, that is H. And then we go along the blue side to F, go along the blue side to I. And then you went along the orange side to G. And then you go along the orange side to J. So triangle E, F, G is similar to triangle H, I, J by side, side, side similarity. They're not congruent sides. They all have just the same ratio or the same scaling factor. Now let's do this last one right over here. So we have an angle that's congruent to another angle right over there."}, {"video_title": "Similar triangle example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "They're not congruent sides. They all have just the same ratio or the same scaling factor. Now let's do this last one right over here. So we have an angle that's congruent to another angle right over there. And we have two sides. And so it might be tempting to use side, angle, side, because we have side, angle, side here. And even the ratios look kind of tempting, because 4 times 2 is 8, 5 times 2 is 10."}, {"video_title": "Similar triangle example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "So we have an angle that's congruent to another angle right over there. And we have two sides. And so it might be tempting to use side, angle, side, because we have side, angle, side here. And even the ratios look kind of tempting, because 4 times 2 is 8, 5 times 2 is 10. But it's tricky here, because they aren't the same corresponding sides. In order to use side, angle, side, the two sides that have the same corresponding ratios, they have to be on either side of the angle. So in this case, they are on either side of the angle."}, {"video_title": "Similar triangle example problems   Similarity   Geometry   Khan Academy.mp3", "Sentence": "And even the ratios look kind of tempting, because 4 times 2 is 8, 5 times 2 is 10. But it's tricky here, because they aren't the same corresponding sides. In order to use side, angle, side, the two sides that have the same corresponding ratios, they have to be on either side of the angle. So in this case, they are on either side of the angle. In this case, the 4 is on one side of the angle, but the 5 is not. So because if this 5 was over here, then we could make an argument for similarity. But with this 5 not being on the other side of the angle, it's not sandwiching the angle with the 4, we can't use side, angle, side."}, {"video_title": "Identifying perfect square form   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "we want to figure out what AX plus B squared is. And I encourage you to pause the video and figure out what that is in terms of capital A and capital B. So let's work through it. This is the same thing as multiplying AX plus B times AX plus B. So let me fill that in. This is AX there, another AX there. I just wrote it in that order to make the color switching a little bit easier."}, {"video_title": "Identifying perfect square form   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "This is the same thing as multiplying AX plus B times AX plus B. So let me fill that in. This is AX there, another AX there. I just wrote it in that order to make the color switching a little bit easier. So AX plus B times AX plus B. Well, what's that going to be equal to? Well, if you take this AX and you multiply it times that AX, you're going to get AX squared, AX, the entire thing squared."}, {"video_title": "Identifying perfect square form   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "I just wrote it in that order to make the color switching a little bit easier. So AX plus B times AX plus B. Well, what's that going to be equal to? Well, if you take this AX and you multiply it times that AX, you're going to get AX squared, AX, the entire thing squared. And then if you take this AX and then multiply it times this B, you're going to get ABX. Then if you take this B and you multiply it times this AX, you're going to get another ABX. A, B, X."}, {"video_title": "Identifying perfect square form   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "Well, if you take this AX and you multiply it times that AX, you're going to get AX squared, AX, the entire thing squared. And then if you take this AX and then multiply it times this B, you're going to get ABX. Then if you take this B and you multiply it times this AX, you're going to get another ABX. A, B, X. And then last but not least, if you take this B and multiply it times the other B, it's going to be plus B squared. And so what are you left with? Well, you're going to be left with A, I'll write it like this, AX squared, which we actually, if we want, well, I'll write it in a different way in a second."}, {"video_title": "Identifying perfect square form   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "A, B, X. And then last but not least, if you take this B and multiply it times the other B, it's going to be plus B squared. And so what are you left with? Well, you're going to be left with A, I'll write it like this, AX squared, which we actually, if we want, well, I'll write it in a different way in a second. And then you have plus, plus two, it's a slightly different color, let me do that other color, plus two ABX and then finally plus B squared. Plus B squared. Now I said I could write it in a slightly different way."}, {"video_title": "Identifying perfect square form   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "Well, you're going to be left with A, I'll write it like this, AX squared, which we actually, if we want, well, I'll write it in a different way in a second. And then you have plus, plus two, it's a slightly different color, let me do that other color, plus two ABX and then finally plus B squared. Plus B squared. Now I said I could write it in a slightly different way. What I could do is just rewrite out AX squared as being the same thing, this is the same thing as A squared, X squared, and then I can write out everything else the same way. Plus two ABX and then plus B squared. Now why did I, what's interesting about doing this?"}, {"video_title": "Identifying perfect square form   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "Now I said I could write it in a slightly different way. What I could do is just rewrite out AX squared as being the same thing, this is the same thing as A squared, X squared, and then I can write out everything else the same way. Plus two ABX and then plus B squared. Now why did I, what's interesting about doing this? Well, now we can see the pattern for the square of any binomial or binomial like this. So for example, if someone were to walk up to you and say, alright, I have a trinomial of the form, let's say they have a trinomial of the form 25X squared plus 20X plus four, and they were to tell you to factor this, well actually, let's just do that. Why don't you pause the video and see if you could factor this as a product of two binomials."}, {"video_title": "Identifying perfect square form   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "Now why did I, what's interesting about doing this? Well, now we can see the pattern for the square of any binomial or binomial like this. So for example, if someone were to walk up to you and say, alright, I have a trinomial of the form, let's say they have a trinomial of the form 25X squared plus 20X plus four, and they were to tell you to factor this, well actually, let's just do that. Why don't you pause the video and see if you could factor this as a product of two binomials. Well, when you look at this, you say, well look, you know, this 25X squared, that looks like a square, a perfect square. 25X squared, that's the same thing as five squared X squared, or you could write it as five X, five X squared. This four here, that's a perfect square."}, {"video_title": "Identifying perfect square form   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "Why don't you pause the video and see if you could factor this as a product of two binomials. Well, when you look at this, you say, well look, you know, this 25X squared, that looks like a square, a perfect square. 25X squared, that's the same thing as five squared X squared, or you could write it as five X, five X squared. This four here, that's a perfect square. That's the same thing as two squared. And let's see, 20 right over here, if we wanted to fit this pattern, we would say that A is five and B is two, and so let's see, what would be two times AB? Well, five times two, AB would be 10, and then two times that would be 20."}, {"video_title": "Identifying perfect square form   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "This four here, that's a perfect square. That's the same thing as two squared. And let's see, 20 right over here, if we wanted to fit this pattern, we would say that A is five and B is two, and so let's see, what would be two times AB? Well, five times two, AB would be 10, and then two times that would be 20. So this right over here is, that is, plus two times five, two times five times two, times two X, times two X, I'll do it in this color, times two X. So you see that this completely matches this pattern here, where A is equal to five, and B is equal to two. Once again, this is AX, whole thing squared, then you have two times A times B, X, we see that there, and then finally you have the B squared."}, {"video_title": "Recognizing quadratic factor methods.mp3", "Sentence": "We have other videos on individual techniques for factoring quadratics, but what I would like to do in this video is get some practice figuring out which technique to use. So I'm gonna write a bunch of quadratics. I encourage you to pause the video, try to see if you can factor that quadratic yourself before I work through it with you. So the first quadratic is six x squared plus three x. So pause and see if you can factor this. So this one might jump out at you that both of these terms here have a common factor. Both of them are divisible by three."}, {"video_title": "Recognizing quadratic factor methods.mp3", "Sentence": "So the first quadratic is six x squared plus three x. So pause and see if you can factor this. So this one might jump out at you that both of these terms here have a common factor. Both of them are divisible by three. Six is divisible by three and so is three. And both of them are divisible by x. So you can factor out a three x."}, {"video_title": "Recognizing quadratic factor methods.mp3", "Sentence": "Both of them are divisible by three. Six is divisible by three and so is three. And both of them are divisible by x. So you can factor out a three x. So if you factor out a three x, six x squared divided by three x, you're gonna have a two x left over there. And then three x divided by three x, you're going to have a one. And that's about as much as we can actually factor."}, {"video_title": "Recognizing quadratic factor methods.mp3", "Sentence": "So you can factor out a three x. So if you factor out a three x, six x squared divided by three x, you're gonna have a two x left over there. And then three x divided by three x, you're going to have a one. And that's about as much as we can actually factor. And you can verify that these two expressions are the same if you distribute the three x, three x times two x is six x squared. Three x times one is three x. And that's all we would do."}, {"video_title": "Recognizing quadratic factor methods.mp3", "Sentence": "And that's about as much as we can actually factor. And you can verify that these two expressions are the same if you distribute the three x, three x times two x is six x squared. Three x times one is three x. And that's all we would do. We would be done. That's all you can do to really factor that. And as we'll see, in this example, trying to factor out a common factor was all we had to do."}, {"video_title": "Recognizing quadratic factor methods.mp3", "Sentence": "And that's all we would do. We would be done. That's all you can do to really factor that. And as we'll see, in this example, trying to factor out a common factor was all we had to do. But as we'll see in future examples, that's usually a good first step. Do all of, check whether the terms have a common factor. And if they do, it never hurts to factor that out."}, {"video_title": "Recognizing quadratic factor methods.mp3", "Sentence": "And as we'll see, in this example, trying to factor out a common factor was all we had to do. But as we'll see in future examples, that's usually a good first step. Do all of, check whether the terms have a common factor. And if they do, it never hurts to factor that out. So let's do another example. Let's say I have the quadratic four x squared minus four x minus 48. Pause the video and try to factor that as much as you can."}, {"video_title": "Recognizing quadratic factor methods.mp3", "Sentence": "And if they do, it never hurts to factor that out. So let's do another example. Let's say I have the quadratic four x squared minus four x minus 48. Pause the video and try to factor that as much as you can. All right, so the first thing you might have noticed is that there is a common factor amongst the terms. All of them are divisible by four. Four is clearly divisible by four, and 48 is also divisible by four."}, {"video_title": "Recognizing quadratic factor methods.mp3", "Sentence": "Pause the video and try to factor that as much as you can. All right, so the first thing you might have noticed is that there is a common factor amongst the terms. All of them are divisible by four. Four is clearly divisible by four, and 48 is also divisible by four. So let's factor a four out. This would be the same thing as four times x squared minus x minus 12. I just divided each of these by four and I factored it out."}, {"video_title": "Recognizing quadratic factor methods.mp3", "Sentence": "Four is clearly divisible by four, and 48 is also divisible by four. So let's factor a four out. This would be the same thing as four times x squared minus x minus 12. I just divided each of these by four and I factored it out. You can distribute the four and verify that these two expressions are the same. Now, are we done? Well, no, we can factor what we have inside the parentheses."}, {"video_title": "Recognizing quadratic factor methods.mp3", "Sentence": "I just divided each of these by four and I factored it out. You can distribute the four and verify that these two expressions are the same. Now, are we done? Well, no, we can factor what we have inside the parentheses. We can factor this further. Now, how would we do that? So over here, the key realization is, all right, I have a one as a coefficient on my second degree term."}, {"video_title": "Recognizing quadratic factor methods.mp3", "Sentence": "Well, no, we can factor what we have inside the parentheses. We can factor this further. Now, how would we do that? So over here, the key realization is, all right, I have a one as a coefficient on my second degree term. I've written it in standard form where I have the second degree, and then if there's a first degree term, then I have my constant term or my zero degree term. And if I have a one coefficient right over here, then I say, okay, are there two numbers whose sum equals the coefficient on the first degree term, on the x term? So are there two numbers that add up to negative one?"}, {"video_title": "Recognizing quadratic factor methods.mp3", "Sentence": "So over here, the key realization is, all right, I have a one as a coefficient on my second degree term. I've written it in standard form where I have the second degree, and then if there's a first degree term, then I have my constant term or my zero degree term. And if I have a one coefficient right over here, then I say, okay, are there two numbers whose sum equals the coefficient on the first degree term, on the x term? So are there two numbers that add up to negative one? You didn't see a one here before, but it's implicit there. Negative x is the same thing as negative one x. So are there two numbers, a plus b, that is equal to negative one, and whose product is equal to negative 12?"}, {"video_title": "Recognizing quadratic factor methods.mp3", "Sentence": "So are there two numbers that add up to negative one? You didn't see a one here before, but it's implicit there. Negative x is the same thing as negative one x. So are there two numbers, a plus b, that is equal to negative one, and whose product is equal to negative 12? This is a technique that we do in other videos. And here, the key is to realize that, hey, maybe we can use it here. So a times b is equal to negative 12."}, {"video_title": "Recognizing quadratic factor methods.mp3", "Sentence": "So are there two numbers, a plus b, that is equal to negative one, and whose product is equal to negative 12? This is a technique that we do in other videos. And here, the key is to realize that, hey, maybe we can use it here. So a times b is equal to negative 12. And there's a couple of key realizations here. It's like, okay, if I have two numbers and their product is going to be a negative, that means one of them, that means they're gonna have different signs. One's going to be positive and one's going to be negative."}, {"video_title": "Recognizing quadratic factor methods.mp3", "Sentence": "So a times b is equal to negative 12. And there's a couple of key realizations here. It's like, okay, if I have two numbers and their product is going to be a negative, that means one of them, that means they're gonna have different signs. One's going to be positive and one's going to be negative. If they had the same sign, then this would be positive. So let's think about the factors of 12, and especially think about them in terms of different sign combinations. So you could think about one and 12, and whether you're thinking about negative one and 12, negative one plus 12 would be positive 11."}, {"video_title": "Recognizing quadratic factor methods.mp3", "Sentence": "One's going to be positive and one's going to be negative. If they had the same sign, then this would be positive. So let's think about the factors of 12, and especially think about them in terms of different sign combinations. So you could think about one and 12, and whether you're thinking about negative one and 12, negative one plus 12 would be positive 11. If you went the other way around, if you went negative 12 and one, it would be negative 11. But either way, that doesn't work. Two and six, negative two and six would be four."}, {"video_title": "Recognizing quadratic factor methods.mp3", "Sentence": "So you could think about one and 12, and whether you're thinking about negative one and 12, negative one plus 12 would be positive 11. If you went the other way around, if you went negative 12 and one, it would be negative 11. But either way, that doesn't work. Two and six, negative two and six would be four. Negative six and two would be negative four, so that doesn't work. Three and four. Let's see, negative three and four would be positive one, but three and negative four works out."}, {"video_title": "Recognizing quadratic factor methods.mp3", "Sentence": "Two and six, negative two and six would be four. Negative six and two would be negative four, so that doesn't work. Three and four. Let's see, negative three and four would be positive one, but three and negative four works out. You add these two together, you take their product, you clearly get negative 12, and then you add them together, you get negative one. So we can write inside the parentheses, so let me write, so it's gonna be four times, so we can factor that as two binomials. The first is going to be x plus, the first is going to be x plus three, and then the next is going to be, we could say x plus negative four, or we could say x minus four."}, {"video_title": "Recognizing quadratic factor methods.mp3", "Sentence": "Let's see, negative three and four would be positive one, but three and negative four works out. You add these two together, you take their product, you clearly get negative 12, and then you add them together, you get negative one. So we can write inside the parentheses, so let me write, so it's gonna be four times, so we can factor that as two binomials. The first is going to be x plus, the first is going to be x plus three, and then the next is going to be, we could say x plus negative four, or we could say x minus four. And we're done. And if any of this seems intimidating to you, I encourage you to watch the videos on introduction to factoring polynomials. The key here is to recognize the method."}, {"video_title": "Recognizing quadratic factor methods.mp3", "Sentence": "The first is going to be x plus, the first is going to be x plus three, and then the next is going to be, we could say x plus negative four, or we could say x minus four. And we're done. And if any of this seems intimidating to you, I encourage you to watch the videos on introduction to factoring polynomials. The key here is to recognize the method. So once again, at first, try to factor out any common factors. We did that in both examples. Then we saw here, hey, if we have a leading one coefficient here on the second degree term, and we have it written in standard form, well, let's think of two numbers that add up to this coefficient, and whose product is equal to the constant term."}, {"video_title": "Recognizing quadratic factor methods.mp3", "Sentence": "The key here is to recognize the method. So once again, at first, try to factor out any common factors. We did that in both examples. Then we saw here, hey, if we have a leading one coefficient here on the second degree term, and we have it written in standard form, well, let's think of two numbers that add up to this coefficient, and whose product is equal to the constant term. And in this case, it was three and negative four. We were able to factor it this way. We proved that in other videos."}, {"video_title": "Recognizing quadratic factor methods.mp3", "Sentence": "Then we saw here, hey, if we have a leading one coefficient here on the second degree term, and we have it written in standard form, well, let's think of two numbers that add up to this coefficient, and whose product is equal to the constant term. And in this case, it was three and negative four. We were able to factor it this way. We proved that in other videos. Let's do another example. We can't get enough practice. And like always, pause the video and see if you can work through it yourself."}, {"video_title": "Recognizing quadratic factor methods.mp3", "Sentence": "We proved that in other videos. Let's do another example. We can't get enough practice. And like always, pause the video and see if you can work through it yourself. Three x squared plus 30 plus 75. All right, I'm assuming you had a go at it. So you might immediately see that all of the terms are divisible by three, so let's factor a three out."}, {"video_title": "Recognizing quadratic factor methods.mp3", "Sentence": "And like always, pause the video and see if you can work through it yourself. Three x squared plus 30 plus 75. All right, I'm assuming you had a go at it. So you might immediately see that all of the terms are divisible by three, so let's factor a three out. So it's gonna be three times x squared plus, whoops, this should be an x here, my apologies. Pause the video again and see if you can do it now that I wrote the actual right thing there. All right, so as you imagine, it's nice to factor out a three first."}, {"video_title": "Recognizing quadratic factor methods.mp3", "Sentence": "So you might immediately see that all of the terms are divisible by three, so let's factor a three out. So it's gonna be three times x squared plus, whoops, this should be an x here, my apologies. Pause the video again and see if you can do it now that I wrote the actual right thing there. All right, so as you imagine, it's nice to factor out a three first. So it's three times x squared plus 10x plus 25. And so you might immediately say, all right, well, let's use the technique we had here. We have a leading one coefficient."}, {"video_title": "Recognizing quadratic factor methods.mp3", "Sentence": "All right, so as you imagine, it's nice to factor out a three first. So it's three times x squared plus 10x plus 25. And so you might immediately say, all right, well, let's use the technique we had here. We have a leading one coefficient. It's written in standard form. Can I think of two numbers that add up to 10? So a plus b is equal to 10."}, {"video_title": "Recognizing quadratic factor methods.mp3", "Sentence": "We have a leading one coefficient. It's written in standard form. Can I think of two numbers that add up to 10? So a plus b is equal to 10. And whose product, a times b, is equal to 25? And this would work. And if you look at the factors of 25, you'd say, all right, well, this thing here is positive."}, {"video_title": "Recognizing quadratic factor methods.mp3", "Sentence": "So a plus b is equal to 10. And whose product, a times b, is equal to 25? And this would work. And if you look at the factors of 25, you'd say, all right, well, this thing here is positive. This is positive, so I'm dealing with two positive numbers. And to get 25, it's either one and 25 or five and five. And five and five match this."}, {"video_title": "Recognizing quadratic factor methods.mp3", "Sentence": "And if you look at the factors of 25, you'd say, all right, well, this thing here is positive. This is positive, so I'm dealing with two positive numbers. And to get 25, it's either one and 25 or five and five. And five and five match this. Five plus five is equal to 10. Five times five is equal to 25. And so just using the exact technique we just did, you would say, okay, this is three times, and the stuff in parentheses would be x plus five times x plus five."}, {"video_title": "Recognizing quadratic factor methods.mp3", "Sentence": "And five and five match this. Five plus five is equal to 10. Five times five is equal to 25. And so just using the exact technique we just did, you would say, okay, this is three times, and the stuff in parentheses would be x plus five times x plus five. Or you could say three times x plus five squared. So some of you might have immediately said, well, I don't have to do that exact technique. I could have immediately recognized this as a perfect square because I have a square constant right over here."}, {"video_title": "Recognizing quadratic factor methods.mp3", "Sentence": "And so just using the exact technique we just did, you would say, okay, this is three times, and the stuff in parentheses would be x plus five times x plus five. Or you could say three times x plus five squared. So some of you might have immediately said, well, I don't have to do that exact technique. I could have immediately recognized this as a perfect square because I have a square constant right over here. And that's a good sign that, hey, maybe I should explore whether this is going to be a perfect square polynomial. So if this is a perfect square, and if I were to take the square root of it, and this coefficient is twice that square root, well, that's a good sign that I'm probably dealing, or that I am dealing with a perfect square. But either way, whether you recognize it as a perfect square or whether you use the technique that we used in the second problem, either one of those would get you to the appropriate answer."}, {"video_title": "Introduction to minimum and maximum points   Functions   Algebra I   Khan Academy.mp3", "Sentence": "And I want to think about the maximum and minimum points on this. So we've already talked a little bit about absolute maximum and absolute minimum points on an interval. And those are pretty obvious. We hit a maximum point right over here, right at the beginning of our interval. Looks like when x is equal to 0, this is the absolute maximum point for the interval. And the absolute minimum point for the interval happens at the other end point. So if this is a, this is b, the absolute minimum point is f of b."}, {"video_title": "Introduction to minimum and maximum points   Functions   Algebra I   Khan Academy.mp3", "Sentence": "We hit a maximum point right over here, right at the beginning of our interval. Looks like when x is equal to 0, this is the absolute maximum point for the interval. And the absolute minimum point for the interval happens at the other end point. So if this is a, this is b, the absolute minimum point is f of b. And the absolute maximum point is f of a. And it looks like a is equal to 0. But you're probably thinking, hey, there are other kind of interesting points right over here."}, {"video_title": "Introduction to minimum and maximum points   Functions   Algebra I   Khan Academy.mp3", "Sentence": "So if this is a, this is b, the absolute minimum point is f of b. And the absolute maximum point is f of a. And it looks like a is equal to 0. But you're probably thinking, hey, there are other kind of interesting points right over here. This point right over here, it isn't the largest. We're not taking on this value right over here. It is definitely not the largest value that the function takes on in that interval."}, {"video_title": "Introduction to minimum and maximum points   Functions   Algebra I   Khan Academy.mp3", "Sentence": "But you're probably thinking, hey, there are other kind of interesting points right over here. This point right over here, it isn't the largest. We're not taking on this value right over here. It is definitely not the largest value that the function takes on in that interval. But relative to the other values around it, it seems like a little bit of a hill. It's larger than the other ones. Locally, it looks like a little bit of a maximum."}, {"video_title": "Introduction to minimum and maximum points   Functions   Algebra I   Khan Academy.mp3", "Sentence": "It is definitely not the largest value that the function takes on in that interval. But relative to the other values around it, it seems like a little bit of a hill. It's larger than the other ones. Locally, it looks like a little bit of a maximum. And so that's why this value right over here would be called, this value right over here, let's call this, let's say this right over here is c. This is c, so this is f of c. We would call f of c is a relative maximum value. And we're saying relative because it's obviously the function takes on other values that are larger than it. But for the x values near c, f of c is larger than all of those."}, {"video_title": "Introduction to minimum and maximum points   Functions   Algebra I   Khan Academy.mp3", "Sentence": "Locally, it looks like a little bit of a maximum. And so that's why this value right over here would be called, this value right over here, let's call this, let's say this right over here is c. This is c, so this is f of c. We would call f of c is a relative maximum value. And we're saying relative because it's obviously the function takes on other values that are larger than it. But for the x values near c, f of c is larger than all of those. Similarly, I can never say that word, if this point right over here is d, f of d looks like a relative minimum point or relative minimum value. f of d is a relative minimum or a local minimum value. Once again, over the whole interval, there's definitely points that are lower."}, {"video_title": "Introduction to minimum and maximum points   Functions   Algebra I   Khan Academy.mp3", "Sentence": "But for the x values near c, f of c is larger than all of those. Similarly, I can never say that word, if this point right over here is d, f of d looks like a relative minimum point or relative minimum value. f of d is a relative minimum or a local minimum value. Once again, over the whole interval, there's definitely points that are lower. And we hit an absolute minimum for the interval at x is equal to b. But this is a relative minimum or a local minimum because it's lower than the, if we look at the x values around d, the function at those values is higher than when we get to d. So let's think about, it's fine for me to say, well, you're at a relative maximum if you hit a larger value of your function than any of the surrounding values. And you're at a minimum if you're at a smaller value than any of the surrounding areas."}, {"video_title": "Introduction to minimum and maximum points   Functions   Algebra I   Khan Academy.mp3", "Sentence": "Once again, over the whole interval, there's definitely points that are lower. And we hit an absolute minimum for the interval at x is equal to b. But this is a relative minimum or a local minimum because it's lower than the, if we look at the x values around d, the function at those values is higher than when we get to d. So let's think about, it's fine for me to say, well, you're at a relative maximum if you hit a larger value of your function than any of the surrounding values. And you're at a minimum if you're at a smaller value than any of the surrounding areas. But how could we write that mathematically? So here, I'll just give you the definition that really is just a more formal way of saying what we just said. So we say that f of c is a relative maximum value if f of c is greater than or equal to f of x for all x."}, {"video_title": "Introduction to minimum and maximum points   Functions   Algebra I   Khan Academy.mp3", "Sentence": "And you're at a minimum if you're at a smaller value than any of the surrounding areas. But how could we write that mathematically? So here, I'll just give you the definition that really is just a more formal way of saying what we just said. So we say that f of c is a relative maximum value if f of c is greater than or equal to f of x for all x. We could just say kind of in a casual way, for all x near c. So we could write it like that. But that's not too rigorous because what does it mean to be near c? And so a more rigorous way of saying it, for all x that's within an open interval of c minus h to c plus h, where h is some value greater than 0."}, {"video_title": "Introduction to minimum and maximum points   Functions   Algebra I   Khan Academy.mp3", "Sentence": "So we say that f of c is a relative maximum value if f of c is greater than or equal to f of x for all x. We could just say kind of in a casual way, for all x near c. So we could write it like that. But that's not too rigorous because what does it mean to be near c? And so a more rigorous way of saying it, for all x that's within an open interval of c minus h to c plus h, where h is some value greater than 0. So does that make sense? Well, let's look at it. So let's construct an open interval."}, {"video_title": "Introduction to minimum and maximum points   Functions   Algebra I   Khan Academy.mp3", "Sentence": "And so a more rigorous way of saying it, for all x that's within an open interval of c minus h to c plus h, where h is some value greater than 0. So does that make sense? Well, let's look at it. So let's construct an open interval. So it looks like for all of the x values in, and you just have to find one open interval. There might be many open intervals where this is true. But if we construct an open interval that looks something like that."}, {"video_title": "Introduction to minimum and maximum points   Functions   Algebra I   Khan Academy.mp3", "Sentence": "So let's construct an open interval. So it looks like for all of the x values in, and you just have to find one open interval. There might be many open intervals where this is true. But if we construct an open interval that looks something like that. So this value right over here is c plus h. That value right over here is c minus h. And you see that over that interval, the function at c, f of c, is definitely greater than or equal to the value of the function over any other part of that open interval. And so you could imagine, I encourage you to pause the video and you could write out what the more formal definition of a relative minimum point would be. Well, we would just write, let's take d as our relative minimum."}, {"video_title": "Introduction to minimum and maximum points   Functions   Algebra I   Khan Academy.mp3", "Sentence": "But if we construct an open interval that looks something like that. So this value right over here is c plus h. That value right over here is c minus h. And you see that over that interval, the function at c, f of c, is definitely greater than or equal to the value of the function over any other part of that open interval. And so you could imagine, I encourage you to pause the video and you could write out what the more formal definition of a relative minimum point would be. Well, we would just write, let's take d as our relative minimum. We can say that f of d is a relative minimum point if f of d is less than or equal to f of x for all x in an interval, in an open interval, between d minus h and d plus h for h is greater than 0. So you can find an interval here. So let's say this is d plus h. This is d minus h. The function over that interval, f of d is always less than or equal to any of the other values."}, {"video_title": "Introduction to minimum and maximum points   Functions   Algebra I   Khan Academy.mp3", "Sentence": "Well, we would just write, let's take d as our relative minimum. We can say that f of d is a relative minimum point if f of d is less than or equal to f of x for all x in an interval, in an open interval, between d minus h and d plus h for h is greater than 0. So you can find an interval here. So let's say this is d plus h. This is d minus h. The function over that interval, f of d is always less than or equal to any of the other values. The f is of all of these other x's in that interval. And that's why we say that it's a relative minimum point. So in everyday language, relative max, if the function takes on a larger value at c, then for the x values around c, and you're at a relative minimum value, if the function takes on a lower value at d, then for the x values near d."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "And on our xy-coordinate plane, we want to show all of the x and y points that satisfy this condition right here. So a good starting point might be to break up this less than or equal to, because we know how to graph y is equal to 4x plus 3. So this thing is the same thing as y could be less than 4x plus 3, or y could be equal to 4x plus 3. That's what less than or equal means. It could be less than or equal. And the reason why I did that on this first example problem is because we know how to graph that. So let's graph that."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "That's what less than or equal means. It could be less than or equal. And the reason why I did that on this first example problem is because we know how to graph that. So let's graph that. I'll try to draw it a little bit neater than that. So that is my vertical axis, my y-axis. This is my x-axis right there."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So let's graph that. I'll try to draw it a little bit neater than that. So that is my vertical axis, my y-axis. This is my x-axis right there. That is the x-axis. And then we know the y-intercept. The y-intercept is 3, so the point 0, 3, 1, 2, 3, is on the line, and we know we have a slope of 4, which means if we go 1 in the x direction, we're going to go up 4 in the y."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "This is my x-axis right there. That is the x-axis. And then we know the y-intercept. The y-intercept is 3, so the point 0, 3, 1, 2, 3, is on the line, and we know we have a slope of 4, which means if we go 1 in the x direction, we're going to go up 4 in the y. So 1, 2, 3, 4. So it's going to be right here. And that's enough to draw a line, but we could even go back in the x direction."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "The y-intercept is 3, so the point 0, 3, 1, 2, 3, is on the line, and we know we have a slope of 4, which means if we go 1 in the x direction, we're going to go up 4 in the y. So 1, 2, 3, 4. So it's going to be right here. And that's enough to draw a line, but we could even go back in the x direction. If we go 1 back in the x direction, we're going to go down 4. 1, 2, 3, 4. So that's also going to be a point on the line."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "And that's enough to draw a line, but we could even go back in the x direction. If we go 1 back in the x direction, we're going to go down 4. 1, 2, 3, 4. So that's also going to be a point on the line. So my best attempt at drawing this line is going to look something like something. This is the hardest part. It's going to look something like that."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So that's also going to be a point on the line. So my best attempt at drawing this line is going to look something like something. This is the hardest part. It's going to look something like that. That is a line. It should be straight. I think you get the idea."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "It's going to look something like that. That is a line. It should be straight. I think you get the idea. That right there is the graph of y is equal to 4x plus 3. So let's think about what it means to be less than. So all of these points satisfy this inequality, but we have more."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "I think you get the idea. That right there is the graph of y is equal to 4x plus 3. So let's think about what it means to be less than. So all of these points satisfy this inequality, but we have more. This is just these points over here. What about all of these where y is less than 4x plus 3? So let's think about what this means."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So all of these points satisfy this inequality, but we have more. This is just these points over here. What about all of these where y is less than 4x plus 3? So let's think about what this means. When, let's pick up some values for x. When x is equal to 0, what does this say? When x is equal to 0, then that means y is going to be less than 0 plus 3. y is less than 3."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So let's think about what this means. When, let's pick up some values for x. When x is equal to 0, what does this say? When x is equal to 0, then that means y is going to be less than 0 plus 3. y is less than 3. When x is equal to negative 1, what is this telling us? 4 times negative 1 is negative 4, plus 3 is negative 1. y would be less than negative 1. When x is equal to 1, what is this telling us?"}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "When x is equal to 0, then that means y is going to be less than 0 plus 3. y is less than 3. When x is equal to negative 1, what is this telling us? 4 times negative 1 is negative 4, plus 3 is negative 1. y would be less than negative 1. When x is equal to 1, what is this telling us? 4 times 1 is 4, plus 3 is 7. So y is going to be less than 7. So let's at least try to plot these."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "When x is equal to 1, what is this telling us? 4 times 1 is 4, plus 3 is 7. So y is going to be less than 7. So let's at least try to plot these. So when x is equal to, let's plot this one first. When x is equal to 0, y is less than 3. So when x is equal to 0, y is less than 3."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So let's at least try to plot these. So when x is equal to, let's plot this one first. When x is equal to 0, y is less than 3. So when x is equal to 0, y is less than 3. So it's all of these points here that I'm shading in in green, satisfy that right there. If I were to look at this one over here, when x is negative 1, y is less than negative 1. So y has to be all of these points down here."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So when x is equal to 0, y is less than 3. So it's all of these points here that I'm shading in in green, satisfy that right there. If I were to look at this one over here, when x is negative 1, y is less than negative 1. So y has to be all of these points down here. When x is equal to 1, y is less than 7. So it's all of these points down here. And in general, you take any point x."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So y has to be all of these points down here. When x is equal to 1, y is less than 7. So it's all of these points down here. And in general, you take any point x. Let's say you take this point x right there. If you evaluate 4x plus 3, you're going to get the point on the line. That is that x times 4 plus 3."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "And in general, you take any point x. Let's say you take this point x right there. If you evaluate 4x plus 3, you're going to get the point on the line. That is that x times 4 plus 3. Now, the y's that satisfy it, it could be equal to that point on the line, or it could be less than. So it's going to go below the line. So if you were to do this for all the possible x's, you would not only get all the points on this line, which we've drawn, you would get all the points below the line."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "That is that x times 4 plus 3. Now, the y's that satisfy it, it could be equal to that point on the line, or it could be less than. So it's going to go below the line. So if you were to do this for all the possible x's, you would not only get all the points on this line, which we've drawn, you would get all the points below the line. So now we have graphed this inequality. It's essentially this line, 4x plus 3, with all of the area below it shaded. Now, if this was just a less than, not less than or equal sign, we would not include the actual line."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So if you were to do this for all the possible x's, you would not only get all the points on this line, which we've drawn, you would get all the points below the line. So now we have graphed this inequality. It's essentially this line, 4x plus 3, with all of the area below it shaded. Now, if this was just a less than, not less than or equal sign, we would not include the actual line. And the convention to do that is to actually make the line a dashed line. This is the situation if we were dealing with just less than 4x plus 3, because in that situation, this wouldn't apply, and we would just have that. So the line itself wouldn't have satisfied it, just the area below it."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Now, if this was just a less than, not less than or equal sign, we would not include the actual line. And the convention to do that is to actually make the line a dashed line. This is the situation if we were dealing with just less than 4x plus 3, because in that situation, this wouldn't apply, and we would just have that. So the line itself wouldn't have satisfied it, just the area below it. Let's do one like that. So let's say we have y is greater than negative x over 2 minus 6. So a good way to start, the way I like to start these problems, is to just graph this equation right here."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So the line itself wouldn't have satisfied it, just the area below it. Let's do one like that. So let's say we have y is greater than negative x over 2 minus 6. So a good way to start, the way I like to start these problems, is to just graph this equation right here. So let me just graph, just for fun, y is equal to, this is the same thing as negative 1 half minus 6. So if we were to graph it, that is my vertical axis, that is my horizontal axis, and our y-intercept is negative 6. So 1, 2, 3, 4, 5, 6."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So a good way to start, the way I like to start these problems, is to just graph this equation right here. So let me just graph, just for fun, y is equal to, this is the same thing as negative 1 half minus 6. So if we were to graph it, that is my vertical axis, that is my horizontal axis, and our y-intercept is negative 6. So 1, 2, 3, 4, 5, 6. So that's my y-intercept. And my slope is negative 1 half. Well, that should be an x there."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So 1, 2, 3, 4, 5, 6. So that's my y-intercept. And my slope is negative 1 half. Well, that should be an x there. Negative 1 half x minus 6. So my slope is negative 1 half, which means when I go 2 to the right, I go down 1. So if I go 2 to the right, I'm going to go down 1."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Well, that should be an x there. Negative 1 half x minus 6. So my slope is negative 1 half, which means when I go 2 to the right, I go down 1. So if I go 2 to the right, I'm going to go down 1. So 2 to the right, down 1. If I go 2 to the left, if I go negative 2, I'm going to go up 1. So negative 2, up 1."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So if I go 2 to the right, I'm going to go down 1. So 2 to the right, down 1. If I go 2 to the left, if I go negative 2, I'm going to go up 1. So negative 2, up 1. So my line is going to look like this. My line is going to look like that. That's my best attempt at drawing the line."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So negative 2, up 1. So my line is going to look like this. My line is going to look like that. That's my best attempt at drawing the line. So that's the line of y is equal to negative 1 half x minus 6. Now, our inequality is not greater than or equal. It's just greater than negative x over 2 minus 6, or greater than negative 1 half x minus 6."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "That's my best attempt at drawing the line. So that's the line of y is equal to negative 1 half x minus 6. Now, our inequality is not greater than or equal. It's just greater than negative x over 2 minus 6, or greater than negative 1 half x minus 6. So using the same logic as before, for any x, let's say that's a particular x we want to pick. If you evaluate negative x over 2 minus 6, you're going to get that point right there. You're going to get the point on the line."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "It's just greater than negative x over 2 minus 6, or greater than negative 1 half x minus 6. So using the same logic as before, for any x, let's say that's a particular x we want to pick. If you evaluate negative x over 2 minus 6, you're going to get that point right there. You're going to get the point on the line. But the y's that satisfy this inequality are the y's greater than that. So it's going to be not that point. In fact, you would draw an open circle there, because you can't include the point of negative 1 half x minus 6."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "You're going to get the point on the line. But the y's that satisfy this inequality are the y's greater than that. So it's going to be not that point. In fact, you would draw an open circle there, because you can't include the point of negative 1 half x minus 6. But it's going to be all the y's greater than that. So it's going to be all the y's greater than that. And that would be true for any x."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "In fact, you would draw an open circle there, because you can't include the point of negative 1 half x minus 6. But it's going to be all the y's greater than that. So it's going to be all the y's greater than that. And that would be true for any x. You take this x, you evaluate negative 1 half or negative x over 2 minus 6, you're going to get this point over here. But the y's that satisfy it are all the y's above that. So all of the y's that satisfy this equation, or all of the coordinates that satisfy the equation, is this entire area above the line."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "And that would be true for any x. You take this x, you evaluate negative 1 half or negative x over 2 minus 6, you're going to get this point over here. But the y's that satisfy it are all the y's above that. So all of the y's that satisfy this equation, or all of the coordinates that satisfy the equation, is this entire area above the line. And we're not going to include the line. So the convention is to make this line into a dashed line. And let me try my best to turn it into a dashed line."}, {"video_title": "Introduction to graphing inequalities   Two-variable linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So all of the y's that satisfy this equation, or all of the coordinates that satisfy the equation, is this entire area above the line. And we're not going to include the line. So the convention is to make this line into a dashed line. And let me try my best to turn it into a dashed line. I'll just erase sections of the line. And hopefully it will look dashed to you. So I'm turning that solid line into a dashed line to show that it's just a boundary, but it's not included in the coordinates that satisfy our inequality."}, {"video_title": "Abstract-ness   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "You'll hear me use the word abstract a lot, so I thought I would actually give you an attempt at a definition, or maybe even more important, an intuition of what abstract means. An abstract, it can be an adjective. You can have an abstract idea. You can have abstract art, or it can be a verb. You can abstract something, abstract the idea from some other idea. And you can even have it as a noun. You can have an abstract, and it tends to, when you use it as a noun, the one that I tend to associate is the abstract of a research paper, which is kind of distills the essence of the research paper, it's kind of a summary of that paper."}, {"video_title": "Abstract-ness   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "You can have abstract art, or it can be a verb. You can abstract something, abstract the idea from some other idea. And you can even have it as a noun. You can have an abstract, and it tends to, when you use it as a noun, the one that I tend to associate is the abstract of a research paper, which is kind of distills the essence of the research paper, it's kind of a summary of that paper. And the one thing you're going to see, regardless of how you, what context you use the word abstract, there's this kind of notion of taking the essence of a real world, of a real world object, whether you use it as a noun, an adjective, or a verb. So over here we have our real world. We have our real world."}, {"video_title": "Abstract-ness   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "You can have an abstract, and it tends to, when you use it as a noun, the one that I tend to associate is the abstract of a research paper, which is kind of distills the essence of the research paper, it's kind of a summary of that paper. And the one thing you're going to see, regardless of how you, what context you use the word abstract, there's this kind of notion of taking the essence of a real world, of a real world object, whether you use it as a noun, an adjective, or a verb. So over here we have our real world. We have our real world. And then over here you have your world of, you have ideas, ideas and concepts. And the general idea behind abstraction, or abstracting something, is you're taking it away from the particular concrete real world, and you're going more in the direction of ideas and concepts. And probably for me, one of the most tangible ways of thinking about abstraction, which is kind of a contradiction in itself, to think of abstraction in a tangible way, is things like geometric shapes."}, {"video_title": "Abstract-ness   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "We have our real world. And then over here you have your world of, you have ideas, ideas and concepts. And the general idea behind abstraction, or abstracting something, is you're taking it away from the particular concrete real world, and you're going more in the direction of ideas and concepts. And probably for me, one of the most tangible ways of thinking about abstraction, which is kind of a contradiction in itself, to think of abstraction in a tangible way, is things like geometric shapes. So if I were to tell you find me some cubes, you might point to a Borg vessel right over there. A Borg vessel. You might point to maybe a pair of dice."}, {"video_title": "Abstract-ness   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And probably for me, one of the most tangible ways of thinking about abstraction, which is kind of a contradiction in itself, to think of abstraction in a tangible way, is things like geometric shapes. So if I were to tell you find me some cubes, you might point to a Borg vessel right over there. A Borg vessel. You might point to maybe a pair of dice. So let me draw the pair of dice. If you are looking for cubes. So you might point to a pair of dice that looks something like that."}, {"video_title": "Abstract-ness   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "You might point to maybe a pair of dice. So let me draw the pair of dice. If you are looking for cubes. So you might point to a pair of dice that looks something like that. You might point to a Rubik's cube. Anything that you could find. There might be a building that looks like a cube."}, {"video_title": "Abstract-ness   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "So you might point to a pair of dice that looks something like that. You might point to a Rubik's cube. Anything that you could find. There might be a building that looks like a cube. There might be a building that looks like a cube. Or maybe there's a box in your house that is a cube. But in your mind, you have a general idea of what a cube is."}, {"video_title": "Abstract-ness   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "There might be a building that looks like a cube. There might be a building that looks like a cube. Or maybe there's a box in your house that is a cube. But in your mind, you have a general idea of what a cube is. You're like, well I know a cube when I see one. And that general idea is essentially distilling the concept, the idea of what a cube is. All of these things are very different."}, {"video_title": "Abstract-ness   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "But in your mind, you have a general idea of what a cube is. You're like, well I know a cube when I see one. And that general idea is essentially distilling the concept, the idea of what a cube is. All of these things are very different. This is some plastic thing that I can hold in my hand. These are these little white things. And they're not even, you know, geometrically even close to being perfect."}, {"video_title": "Abstract-ness   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "All of these things are very different. This is some plastic thing that I can hold in my hand. These are these little white things. And they're not even, you know, geometrically even close to being perfect. They have these little divots on the side right over there. This is a large Borg vessel that, you know, it doesn't even exist yet. Obviously it's a fictional thing."}, {"video_title": "Abstract-ness   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And they're not even, you know, geometrically even close to being perfect. They have these little divots on the side right over there. This is a large Borg vessel that, you know, it doesn't even exist yet. Obviously it's a fictional thing. But they all have this cubeness to them. And this is one of the fun things of geometry is to really distill the essence of these real world shapes. And we do have a definition in geometry, which is an object like this."}, {"video_title": "Abstract-ness   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Obviously it's a fictional thing. But they all have this cubeness to them. And this is one of the fun things of geometry is to really distill the essence of these real world shapes. And we do have a definition in geometry, which is an object like this. I guess you could have an object like this, where every side has the exact same length. So if this is length 1, then that would be length 1. That is length 1."}, {"video_title": "Abstract-ness   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And we do have a definition in geometry, which is an object like this. I guess you could have an object like this, where every side has the exact same length. So if this is length 1, then that would be length 1. That is length 1. But it doesn't have to be. Whatever the length this side is in this dimension, then in this dimension it's going to be that length. And in that dimension it's going to be in that length."}, {"video_title": "Abstract-ness   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "That is length 1. But it doesn't have to be. Whatever the length this side is in this dimension, then in this dimension it's going to be that length. And in that dimension it's going to be in that length. And I'm not giving you the rigorous geometric definition. But I'm just trying to highlight that there is a pure idea of what a cube is. And in the real world, there is nothing that is actually a perfect cube."}, {"video_title": "Abstract-ness   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And in that dimension it's going to be in that length. And I'm not giving you the rigorous geometric definition. But I'm just trying to highlight that there is a pure idea of what a cube is. And in the real world, there is nothing that is actually a perfect cube. If you were to get really, really close to these die, if you were to really try to measure exactly their measurements, they won't be exactly the same measurement. But the abstract idea of a cube, this is completely the same length as this. And this, and actually this, and all of the edges are going to have the exact same length."}, {"video_title": "Abstract-ness   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And in the real world, there is nothing that is actually a perfect cube. If you were to get really, really close to these die, if you were to really try to measure exactly their measurements, they won't be exactly the same measurement. But the abstract idea of a cube, this is completely the same length as this. And this, and actually this, and all of the edges are going to have the exact same length. So this is going from the concrete, the specific, from the real world, if you consider the 24th or 25th century the real world, to going to the idea behind it, the general idea. And you've probably also heard the word abstract in terms of art, like abstract art. So this is abstract art."}, {"video_title": "Abstract-ness   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And this, and actually this, and all of the edges are going to have the exact same length. So this is going from the concrete, the specific, from the real world, if you consider the 24th or 25th century the real world, to going to the idea behind it, the general idea. And you've probably also heard the word abstract in terms of art, like abstract art. So this is abstract art. And it's the same general idea. If you look it up on a dictionary, you're going to find 20 definitions of the word abstract. But it's all essentially trying to say the same thing."}, {"video_title": "Abstract-ness   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "So this is abstract art. And it's the same general idea. If you look it up on a dictionary, you're going to find 20 definitions of the word abstract. But it's all essentially trying to say the same thing. Abstract art is art that is not focused on trying to paint reality the exact way that reality exists. If you look at a lot of Renaissance art, they're skilled at painting figures exactly how they look in the real world. But the abstract artists, sometimes they're not even trying to even represent anything that's in the real world."}, {"video_title": "Abstract-ness   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "But it's all essentially trying to say the same thing. Abstract art is art that is not focused on trying to paint reality the exact way that reality exists. If you look at a lot of Renaissance art, they're skilled at painting figures exactly how they look in the real world. But the abstract artists, sometimes they're not even trying to even represent anything that's in the real world. They're trying to represent a raw idea, or they're trying to represent a raw expression of color, and form, and texture. And this is a Jackson Pollock painting right over here. And I pronounced many, many things wrong."}, {"video_title": "Abstract-ness   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "But the abstract artists, sometimes they're not even trying to even represent anything that's in the real world. They're trying to represent a raw idea, or they're trying to represent a raw expression of color, and form, and texture. And this is a Jackson Pollock painting right over here. And I pronounced many, many things wrong. This was taken by our own Stephen Zucker, art historian. And as you can see, it's not clear. Jackson Pollock is not trying to paint a dog, or a horse, or anything like that."}, {"video_title": "Abstract-ness   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And I pronounced many, many things wrong. This was taken by our own Stephen Zucker, art historian. And as you can see, it's not clear. Jackson Pollock is not trying to paint a dog, or a horse, or anything like that. He is painting something that is devoid. It's completely independent of anything that we actually see in physical reality. And the word abstraction, it doesn't just apply to pure geometry and art."}, {"video_title": "Abstract-ness   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Jackson Pollock is not trying to paint a dog, or a horse, or anything like that. He is painting something that is devoid. It's completely independent of anything that we actually see in physical reality. And the word abstraction, it doesn't just apply to pure geometry and art. It applies to almost everything that we do on a daily basis. When we even talk about things, when we even use words, or use symbols, we're essentially abstracting away. We're abstracting the essence of something that actually exists in physical reality."}, {"video_title": "Abstract-ness   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And the word abstraction, it doesn't just apply to pure geometry and art. It applies to almost everything that we do on a daily basis. When we even talk about things, when we even use words, or use symbols, we're essentially abstracting away. We're abstracting the essence of something that actually exists in physical reality. So if I use the word dog, it's a set of symbols. This set of symbols represents something in our minds that we associate with dog. We have in our minds kind of the qualities of what a dog actually is."}, {"video_title": "Abstract-ness   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "We're abstracting the essence of something that actually exists in physical reality. So if I use the word dog, it's a set of symbols. This set of symbols represents something in our minds that we associate with dog. We have in our minds kind of the qualities of what a dog actually is. It has four legs, and floppy ears, and you enjoy petting it, and they're man's, or I guess people's, best friend. You imagine this thing called a dog. And it has the essence of dogness, even though when you actually look at dogs in the real world, they look very, very, very different types of animals, whether you're looking at a Great Dane or a super small poodle."}, {"video_title": "Abstract-ness   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "We have in our minds kind of the qualities of what a dog actually is. It has four legs, and floppy ears, and you enjoy petting it, and they're man's, or I guess people's, best friend. You imagine this thing called a dog. And it has the essence of dogness, even though when you actually look at dogs in the real world, they look very, very, very different types of animals, whether you're looking at a Great Dane or a super small poodle. But we recognize there's an essence of those particulars that we can abstract away and say this is a dog. And then we abstract it even more by representing these letter symbols that tend to conjure up this image, even when we write something as simple as a number. So if I write the number 5, we use it so frequently that we just, to us, a number 5 seems kind of like a concrete thing."}, {"video_title": "Abstract-ness   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And it has the essence of dogness, even though when you actually look at dogs in the real world, they look very, very, very different types of animals, whether you're looking at a Great Dane or a super small poodle. But we recognize there's an essence of those particulars that we can abstract away and say this is a dog. And then we abstract it even more by representing these letter symbols that tend to conjure up this image, even when we write something as simple as a number. So if I write the number 5, we use it so frequently that we just, to us, a number 5 seems kind of like a concrete thing. But it's so abstract. It's just a quantity of things. I could have symbolized it like that."}, {"video_title": "Abstract-ness   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "So if I write the number 5, we use it so frequently that we just, to us, a number 5 seems kind of like a concrete thing. But it's so abstract. It's just a quantity of things. I could have symbolized it like that. I could have symbolized 5 like that. I could have symbolized 5 in Roman numerals like that. I can symbolize it like that."}, {"video_title": "Abstract-ness   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "I could have symbolized it like that. I could have symbolized 5 like that. I could have symbolized 5 in Roman numerals like that. I can symbolize it like that. And in all of these cases, it's the idea of a quantity of five things. You could say point me to a 5. Someone would probably point you to something like that."}, {"video_title": "Abstract-ness   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "I can symbolize it like that. And in all of these cases, it's the idea of a quantity of five things. You could say point me to a 5. Someone would probably point you to something like that. But they're still pointing you to the symbol of 5. But it's still a very, very abstract idea. So hopefully this gives you an appreciation for what abstract means."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "In this video, I want to do a bunch of examples of factoring a second degree polynomial, which is often called a quadratic. Sometimes a quadratic polynomial, or just a quadratic itself, or quadratic expression. But all it means is a second degree polynomial. So something that's going to have a variable raised to the second power. In this case, and all of the examples will do, it'll be x. So let's say I have the quadratic expression, x squared plus 10x plus 9. And I want to factor it into the product of two binomials."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "So something that's going to have a variable raised to the second power. In this case, and all of the examples will do, it'll be x. So let's say I have the quadratic expression, x squared plus 10x plus 9. And I want to factor it into the product of two binomials. How do we do that? Well, let's just think about what happens if we were to take x plus a and multiply that by x plus b. If we were to multiply these two things, what happens?"}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "And I want to factor it into the product of two binomials. How do we do that? Well, let's just think about what happens if we were to take x plus a and multiply that by x plus b. If we were to multiply these two things, what happens? Well, we have a little bit of experience doing this. This will be x times x, which is x squared, plus x times b, which is bx, plus a times x, plus a times b, plus ab. Or if we want to add these two in the middle right here, because they have the same, they're both coefficients of x, we could write this as x squared plus, I could write it as b plus a, or a plus bx, plus ab."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "If we were to multiply these two things, what happens? Well, we have a little bit of experience doing this. This will be x times x, which is x squared, plus x times b, which is bx, plus a times x, plus a times b, plus ab. Or if we want to add these two in the middle right here, because they have the same, they're both coefficients of x, we could write this as x squared plus, I could write it as b plus a, or a plus bx, plus ab. So in general, if we assume that this is the product of two binomials, we see that this middle coefficient on the x term, or you could say the first degree coefficient there, that's going to be the sum of our a and b. This is going to be the sum of our a and b. And then the constant term is going to be the product of our a and b."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "Or if we want to add these two in the middle right here, because they have the same, they're both coefficients of x, we could write this as x squared plus, I could write it as b plus a, or a plus bx, plus ab. So in general, if we assume that this is the product of two binomials, we see that this middle coefficient on the x term, or you could say the first degree coefficient there, that's going to be the sum of our a and b. This is going to be the sum of our a and b. And then the constant term is going to be the product of our a and b. Notice, this would map to this, and this would map to this. And of course, this is the same thing as this. So can we somehow pattern match this to that?"}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "And then the constant term is going to be the product of our a and b. Notice, this would map to this, and this would map to this. And of course, this is the same thing as this. So can we somehow pattern match this to that? Is there some a and b where a plus b is equal to 10, and a times b is equal to 9? Well, let's just think about it a little bit. What are the factors of 9?"}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "So can we somehow pattern match this to that? Is there some a and b where a plus b is equal to 10, and a times b is equal to 9? Well, let's just think about it a little bit. What are the factors of 9? What are the things that a and b could be equal to? And we're assuming that everything is an integer, and normally when we're factoring, especially when we're beginning to factor, we're dealing with integer numbers. So what are the factors of 9?"}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "What are the factors of 9? What are the things that a and b could be equal to? And we're assuming that everything is an integer, and normally when we're factoring, especially when we're beginning to factor, we're dealing with integer numbers. So what are the factors of 9? They're 1, 3, and 9. So this could be a 3 and a 3, or it could be a 1 and a 9. Now, if it's a 3 and a 3, then you'll have 3 plus 3."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "So what are the factors of 9? They're 1, 3, and 9. So this could be a 3 and a 3, or it could be a 1 and a 9. Now, if it's a 3 and a 3, then you'll have 3 plus 3. That doesn't equal 10. But if it's a 1 and a 9, 1 times 9 is 9. 1 plus 9 is 10."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "Now, if it's a 3 and a 3, then you'll have 3 plus 3. That doesn't equal 10. But if it's a 1 and a 9, 1 times 9 is 9. 1 plus 9 is 10. So it does work. So a could be equal to 1, and b could be equal to 9. So we could factor this as being x plus 1 times x plus 9."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "1 plus 9 is 10. So it does work. So a could be equal to 1, and b could be equal to 9. So we could factor this as being x plus 1 times x plus 9. And if you multiply these two out using the skills we developed in the last few videos, you'll see that it is indeed x squared plus 10x plus 9. So when you see something like this, when the coefficient on the x squared term, or the leading coefficient on this quadratic, is a 1, you can just say, all right, what two numbers add up to this coefficient right here? And what two numbers add up?"}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "So we could factor this as being x plus 1 times x plus 9. And if you multiply these two out using the skills we developed in the last few videos, you'll see that it is indeed x squared plus 10x plus 9. So when you see something like this, when the coefficient on the x squared term, or the leading coefficient on this quadratic, is a 1, you can just say, all right, what two numbers add up to this coefficient right here? And what two numbers add up? And those same two numbers, when you take their product, have to be equal to 9. And of course, this has to be in standard form. Or if it's not in standard form, you should put it in that form so that you can always say, OK, whatever's on the first degree coefficient, my a and b have to add to that."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "And what two numbers add up? And those same two numbers, when you take their product, have to be equal to 9. And of course, this has to be in standard form. Or if it's not in standard form, you should put it in that form so that you can always say, OK, whatever's on the first degree coefficient, my a and b have to add to that. Whatever's my constant term, my a times b, the product has to be that. Let's do several more examples. And I think the more examples we do, the more sense this will make."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "Or if it's not in standard form, you should put it in that form so that you can always say, OK, whatever's on the first degree coefficient, my a and b have to add to that. Whatever's my constant term, my a times b, the product has to be that. Let's do several more examples. And I think the more examples we do, the more sense this will make. Let's say we had x squared plus 10x plus, well, I already did 10x. Let's do a different number. x squared plus 15x plus 50."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "And I think the more examples we do, the more sense this will make. Let's say we had x squared plus 10x plus, well, I already did 10x. Let's do a different number. x squared plus 15x plus 50. We want to factor this. Well, same drill. We have an x squared term, x squared term."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "x squared plus 15x plus 50. We want to factor this. Well, same drill. We have an x squared term, x squared term. We have a first degree term. This should be, this right here, should be the sum of two numbers, and then this term, the constant term right here, should be the product of two numbers. So we need to think of two numbers that when I multiply them, I get 50, and when I add them, I get 15."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "We have an x squared term, x squared term. We have a first degree term. This should be, this right here, should be the sum of two numbers, and then this term, the constant term right here, should be the product of two numbers. So we need to think of two numbers that when I multiply them, I get 50, and when I add them, I get 15. And this is going to be a bit of an art that you're going to develop, but the more practice you do, you're going to see that it'll start to come naturally. So what could a and b be? Let's think about the factors of 50."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "So we need to think of two numbers that when I multiply them, I get 50, and when I add them, I get 15. And this is going to be a bit of an art that you're going to develop, but the more practice you do, you're going to see that it'll start to come naturally. So what could a and b be? Let's think about the factors of 50. It could be 1 times 50, 2 times 25. See, 4 doesn't go into 50. It could be 5 times 10."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "Let's think about the factors of 50. It could be 1 times 50, 2 times 25. See, 4 doesn't go into 50. It could be 5 times 10. I think that's all of them. Let's try out these numbers and see if any of these add up to 15. So 1 plus 50 does not add up to 15."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "It could be 5 times 10. I think that's all of them. Let's try out these numbers and see if any of these add up to 15. So 1 plus 50 does not add up to 15. 2 plus 25 does not add up to 15. But 5 plus 10 does add up to 15. So this could be 5 plus 10, and this could be 5 times 10."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "So 1 plus 50 does not add up to 15. 2 plus 25 does not add up to 15. But 5 plus 10 does add up to 15. So this could be 5 plus 10, and this could be 5 times 10. So if we were to factor this, this would be equal to x plus 5 times x plus 10. And multiply it out. I encourage you to multiply this out and see that this is indeed x squared plus 15x plus 10."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "So this could be 5 plus 10, and this could be 5 times 10. So if we were to factor this, this would be equal to x plus 5 times x plus 10. And multiply it out. I encourage you to multiply this out and see that this is indeed x squared plus 15x plus 10. In fact, let's do it. x times x, x squared. x times 10 plus 10x."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "I encourage you to multiply this out and see that this is indeed x squared plus 15x plus 10. In fact, let's do it. x times x, x squared. x times 10 plus 10x. 5 times x plus 5x. 5 times 10 plus 50. Notice the 5 times 10 gave us the 50."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "x times 10 plus 10x. 5 times x plus 5x. 5 times 10 plus 50. Notice the 5 times 10 gave us the 50. The 5x plus the 10x is giving us the 15x in between. So it's x squared plus 15x plus 50. x squared plus 15x plus 50. Let's up the stakes a little bit, introduce some negative signs in here."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "Notice the 5 times 10 gave us the 50. The 5x plus the 10x is giving us the 15x in between. So it's x squared plus 15x plus 50. x squared plus 15x plus 50. Let's up the stakes a little bit, introduce some negative signs in here. Let's say I had x squared minus 11x plus 24. Now, it's the exact same principle. I need to think of two numbers that when I add them need to be equal to negative 11. a plus b need to be equal to negative 11."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "Let's up the stakes a little bit, introduce some negative signs in here. Let's say I had x squared minus 11x plus 24. Now, it's the exact same principle. I need to think of two numbers that when I add them need to be equal to negative 11. a plus b need to be equal to negative 11. And a times b need to be equal to 24. Now, there's something for you to think about. If when I multiply both of these numbers, I'm getting a positive number."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "I need to think of two numbers that when I add them need to be equal to negative 11. a plus b need to be equal to negative 11. And a times b need to be equal to 24. Now, there's something for you to think about. If when I multiply both of these numbers, I'm getting a positive number. I'm getting a 24. That means that both of these need to be positive or both of these need to be negative. That's the only way I'm going to get a positive number here."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "If when I multiply both of these numbers, I'm getting a positive number. I'm getting a 24. That means that both of these need to be positive or both of these need to be negative. That's the only way I'm going to get a positive number here. Now, if when I add them I get a negative number, if these were positive, there's no way I can add two positive numbers and get a negative number. So the fact that their sum is negative and the fact that their product is positive tells me that both a and b are negative. a and b have to be negative."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "That's the only way I'm going to get a positive number here. Now, if when I add them I get a negative number, if these were positive, there's no way I can add two positive numbers and get a negative number. So the fact that their sum is negative and the fact that their product is positive tells me that both a and b are negative. a and b have to be negative. Remember, one can't be negative and the other one can't be positive because then the product would be negative. And they both can't be positive because then this, when you add them, it would get you a positive number. So let's just think about what a and b can be."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "a and b have to be negative. Remember, one can't be negative and the other one can't be positive because then the product would be negative. And they both can't be positive because then this, when you add them, it would get you a positive number. So let's just think about what a and b can be. So two negative numbers. So let's think about the factors of 24. And we'll kind of have to think of the negative factors."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "So let's just think about what a and b can be. So two negative numbers. So let's think about the factors of 24. And we'll kind of have to think of the negative factors. But let me see, it could be 1 times 24, 2 times 11, 3 times 8, or 4 times 6. Now, which of these, when I multiply these, well, obviously when I multiply 1 times 24 I get 24. When I get 2 times 12 I get 24."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "And we'll kind of have to think of the negative factors. But let me see, it could be 1 times 24, 2 times 11, 3 times 8, or 4 times 6. Now, which of these, when I multiply these, well, obviously when I multiply 1 times 24 I get 24. When I get 2 times 12 I get 24. So we know that all of these, the products are 24. But which two of these, which two factors, when I add them, should I get 11? And then we could say, let's take the negative of both of those."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "When I get 2 times 12 I get 24. So we know that all of these, the products are 24. But which two of these, which two factors, when I add them, should I get 11? And then we could say, let's take the negative of both of those. So when you look at these, 3 and 8 jump out. 3 times 8 is equal to 24. 3 plus 8 is equal to 11."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "And then we could say, let's take the negative of both of those. So when you look at these, 3 and 8 jump out. 3 times 8 is equal to 24. 3 plus 8 is equal to 11. But that doesn't quite work out, right? Because we have a negative 11 here. But what if we did negative 3 and negative 8?"}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "3 plus 8 is equal to 11. But that doesn't quite work out, right? Because we have a negative 11 here. But what if we did negative 3 and negative 8? Negative 3 times negative 8 is equal to positive 24. Negative 3 minus 11, or sorry, negative 3 plus negative 8 is equal to negative 11. So negative 3 and negative 8 work."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "But what if we did negative 3 and negative 8? Negative 3 times negative 8 is equal to positive 24. Negative 3 minus 11, or sorry, negative 3 plus negative 8 is equal to negative 11. So negative 3 and negative 8 work. So if we factor this, this is going to x squared minus 11x plus 24 is going to be equal to x minus 3 times x minus 8. Let's do another one like that. Actually, let's mix it up a little bit."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "So negative 3 and negative 8 work. So if we factor this, this is going to x squared minus 11x plus 24 is going to be equal to x minus 3 times x minus 8. Let's do another one like that. Actually, let's mix it up a little bit. Let's say I had x squared plus 5x minus 14. So here we have a different situation. The product of my two numbers is negative."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "Actually, let's mix it up a little bit. Let's say I had x squared plus 5x minus 14. So here we have a different situation. The product of my two numbers is negative. a times b is equal to negative 14. My product is negative. That tells me that one of them is positive and one of them is negative."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "The product of my two numbers is negative. a times b is equal to negative 14. My product is negative. That tells me that one of them is positive and one of them is negative. And when I add the two, a plus b, I get it being equal to 5. So let's think about the factors of 14 and what combinations of them, when I add them, if one is positive and one is negative, or I'm really kind of taking the difference of the two, do I get 5? So if I take 1 and 14, I'm just going to try out things."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "That tells me that one of them is positive and one of them is negative. And when I add the two, a plus b, I get it being equal to 5. So let's think about the factors of 14 and what combinations of them, when I add them, if one is positive and one is negative, or I'm really kind of taking the difference of the two, do I get 5? So if I take 1 and 14, I'm just going to try out things. Negative 1 plus 14 is negative 13. So let me write all of the combinations that I could do. And eventually your brain will just zone in on it."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "So if I take 1 and 14, I'm just going to try out things. Negative 1 plus 14 is negative 13. So let me write all of the combinations that I could do. And eventually your brain will just zone in on it. So you could have negative 1 plus 14 is equal to 13. And 1 plus negative 14 is equal to negative 13. So those don't work."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "And eventually your brain will just zone in on it. So you could have negative 1 plus 14 is equal to 13. And 1 plus negative 14 is equal to negative 13. So those don't work. That doesn't equal 5. Now what about 2 and 7? If I do negative 2 plus 7, that is equal to 5."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "So those don't work. That doesn't equal 5. Now what about 2 and 7? If I do negative 2 plus 7, that is equal to 5. We're done. That worked. I mean, we could have tried 2 plus negative 7, but that would have equaled negative 5."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "If I do negative 2 plus 7, that is equal to 5. We're done. That worked. I mean, we could have tried 2 plus negative 7, but that would have equaled negative 5. So that wouldn't have worked. But negative 2 plus 7 works. And negative 2 times 7 is negative 14."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "I mean, we could have tried 2 plus negative 7, but that would have equaled negative 5. So that wouldn't have worked. But negative 2 plus 7 works. And negative 2 times 7 is negative 14. So there we have it. We know it's x minus 2 times x plus 7. That's pretty neat."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "And negative 2 times 7 is negative 14. So there we have it. We know it's x minus 2 times x plus 7. That's pretty neat. Negative 2 times 7 is negative 14. Negative 2 plus 7 is positive 5. Let's do several more of these, just to really get well honed the skill."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "That's pretty neat. Negative 2 times 7 is negative 14. Negative 2 plus 7 is positive 5. Let's do several more of these, just to really get well honed the skill. So let's say we have x squared minus x minus 56. So the product of the two numbers have to be minus 56, have to be negative 56. And their difference, because 1 is going to be positive and 1 is going to be negative, their difference has to be negative 1."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "Let's do several more of these, just to really get well honed the skill. So let's say we have x squared minus x minus 56. So the product of the two numbers have to be minus 56, have to be negative 56. And their difference, because 1 is going to be positive and 1 is going to be negative, their difference has to be negative 1. And the numbers that immediately jump out in my brain, and I don't know if they jump out in your brain, we just learned this in the times tables, 56 is 8 times 7. I mean, there's other numbers. It's also 28 times 2."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "And their difference, because 1 is going to be positive and 1 is going to be negative, their difference has to be negative 1. And the numbers that immediately jump out in my brain, and I don't know if they jump out in your brain, we just learned this in the times tables, 56 is 8 times 7. I mean, there's other numbers. It's also 28 times 2. It's all sorts of things. But 8 times 7 really jumped out into my brain because they're very close to each other. We need numbers that are very close to each other."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "It's also 28 times 2. It's all sorts of things. But 8 times 7 really jumped out into my brain because they're very close to each other. We need numbers that are very close to each other. And one of these has to be positive and one of these has to be negative. Now, the fact that when their sum is negative tells me that the larger of these two should probably be negative. So if we take negative 8 times 7, that's equal to negative 56, and then if we take negative 8 plus 7, that is equal to negative 1, which is exactly the coefficient right there."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "We need numbers that are very close to each other. And one of these has to be positive and one of these has to be negative. Now, the fact that when their sum is negative tells me that the larger of these two should probably be negative. So if we take negative 8 times 7, that's equal to negative 56, and then if we take negative 8 plus 7, that is equal to negative 1, which is exactly the coefficient right there. So when I factor this, this is going to be x minus 8 times x plus 7. This is often one of the hardest concepts people learn in algebra because it is a bit of an art. You have to look at all of the factors here, play with the positive and negative signs, see which of those factors, when 1 is positive and 1 is negative, add up to the coefficient on the x term."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "So if we take negative 8 times 7, that's equal to negative 56, and then if we take negative 8 plus 7, that is equal to negative 1, which is exactly the coefficient right there. So when I factor this, this is going to be x minus 8 times x plus 7. This is often one of the hardest concepts people learn in algebra because it is a bit of an art. You have to look at all of the factors here, play with the positive and negative signs, see which of those factors, when 1 is positive and 1 is negative, add up to the coefficient on the x term. But as you do more and more practice, you'll see that it'll become a bit of second nature. Now let's step up the stakes a little bit more. Let's say we had negative x squared."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "You have to look at all of the factors here, play with the positive and negative signs, see which of those factors, when 1 is positive and 1 is negative, add up to the coefficient on the x term. But as you do more and more practice, you'll see that it'll become a bit of second nature. Now let's step up the stakes a little bit more. Let's say we had negative x squared. Everything we've done so far had a positive coefficient, a positive 1 coefficient on the x squared term. But let's say we had a negative x squared minus 5x plus 24. How do we do this?"}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "Let's say we had negative x squared. Everything we've done so far had a positive coefficient, a positive 1 coefficient on the x squared term. But let's say we had a negative x squared minus 5x plus 24. How do we do this? Well, the easiest way I can think of doing it is factor out a negative 1, and then it becomes just like the problems we've been doing before. So this is the same thing as negative 1 times positive x squared plus 5x minus 24. I just factored a negative 1 out."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "How do we do this? Well, the easiest way I can think of doing it is factor out a negative 1, and then it becomes just like the problems we've been doing before. So this is the same thing as negative 1 times positive x squared plus 5x minus 24. I just factored a negative 1 out. You can multiply negative 1 times all of these, and you'll see it becomes this. Or you could factor the negative 1 out and divide all of these by negative 1, and you get that right there. Now, same game as before."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "I just factored a negative 1 out. You can multiply negative 1 times all of these, and you'll see it becomes this. Or you could factor the negative 1 out and divide all of these by negative 1, and you get that right there. Now, same game as before. I need two numbers that when I take their product, I get negative 24. So one will be positive, one will be negative. And when I take their sum, it's going to be 5."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "Now, same game as before. I need two numbers that when I take their product, I get negative 24. So one will be positive, one will be negative. And when I take their sum, it's going to be 5. So let's think about 24. Is 1 and 24. Let's see, if this is negative 1 and 24, it's negative 23."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "And when I take their sum, it's going to be 5. So let's think about 24. Is 1 and 24. Let's see, if this is negative 1 and 24, it's negative 23. Otherwise, if it's negative 1 and 24, it would be positive 23. If it was the other way around, it would be negative 23. It doesn't work."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "Let's see, if this is negative 1 and 24, it's negative 23. Otherwise, if it's negative 1 and 24, it would be positive 23. If it was the other way around, it would be negative 23. It doesn't work. What about 2 and 12? Well, if this is negative, if the 2 is negative, remember, one of these have to be negative. If the 2 is negative, their sum would be 10."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "It doesn't work. What about 2 and 12? Well, if this is negative, if the 2 is negative, remember, one of these have to be negative. If the 2 is negative, their sum would be 10. If the 12 is negative, their sum would be negative 10. Still doesn't work. 3 and 8."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "If the 2 is negative, their sum would be 10. If the 12 is negative, their sum would be negative 10. Still doesn't work. 3 and 8. If the 3 is negative, their sum will be 5. So it works. So if we pick negative 3 and 8, negative 3 and 8 work."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "3 and 8. If the 3 is negative, their sum will be 5. So it works. So if we pick negative 3 and 8, negative 3 and 8 work. So if we use it because negative 3 plus 8 is 5, negative 3 times 8 is negative 24. So this is going to be equal to, can't forget that negative 1 out front. And then we factor the inside."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "So if we pick negative 3 and 8, negative 3 and 8 work. So if we use it because negative 3 plus 8 is 5, negative 3 times 8 is negative 24. So this is going to be equal to, can't forget that negative 1 out front. And then we factor the inside. Negative 1 times x minus 3 times x plus 8. And if you really wanted to, you could multiply the negative 1 times this. You would get 3 minus x if you did, or you don't have to."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "And then we factor the inside. Negative 1 times x minus 3 times x plus 8. And if you really wanted to, you could multiply the negative 1 times this. You would get 3 minus x if you did, or you don't have to. Let's do one more of these. The more practice, the better, I think. All right, let's say I had negative x squared plus 18x minus 72."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "You would get 3 minus x if you did, or you don't have to. Let's do one more of these. The more practice, the better, I think. All right, let's say I had negative x squared plus 18x minus 72. So once again, I like to factor out the negative 1. So this is equal to negative 1 times x squared minus 18x plus 72. Now we just have to think of two numbers that when I multiply them, I get positive 72."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "All right, let's say I had negative x squared plus 18x minus 72. So once again, I like to factor out the negative 1. So this is equal to negative 1 times x squared minus 18x plus 72. Now we just have to think of two numbers that when I multiply them, I get positive 72. So they have to be the same sign. And that makes it easier in our head, at least in my head. When I multiply them, I get positive 72."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "Now we just have to think of two numbers that when I multiply them, I get positive 72. So they have to be the same sign. And that makes it easier in our head, at least in my head. When I multiply them, I get positive 72. When I add them, I get negative 18. So if they're the same sign and their sum is a negative number, they both must be negative. So they're both negative."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "When I multiply them, I get positive 72. When I add them, I get negative 18. So if they're the same sign and their sum is a negative number, they both must be negative. So they're both negative. And we could go through all of the factors of 72, but the one that springs up, maybe you think of 8 times 9, but 8 times 9, or negative 8 minus 9, or negative 8 plus negative 9 doesn't work. That turns into 17. That was close."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "So they're both negative. And we could go through all of the factors of 72, but the one that springs up, maybe you think of 8 times 9, but 8 times 9, or negative 8 minus 9, or negative 8 plus negative 9 doesn't work. That turns into 17. That was close. Let me show you that. Negative 9 plus negative 8, that is equal to negative 17. Close, but no cigar."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "That was close. Let me show you that. Negative 9 plus negative 8, that is equal to negative 17. Close, but no cigar. So what other ones are? We have 6 and 12. That actually seems pretty good."}, {"video_title": "More examples of factoring quadratics with a leading coefficient of 1   Algebra II   Khan Academy.mp3", "Sentence": "Close, but no cigar. So what other ones are? We have 6 and 12. That actually seems pretty good. If we have negative 6 plus negative 12, that is equal to negative 18. Notice, it's a bit of an art. You have to try the different factors here."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "And in particular, I want to focus on quadratics that don't have a 1 as a leading coefficient. For example, if I wanted to factor 4x squared plus 25x minus 21. Everything we've factored so far, or all of the quadratics we've factored so far, had either a 1 or a negative 1 where this 4 is sitting. All of a sudden now we have this 4 here. So what I'm going to teach you is a technique called factoring by grouping. And it's a little bit more involved than what we've learned before, but it's a neat trick. But to some degree, it'll become obsolete once you learn the quadratic formula, because frankly, the quadratic formula is a lot easier."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "All of a sudden now we have this 4 here. So what I'm going to teach you is a technique called factoring by grouping. And it's a little bit more involved than what we've learned before, but it's a neat trick. But to some degree, it'll become obsolete once you learn the quadratic formula, because frankly, the quadratic formula is a lot easier. But this is how it goes. I'll show you the technique, and then at the end of this video, I'll actually show you why it works. So what we need to do here is we need to think of two numbers."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "But to some degree, it'll become obsolete once you learn the quadratic formula, because frankly, the quadratic formula is a lot easier. But this is how it goes. I'll show you the technique, and then at the end of this video, I'll actually show you why it works. So what we need to do here is we need to think of two numbers. We're going to think of two numbers, a and b, where a times b is equal to 4 times negative 21. So a times b is going to be equal to 4 times negative 21, which is equal to negative 84. And those same two numbers, a and b, a plus b, need to be equal to 25."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "So what we need to do here is we need to think of two numbers. We're going to think of two numbers, a and b, where a times b is equal to 4 times negative 21. So a times b is going to be equal to 4 times negative 21, which is equal to negative 84. And those same two numbers, a and b, a plus b, need to be equal to 25. They need to be equal to 25. Let me be very clear. This is the 25, so they need to be equal to 25."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "And those same two numbers, a and b, a plus b, need to be equal to 25. They need to be equal to 25. Let me be very clear. This is the 25, so they need to be equal to 25. This is where the 4 is, so they need to be equal to 4 times negative 21. That's a negative 21. So what two numbers are there that would do this?"}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "This is the 25, so they need to be equal to 25. This is where the 4 is, so they need to be equal to 4 times negative 21. That's a negative 21. So what two numbers are there that would do this? Well, we have to look at the factors of negative 84. And once again, one of these are going to have to be positive. The other ones are going to have to be negative, because their product is negative."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "So what two numbers are there that would do this? Well, we have to look at the factors of negative 84. And once again, one of these are going to have to be positive. The other ones are going to have to be negative, because their product is negative. So let's think about the different factors that might work. 4 and negative 21 look tantalizing, but when you add them, you get negative 17. Or if you had negative 4 and 21, you'd get positive 17."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "The other ones are going to have to be negative, because their product is negative. So let's think about the different factors that might work. 4 and negative 21 look tantalizing, but when you add them, you get negative 17. Or if you had negative 4 and 21, you'd get positive 17. It doesn't work. Let's try some other combinations. 1 and 84, too far apart when you take their difference, because that's essentially what you're going to do if one is negative and one is positive."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "Or if you had negative 4 and 21, you'd get positive 17. It doesn't work. Let's try some other combinations. 1 and 84, too far apart when you take their difference, because that's essentially what you're going to do if one is negative and one is positive. Too far apart. Let's see. You could do 2 and 42."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "1 and 84, too far apart when you take their difference, because that's essentially what you're going to do if one is negative and one is positive. Too far apart. Let's see. You could do 2 and 42. Once again, too far apart. Negative 2 plus 42 is 40. 2 plus negative 2 is negative 40."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "You could do 2 and 42. Once again, too far apart. Negative 2 plus 42 is 40. 2 plus negative 2 is negative 40. Too far apart. 3 goes into 84. 3 goes into 8."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "2 plus negative 2 is negative 40. Too far apart. 3 goes into 84. 3 goes into 8. 2 times 3 is 6. 8 minus 6 is 2. Bring down the 4."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "3 goes into 8. 2 times 3 is 6. 8 minus 6 is 2. Bring down the 4. It goes exactly 8 times. So 3 and 28. This seems interesting."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "Bring down the 4. It goes exactly 8 times. So 3 and 28. This seems interesting. 3 and 28. And remember, one of these has to be negative. So if we have negative 3 plus 28, that is equal to 25."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "This seems interesting. 3 and 28. And remember, one of these has to be negative. So if we have negative 3 plus 28, that is equal to 25. Now, we found our two numbers, but it's not going to be quite as simple of an operation as what we did when this wasn't a 1 or when this was a 1 or a negative 1. What we're going to do now is split up this term right here. We're going to split this up into negative... We're going to split it up into positive 28x minus 3x."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "So if we have negative 3 plus 28, that is equal to 25. Now, we found our two numbers, but it's not going to be quite as simple of an operation as what we did when this wasn't a 1 or when this was a 1 or a negative 1. What we're going to do now is split up this term right here. We're going to split this up into negative... We're going to split it up into positive 28x minus 3x. We're just going to split that term. That term is that term right there. And of course, you have your minus 21 there."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "We're going to split this up into negative... We're going to split it up into positive 28x minus 3x. We're just going to split that term. That term is that term right there. And of course, you have your minus 21 there. And you have your 4x squared over here. Now, you might say, how did you pick the 28 to go here and the negative 3 to go there? And it actually does matter."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "And of course, you have your minus 21 there. And you have your 4x squared over here. Now, you might say, how did you pick the 28 to go here and the negative 3 to go there? And it actually does matter. The way I thought about it is 3 or negative 3 and 21 or negative 21, they have some common factors in particular. They have the factor 3 in common. And 28 and 4 have some common factors."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "And it actually does matter. The way I thought about it is 3 or negative 3 and 21 or negative 21, they have some common factors in particular. They have the factor 3 in common. And 28 and 4 have some common factors. So I kind of grouped the 28 on the side of the 4. And you're going to see what I mean in a second. If we literally group these, so that term becomes 4x squared plus 28x."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "And 28 and 4 have some common factors. So I kind of grouped the 28 on the side of the 4. And you're going to see what I mean in a second. If we literally group these, so that term becomes 4x squared plus 28x. And then this side over here in pink, well, I could say it's plus negative 3x minus 21. Once again, I picked these. I grouped the negative 3 with the 21 or the negative 21 because they're both divisible by 3."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "If we literally group these, so that term becomes 4x squared plus 28x. And then this side over here in pink, well, I could say it's plus negative 3x minus 21. Once again, I picked these. I grouped the negative 3 with the 21 or the negative 21 because they're both divisible by 3. And I grouped the 28 with the 4 because they're both divisible by 4. And now, in each of these groups, we factor as much out as we can. So both of these terms are divisible by 4x."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "I grouped the negative 3 with the 21 or the negative 21 because they're both divisible by 3. And I grouped the 28 with the 4 because they're both divisible by 4. And now, in each of these groups, we factor as much out as we can. So both of these terms are divisible by 4x. So this orange term is equal to 4x times x. 4x squared divided by 4x is just x. Plus 28x divided by 4x is just 7."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "So both of these terms are divisible by 4x. So this orange term is equal to 4x times x. 4x squared divided by 4x is just x. Plus 28x divided by 4x is just 7. Now, this second term, remember, you factor out everything that you can factor out. Well, both of these terms are divisible by 3 or negative 3. So let's factor out a negative 3."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "Plus 28x divided by 4x is just 7. Now, this second term, remember, you factor out everything that you can factor out. Well, both of these terms are divisible by 3 or negative 3. So let's factor out a negative 3. And this becomes x plus 7. And now, something might pop out at you. We have x plus 7 times 4x plus x plus 7 times negative 3."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "So let's factor out a negative 3. And this becomes x plus 7. And now, something might pop out at you. We have x plus 7 times 4x plus x plus 7 times negative 3. So we can factor out an x plus 7. We can factor out an x plus 7. This might not be completely obvious."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "We have x plus 7 times 4x plus x plus 7 times negative 3. So we can factor out an x plus 7. We can factor out an x plus 7. This might not be completely obvious. You're probably not used to factoring out an entire binomial, but you could view this as like a. Or if you have 4xa minus 3a, you would be able to factor out an a. And I could just leave this as a minus sign."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "This might not be completely obvious. You're probably not used to factoring out an entire binomial, but you could view this as like a. Or if you have 4xa minus 3a, you would be able to factor out an a. And I could just leave this as a minus sign. Let me delete this plus right here because it's just minus 3. It's just minus 3. Plus negative 3, same thing as minus 3."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "And I could just leave this as a minus sign. Let me delete this plus right here because it's just minus 3. It's just minus 3. Plus negative 3, same thing as minus 3. So what can we do here? We have x plus 7 times 4x. We have an x plus 7 times negative 3."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "Plus negative 3, same thing as minus 3. So what can we do here? We have x plus 7 times 4x. We have an x plus 7 times negative 3. Let's factor out the x plus 7. We get x plus 7 times 4x minus 3. Minus that 3 right there."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "We have an x plus 7 times negative 3. Let's factor out the x plus 7. We get x plus 7 times 4x minus 3. Minus that 3 right there. And we've factored our binomial. Sorry, we've factored our quadratic by grouping, and we factored it into two binomials. Let's do another example of that and it's a little bit involved, but once you get the hang of it, it's kind of fun."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "Minus that 3 right there. And we've factored our binomial. Sorry, we've factored our quadratic by grouping, and we factored it into two binomials. Let's do another example of that and it's a little bit involved, but once you get the hang of it, it's kind of fun. So let's say we want to factor 6x squared plus 7x plus 1. Same drill. We want to find a times b that is equal to 1 times 6, and we want to find an a plus b needs to be equal to 7."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "Let's do another example of that and it's a little bit involved, but once you get the hang of it, it's kind of fun. So let's say we want to factor 6x squared plus 7x plus 1. Same drill. We want to find a times b that is equal to 1 times 6, and we want to find an a plus b needs to be equal to 7. This is a little bit more straightforward. The obvious one is 1 and 6. 1 times 6 is 6."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "We want to find a times b that is equal to 1 times 6, and we want to find an a plus b needs to be equal to 7. This is a little bit more straightforward. The obvious one is 1 and 6. 1 times 6 is 6. 1 plus 6 is 7. So we have a is equal to 1. Let me not even assign them."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "1 times 6 is 6. 1 plus 6 is 7. So we have a is equal to 1. Let me not even assign them. The numbers here are 1 and 6. Now we want to split this into a 1x and a 6x, but we want to group it so it's on the side of something that it shares a factor with. So we're going to have a 6x squared here plus, and so I'm going to put the 6x first because 6 and 6 share a factor."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "Let me not even assign them. The numbers here are 1 and 6. Now we want to split this into a 1x and a 6x, but we want to group it so it's on the side of something that it shares a factor with. So we're going to have a 6x squared here plus, and so I'm going to put the 6x first because 6 and 6 share a factor. Then we're going to have plus 1x. 6x plus 1x is 7x. That was the whole point."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "So we're going to have a 6x squared here plus, and so I'm going to put the 6x first because 6 and 6 share a factor. Then we're going to have plus 1x. 6x plus 1x is 7x. That was the whole point. They had to add up to 7. Then we have the final plus 1 there. Now in each of these groups, we can factor out as much as we like."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "That was the whole point. They had to add up to 7. Then we have the final plus 1 there. Now in each of these groups, we can factor out as much as we like. So in this first group, let's factor out a 6x. So this first group becomes 6x times 6x squared divided by 6x is just an x. 6x divided by 6x is just a 1."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "Now in each of these groups, we can factor out as much as we like. So in this first group, let's factor out a 6x. So this first group becomes 6x times 6x squared divided by 6x is just an x. 6x divided by 6x is just a 1. Then in the second group, we're going to have a plus here, but this second group, we just literally have an x plus 1, or we could even write a 1 times an x plus 1. You can imagine I just factored out a 1, so to speak. Now I have 6x times x plus 1 plus 1 times x plus 1."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "6x divided by 6x is just a 1. Then in the second group, we're going to have a plus here, but this second group, we just literally have an x plus 1, or we could even write a 1 times an x plus 1. You can imagine I just factored out a 1, so to speak. Now I have 6x times x plus 1 plus 1 times x plus 1. Well, I can factor out the x plus 1. If I factor out an x plus 1, that's equal to x plus 1 times 6x plus that 1. I'm just doing the distributive property in reverse."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "Now I have 6x times x plus 1 plus 1 times x plus 1. Well, I can factor out the x plus 1. If I factor out an x plus 1, that's equal to x plus 1 times 6x plus that 1. I'm just doing the distributive property in reverse. So hopefully you didn't find that too bad. Now I'm going to actually explain why this little magical system actually works. Why it actually works."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "I'm just doing the distributive property in reverse. So hopefully you didn't find that too bad. Now I'm going to actually explain why this little magical system actually works. Why it actually works. Let me take an example. Let's say I have, well, I'll do it in very general terms. Let's add ax plus b times cx."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "Why it actually works. Let me take an example. Let's say I have, well, I'll do it in very general terms. Let's add ax plus b times cx. Actually, I don't want to use, well, I'm afraid to use the a's and the b's. I think that'll confuse you because I use a's and b's here, and they won't be the same thing. Let me use completely different letters."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "Let's add ax plus b times cx. Actually, I don't want to use, well, I'm afraid to use the a's and the b's. I think that'll confuse you because I use a's and b's here, and they won't be the same thing. Let me use completely different letters. Let's say I have fx plus g times hx plus, I'll use j instead of i. You'll learn in the future why I don't like using i as a variable. So what is this going to be equal to?"}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "Let me use completely different letters. Let's say I have fx plus g times hx plus, I'll use j instead of i. You'll learn in the future why I don't like using i as a variable. So what is this going to be equal to? Well, it's going to be fx times hx, which is fhx, and then fx times j, so plus fjx. Then we're going to have g times hx, so plus ghx, and then g times j, plus gj. Or if we add these two middle terms, if you add the two middle terms, you have fh times x plus, add these two terms, fj plus ghx, plus gj."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "So what is this going to be equal to? Well, it's going to be fx times hx, which is fhx, and then fx times j, so plus fjx. Then we're going to have g times hx, so plus ghx, and then g times j, plus gj. Or if we add these two middle terms, if you add the two middle terms, you have fh times x plus, add these two terms, fj plus ghx, plus gj. Now, what did I do here? Well, remember, in all of these problems where you have a non-1 or non-negative-1 coefficient, we look for two numbers that add up to this whose product is equal to the product of that times that. Well, here we have two numbers that add up."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "Or if we add these two middle terms, if you add the two middle terms, you have fh times x plus, add these two terms, fj plus ghx, plus gj. Now, what did I do here? Well, remember, in all of these problems where you have a non-1 or non-negative-1 coefficient, we look for two numbers that add up to this whose product is equal to the product of that times that. Well, here we have two numbers that add up. Let's say that a is equal to fj. Let's say that a is equal to fj. That is a, and b is equal to gh."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "Well, here we have two numbers that add up. Let's say that a is equal to fj. Let's say that a is equal to fj. That is a, and b is equal to gh. So a plus b is going to be equal to that middle coefficient. a plus b is going to be equal to that middle coefficient there. And then what is a times b?"}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "That is a, and b is equal to gh. So a plus b is going to be equal to that middle coefficient. a plus b is going to be equal to that middle coefficient there. And then what is a times b? a times b is going to be equal to fj times gh, which we could just reorder these terms. We're just multiplying a bunch of terms, so that could be rewritten as f times h times g times j. These are all the same things."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "And then what is a times b? a times b is going to be equal to fj times gh, which we could just reorder these terms. We're just multiplying a bunch of terms, so that could be rewritten as f times h times g times j. These are all the same things. Well, what is fh times gj? This is equal to fh times gj. Well, this is equal to the first coefficient times the constant term."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "These are all the same things. Well, what is fh times gj? This is equal to fh times gj. Well, this is equal to the first coefficient times the constant term. So if a plus b will be equal to the middle coefficient, then a times b will equal the first coefficient times the constant term. So that's why this whole factoring by grouping even works, or how we're able to figure out what a and b even are. Now I'm going to close up with something slightly different, but just to make sure that you have a well-rounded education in factoring things."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "Well, this is equal to the first coefficient times the constant term. So if a plus b will be equal to the middle coefficient, then a times b will equal the first coefficient times the constant term. So that's why this whole factoring by grouping even works, or how we're able to figure out what a and b even are. Now I'm going to close up with something slightly different, but just to make sure that you have a well-rounded education in factoring things. What I want to do is teach you to factor things a little bit more completely. This is a little bit of an add-on. I was going to make a whole video on this, but I think on some level it might be a little obvious for you."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "Now I'm going to close up with something slightly different, but just to make sure that you have a well-rounded education in factoring things. What I want to do is teach you to factor things a little bit more completely. This is a little bit of an add-on. I was going to make a whole video on this, but I think on some level it might be a little obvious for you. So let's say we had 2... Let me get a good one here. Let's say we had negative x to the third plus 17x squared minus 70. Now, I have 70x."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "I was going to make a whole video on this, but I think on some level it might be a little obvious for you. So let's say we had 2... Let me get a good one here. Let's say we had negative x to the third plus 17x squared minus 70. Now, I have 70x. Now, immediately you say, gee, this isn't even a quadratic. I don't know how to solve something like this. It has an x to the third power."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "Now, I have 70x. Now, immediately you say, gee, this isn't even a quadratic. I don't know how to solve something like this. It has an x to the third power. The first thing you should realize is that every term here is divisible by x. So let's factor out an x, or even better, let's factor out a negative x. So if you factor out a negative x, this is equal to negative x times..."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "It has an x to the third power. The first thing you should realize is that every term here is divisible by x. So let's factor out an x, or even better, let's factor out a negative x. So if you factor out a negative x, this is equal to negative x times... Negative x to the third divided by negative x is x squared. 17x squared divided by negative x is negative 17x. Negative 70x divided by negative x is positive 70."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "So if you factor out a negative x, this is equal to negative x times... Negative x to the third divided by negative x is x squared. 17x squared divided by negative x is negative 17x. Negative 70x divided by negative x is positive 70. These cancel out. And now you have something that might look a little bit familiar. We have just a standard quadratic where the leading coefficient is a 1, so we just have to find 2 numbers whose product is 70 and that add up to negative 17."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "Negative 70x divided by negative x is positive 70. These cancel out. And now you have something that might look a little bit familiar. We have just a standard quadratic where the leading coefficient is a 1, so we just have to find 2 numbers whose product is 70 and that add up to negative 17. And the numbers that immediately jumped into my head are negative 10 and negative 7. You take their product, you get 70, you add them up, you get negative 17. So this part right here is going to be x minus 10 times x minus 7."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "We have just a standard quadratic where the leading coefficient is a 1, so we just have to find 2 numbers whose product is 70 and that add up to negative 17. And the numbers that immediately jumped into my head are negative 10 and negative 7. You take their product, you get 70, you add them up, you get negative 17. So this part right here is going to be x minus 10 times x minus 7. And of course, you have that leading negative x. The general idea here is just see if there's anything you can factor out and then it will get into a form that you might recognize. Hopefully you found this helpful."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "So this part right here is going to be x minus 10 times x minus 7. And of course, you have that leading negative x. The general idea here is just see if there's anything you can factor out and then it will get into a form that you might recognize. Hopefully you found this helpful. Now, I want to reiterate what I showed you at the beginning of this video. I think it's a really cool trick, so to speak, to be able to factor things that have a non-1 or non-negative 1 leading coefficient. But to some degree, you're going to find out easier ways to do this, especially with the quadratic formula, in not too long."}, {"video_title": "Multiplying negative and positive fractions   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "So we just, you might say, I don't see a fraction here. This looks like an integer. You just have to remind yourself that the negative seven can be rewritten as negative seven over one times three over 49. Now we can multiply the numerators. So the numerator is gonna be negative seven times three, and the denominator is going to be one times 49. One times 49. And this is going to be equal to, seven times three is 21, and I have a negative, and one of their signs is negative."}, {"video_title": "Multiplying negative and positive fractions   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "Now we can multiply the numerators. So the numerator is gonna be negative seven times three, and the denominator is going to be one times 49. One times 49. And this is going to be equal to, seven times three is 21, and I have a negative, and one of their signs is negative. So a negative times a positive is gonna be a negative. So this is gonna be negative 21. You could view this as negative seven plus negative seven plus negative seven."}, {"video_title": "Multiplying negative and positive fractions   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "And this is going to be equal to, seven times three is 21, and I have a negative, and one of their signs is negative. So a negative times a positive is gonna be a negative. So this is gonna be negative 21. You could view this as negative seven plus negative seven plus negative seven. And that's going to be over, that's going to be over 49. And this is the correct value, but we can simplify it more because 21 and 49 both share seven as a factor. That's their greatest common factor."}, {"video_title": "Multiplying negative and positive fractions   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "You could view this as negative seven plus negative seven plus negative seven. And that's going to be over, that's going to be over 49. And this is the correct value, but we can simplify it more because 21 and 49 both share seven as a factor. That's their greatest common factor. So let's divide both the numerator and the denominator by seven. Divide the numerator and the denominator by seven. And so this gets us negative three in the numerator, and in the denominator we have seven."}, {"video_title": "Multiplying negative and positive fractions   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "That's their greatest common factor. So let's divide both the numerator and the denominator by seven. Divide the numerator and the denominator by seven. And so this gets us negative three in the numerator, and in the denominator we have seven. So we could view it as negative three over seven, or you could even view it as negative three sevenths. Let's do another one. Let's take five ninths, five ninths times, I'll switch colors more in this one."}, {"video_title": "Multiplying negative and positive fractions   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "And so this gets us negative three in the numerator, and in the denominator we have seven. So we could view it as negative three over seven, or you could even view it as negative three sevenths. Let's do another one. Let's take five ninths, five ninths times, I'll switch colors more in this one. That was a little monotonous going all red there. Five ninths times three over 15. So this is going to be equal to, we multiplied the numerators, so it's going to be five times three, five times three in the numerator."}, {"video_title": "Multiplying negative and positive fractions   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "Let's take five ninths, five ninths times, I'll switch colors more in this one. That was a little monotonous going all red there. Five ninths times three over 15. So this is going to be equal to, we multiplied the numerators, so it's going to be five times three, five times three in the numerator. And the denominator's going to be nine times 15. Nine times 15. Nine times 15."}, {"video_title": "Multiplying negative and positive fractions   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "So this is going to be equal to, we multiplied the numerators, so it's going to be five times three, five times three in the numerator. And the denominator's going to be nine times 15. Nine times 15. Nine times 15. We could multiply them out, but just leaving it like this, you see that there's already common factors in the numerator and the denominator. Both the numerator and the denominator, they're both divisible by five, and they're both divisible by three, which essentially tells us that they're divisible by 15. So we can divide the numerator and the denominator by 15."}, {"video_title": "Multiplying negative and positive fractions   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "Nine times 15. We could multiply them out, but just leaving it like this, you see that there's already common factors in the numerator and the denominator. Both the numerator and the denominator, they're both divisible by five, and they're both divisible by three, which essentially tells us that they're divisible by 15. So we can divide the numerator and the denominator by 15. So divide the numerator by 15, which is just like dividing by five and then dividing by three. So we'll just divide by 15. Divide by 15."}, {"video_title": "Multiplying negative and positive fractions   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "So we can divide the numerator and the denominator by 15. So divide the numerator by 15, which is just like dividing by five and then dividing by three. So we'll just divide by 15. Divide by 15. And this is going to be equal to, well, five times three is 15. Divided by 15, you get one in the numerator. And in the denominator, nine times 15 divided by 15, well, that's just going to be nine."}, {"video_title": "Multiplying negative and positive fractions   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "Divide by 15. And this is going to be equal to, well, five times three is 15. Divided by 15, you get one in the numerator. And in the denominator, nine times 15 divided by 15, well, that's just going to be nine. So it's equal to one ninth. Let's do another one. What would negative five ninths times negative three over 15 be?"}, {"video_title": "Multiplying negative and positive fractions   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "And in the denominator, nine times 15 divided by 15, well, that's just going to be nine. So it's equal to one ninth. Let's do another one. What would negative five ninths times negative three over 15 be? Well, we've already figured out what positive five ninths times positive three over 15 would be. So now we just have to care about the sign. If we were just multiplying the two positives, it would be one over nine."}, {"video_title": "Multiplying negative and positive fractions   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "What would negative five ninths times negative three over 15 be? Well, we've already figured out what positive five ninths times positive three over 15 would be. So now we just have to care about the sign. If we were just multiplying the two positives, it would be one over nine. But now we have to think about the fact that we're multiplying a negative times a negative. Now we remember, when you multiply a negative times a negative, it's a positive. The only way that you get a negative is if one of those two numbers that you're taking the product of is negative, not two."}, {"video_title": "Multiplying negative and positive fractions   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "If we were just multiplying the two positives, it would be one over nine. But now we have to think about the fact that we're multiplying a negative times a negative. Now we remember, when you multiply a negative times a negative, it's a positive. The only way that you get a negative is if one of those two numbers that you're taking the product of is negative, not two. If both are positive, it's positive. If both are negative, it's positive. Let's do one more example."}, {"video_title": "Multiplying negative and positive fractions   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "The only way that you get a negative is if one of those two numbers that you're taking the product of is negative, not two. If both are positive, it's positive. If both are negative, it's positive. Let's do one more example. Let's take five. I'm using the number five a lot. So let's do three over two, just to show that this would work with improper fractions."}, {"video_title": "Multiplying negative and positive fractions   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "Let's do one more example. Let's take five. I'm using the number five a lot. So let's do three over two, just to show that this would work with improper fractions. Three over two times negative negative seven over 10. Negative seven over 10. I'm arbitrarily picking colors."}, {"video_title": "Multiplying negative and positive fractions   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "So let's do three over two, just to show that this would work with improper fractions. Three over two times negative negative seven over 10. Negative seven over 10. I'm arbitrarily picking colors. And so our numerator is gonna be three times negative seven. Three times negative seven. And our denominator is going to be two times 10."}, {"video_title": "Multiplying negative and positive fractions   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "I'm arbitrarily picking colors. And so our numerator is gonna be three times negative seven. Three times negative seven. And our denominator is going to be two times 10. Two times 10. Two times 10. So this is going to be the numerator."}, {"video_title": "Multiplying negative and positive fractions   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "And our denominator is going to be two times 10. Two times 10. Two times 10. So this is going to be the numerator. Positive times a negative is a negative. Three times seven. Negative seven is negative 21."}, {"video_title": "Multiplying negative and positive fractions   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "So this is going to be the numerator. Positive times a negative is a negative. Three times seven. Negative seven is negative 21. Negative 21. And the denominator, two times 10, well that is just 20. So this is negative 21 over 20, and you really can't simplify this any further."}, {"video_title": "Adding and subtracting negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "So right now, I'm just subtracting a positive number from another positive number. But you might already see that I'm subtracting a larger number from a smaller number. So I'm probably, or I will definitely, end up with a negative number. But let's just think about this a little bit. And I'm going to do it with a number line. So there's my number line right over there. Now, this is 0."}, {"video_title": "Adding and subtracting negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "But let's just think about this a little bit. And I'm going to do it with a number line. So there's my number line right over there. Now, this is 0. This is 1. This is 2. This is negative 1."}, {"video_title": "Adding and subtracting negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "Now, this is 0. This is 1. This is 2. This is negative 1. This is negative 2. We could view this as starting at 2. So this is 2 right over here."}, {"video_title": "Adding and subtracting negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "This is negative 1. This is negative 2. We could view this as starting at 2. So this is 2 right over here. And then we're going to subtract 3 from that 2. So we're going to move 3 to the left on the number line. So we're going to move 3 to the left."}, {"video_title": "Adding and subtracting negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "So this is 2 right over here. And then we're going to subtract 3 from that 2. So we're going to move 3 to the left on the number line. So we're going to move 3 to the left. 1, 2, 3. And that gets us to negative 1. This is equal to negative 1."}, {"video_title": "Adding and subtracting negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "So we're going to move 3 to the left. 1, 2, 3. And that gets us to negative 1. This is equal to negative 1. Now, let's mix it up a little bit more. Let's imagine what would happen if we had negative 2. Negative 2 minus 3."}, {"video_title": "Adding and subtracting negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "This is equal to negative 1. Now, let's mix it up a little bit more. Let's imagine what would happen if we had negative 2. Negative 2 minus 3. So this was positive 2 minus 3. Now, let's think about negative 2 minus 3. So once again, let's draw our number line."}, {"video_title": "Adding and subtracting negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "Negative 2 minus 3. So this was positive 2 minus 3. Now, let's think about negative 2 minus 3. So once again, let's draw our number line. And I'll put 0 over here. So this is 0. This is 1."}, {"video_title": "Adding and subtracting negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "So once again, let's draw our number line. And I'll put 0 over here. So this is 0. This is 1. This is negative 1, negative 2, negative 3, negative 4, negative 5, negative 6. And I could keep going. But we're starting at negative 2."}, {"video_title": "Adding and subtracting negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "This is 1. This is negative 1, negative 2, negative 3, negative 4, negative 5, negative 6. And I could keep going. But we're starting at negative 2. And then we're subtracting 3 again. So once again, we're going to move 3 to the left of negative 2. So we go 1, 2, 3."}, {"video_title": "Adding and subtracting negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "But we're starting at negative 2. And then we're subtracting 3 again. So once again, we're going to move 3 to the left of negative 2. So we go 1, 2, 3. We end up at negative 5. So this is negative 5. So notice, in both situations, we subtracted 3."}, {"video_title": "Adding and subtracting negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "So we go 1, 2, 3. We end up at negative 5. So this is negative 5. So notice, in both situations, we subtracted 3. We moved 3 to the left on the number line. It's just here we started 2 to the right of 0. Here we started 2 to the left of 0."}, {"video_title": "Adding and subtracting negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "So notice, in both situations, we subtracted 3. We moved 3 to the left on the number line. It's just here we started 2 to the right of 0. Here we started 2 to the left of 0. This is negative 2. Let's do another example with these same numbers. Let's imagine negative 2 plus 3."}, {"video_title": "Adding and subtracting negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "Here we started 2 to the left of 0. This is negative 2. Let's do another example with these same numbers. Let's imagine negative 2 plus 3. I encourage you to pause this video and try to think about this on your own. So we draw the number line again. I could draw a straighter number line than that."}, {"video_title": "Adding and subtracting negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "Let's imagine negative 2 plus 3. I encourage you to pause this video and try to think about this on your own. So we draw the number line again. I could draw a straighter number line than that. So draw the number line again. And let's say that this is negative 2, negative 1, 0, 1, and 2 again. We're starting at negative 2."}, {"video_title": "Adding and subtracting negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "I could draw a straighter number line than that. So draw the number line again. And let's say that this is negative 2, negative 1, 0, 1, and 2 again. We're starting at negative 2. We're starting 2 to the left of 0. So we're starting at negative 2. And we're going to add 3."}, {"video_title": "Adding and subtracting negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "We're starting at negative 2. We're starting 2 to the left of 0. So we're starting at negative 2. And we're going to add 3. So we're going to go 3 to the right now. 1, 2, 3. And we end up at positive 1."}, {"video_title": "Adding and subtracting negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "And we're going to add 3. So we're going to go 3 to the right now. 1, 2, 3. And we end up at positive 1. Now let's think about 2. So positive 2. And we're going to subtract a negative 3."}, {"video_title": "Adding and subtracting negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "And we end up at positive 1. Now let's think about 2. So positive 2. And we're going to subtract a negative 3. So we're going to subtract a negative 3. In other videos, we've already talked about this. In fact, there's a video explaining why this actually makes sense."}, {"video_title": "Adding and subtracting negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "And we're going to subtract a negative 3. So we're going to subtract a negative 3. In other videos, we've already talked about this. In fact, there's a video explaining why this actually makes sense. But when you subtract a negative, this is the same thing as adding the positive. So 2 minus negative 3 is the exact same thing as 2 plus positive 3. These two statements are equivalent."}, {"video_title": "Adding and subtracting negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "In fact, there's a video explaining why this actually makes sense. But when you subtract a negative, this is the same thing as adding the positive. So 2 minus negative 3 is the exact same thing as 2 plus positive 3. These two statements are equivalent. And this just boils down to this right over here is just going to be 5. Now let's mix it up a little bit more. Let's imagine negative 2 minus negative 3."}, {"video_title": "Adding and subtracting negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "These two statements are equivalent. And this just boils down to this right over here is just going to be 5. Now let's mix it up a little bit more. Let's imagine negative 2 minus negative 3. Now this might seem really intimidating. I have all of these negatives in place here. But you just have to remember, subtracting a negative like this is going to get you a positive."}, {"video_title": "Adding and subtracting negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "Let's imagine negative 2 minus negative 3. Now this might seem really intimidating. I have all of these negatives in place here. But you just have to remember, subtracting a negative like this is going to get you a positive. So this is the exact same thing as negative 2 plus 3. And negative 2 plus 3, we've already seen it right over here. You start at negative 2."}, {"video_title": "Adding and subtracting negative numbers   Pre-Algebra   Khan Academy.mp3", "Sentence": "But you just have to remember, subtracting a negative like this is going to get you a positive. So this is the exact same thing as negative 2 plus 3. And negative 2 plus 3, we've already seen it right over here. You start at negative 2. You start 2 to the left of 0. And then we're going to go 3 to the right. We're adding 3."}, {"video_title": "Checking ordered pair solutions to equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "Four x minus one is equal to three y plus five. Now when we look at an ordered pair and we want to figure out whether it's a solution, we just have to remind ourselves that in these ordered pairs, the convention, the standard is, is that the first coordinate is the x coordinate and the second coordinate is the y coordinate. So they're gonna, if this is a solution, if this ordered pair is a solution, that means that if x is equal to three and y is equal to two, that that would satisfy this equation up here. So let's try that out. So we have four times x. Well, we're saying x needs to be equal to three minus one is going to be equal to three times y. Well, if this ordered pair is a solution, then y is going to be equal to two."}, {"video_title": "Checking ordered pair solutions to equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "So let's try that out. So we have four times x. Well, we're saying x needs to be equal to three minus one is going to be equal to three times y. Well, if this ordered pair is a solution, then y is going to be equal to two. So three times y, y is two, plus five. Notice all I did is wherever I saw the x, I substituted it with three, wherever I saw the y, I substituted it with two. Now let's see if this is true."}, {"video_title": "Checking ordered pair solutions to equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "Well, if this ordered pair is a solution, then y is going to be equal to two. So three times y, y is two, plus five. Notice all I did is wherever I saw the x, I substituted it with three, wherever I saw the y, I substituted it with two. Now let's see if this is true. Four times three is 12 minus one. Is this really the same thing as three times two, which is six plus five? See, 12 minus one is 11."}, {"video_title": "Checking ordered pair solutions to equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "Now let's see if this is true. Four times three is 12 minus one. Is this really the same thing as three times two, which is six plus five? See, 12 minus one is 11. Six plus five is also 11. This is true. 11 equals 11."}, {"video_title": "Checking ordered pair solutions to equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "See, 12 minus one is 11. Six plus five is also 11. This is true. 11 equals 11. This pair, three comma two, does satisfy this equation. Now let's see whether this one does. Two comma three."}, {"video_title": "Checking ordered pair solutions to equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "11 equals 11. This pair, three comma two, does satisfy this equation. Now let's see whether this one does. Two comma three. So this is saying when x is equal to two, y would be equal to three for this equation. Let's see if that's true. So four times x, we're now going to see if when x is two, y can be three."}, {"video_title": "Checking ordered pair solutions to equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "Two comma three. So this is saying when x is equal to two, y would be equal to three for this equation. Let's see if that's true. So four times x, we're now going to see if when x is two, y can be three. So four times x, four times two, minus one is equal to three times y. Now y we're testing to see if it can be three. Three times three plus five."}, {"video_title": "Checking ordered pair solutions to equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "So four times x, we're now going to see if when x is two, y can be three. So four times x, four times two, minus one is equal to three times y. Now y we're testing to see if it can be three. Three times three plus five. Let's see if this is true. Four times two is eight minus one. Is this equal to three times three?"}, {"video_title": "Checking ordered pair solutions to equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "Three times three plus five. Let's see if this is true. Four times two is eight minus one. Is this equal to three times three? So that's nine plus five. So is seven equal to 14? No, clearly seven is not equal to 14."}, {"video_title": "Checking ordered pair solutions to equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "Is this equal to three times three? So that's nine plus five. So is seven equal to 14? No, clearly seven is not equal to 14. So these things are not equal to each other. So this is not a solution. When x equals two, y cannot be equal to three and satisfy this equation."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "But he has a little bit of a conundrum in throwing the brunch. He wants to figure out how many cupcakes, how many cupcakes should he order? He doesn't want to waste any, but he wants to make sure that everyone has enough to eat. And you say, well, what's the problem here? And he says, well, I know adults eat a different number of cupcakes than children eat. And I know that in my kingdom, an adult will always eat the same amount and a child will always eat the same amount. And so you say, king, well, what information can you give me?"}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "And you say, well, what's the problem here? And he says, well, I know adults eat a different number of cupcakes than children eat. And I know that in my kingdom, an adult will always eat the same amount and a child will always eat the same amount. And so you say, king, well, what information can you give me? I might be able to help you out a little bit. You're feeling very confident after this Trolls situation. And he says, well, I know at the last party, we had 500 adults, 500 adults, and we had 200 children, 200 children."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "And so you say, king, well, what information can you give me? I might be able to help you out a little bit. You're feeling very confident after this Trolls situation. And he says, well, I know at the last party, we had 500 adults, 500 adults, and we had 200 children, 200 children. And combined, combined, they ate 2,900 cupcakes, 2,900 cupcakes. And you say, okay, that's interesting, but I think I'll need a little bit more information. Have you thrown parties before then?"}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "And he says, well, I know at the last party, we had 500 adults, 500 adults, and we had 200 children, 200 children. And combined, combined, they ate 2,900 cupcakes, 2,900 cupcakes. And you say, okay, that's interesting, but I think I'll need a little bit more information. Have you thrown parties before then? The king says, of course I have, I like to throw parties. Well, what happened at the party before that? And he says, well, there we also had 500 adults, 500 adults, and we had 300 children, 300 children."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "Have you thrown parties before then? The king says, of course I have, I like to throw parties. Well, what happened at the party before that? And he says, well, there we also had 500 adults, 500 adults, and we had 300 children, 300 children. And you say, well, how many cupcakes were eaten at that party? He says, well, we know, it was 3,100, 3,100 cupcakes. And so you get a tingling feeling that a little bit of algebra might apply over here."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "And he says, well, there we also had 500 adults, 500 adults, and we had 300 children, 300 children. And you say, well, how many cupcakes were eaten at that party? He says, well, we know, it was 3,100, 3,100 cupcakes. And so you get a tingling feeling that a little bit of algebra might apply over here. And you say, well, let me see, what do we need to figure out? We need to figure out how many, the number of cupcakes on average that an adult will eat. So number of cupcakes, but for an adult, and we also need to figure out the number for, number of cupcakes for a child."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "And so you get a tingling feeling that a little bit of algebra might apply over here. And you say, well, let me see, what do we need to figure out? We need to figure out how many, the number of cupcakes on average that an adult will eat. So number of cupcakes, but for an adult, and we also need to figure out the number for, number of cupcakes for a child. So these are the two things, these are the two things that we need to figure out, because then we can know how many adults and children are coming to the next brunch that are being held in your honor, and get the exact right number of cupcakes. So those are the things you're trying to figure out, and we don't know what those things are. Let's define some variables that represent those things."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "So number of cupcakes, but for an adult, and we also need to figure out the number for, number of cupcakes for a child. So these are the two things, these are the two things that we need to figure out, because then we can know how many adults and children are coming to the next brunch that are being held in your honor, and get the exact right number of cupcakes. So those are the things you're trying to figure out, and we don't know what those things are. Let's define some variables that represent those things. So let's let, well, let's do A for adults. Let's let A equal the number of cupcakes that on average each adult eats. And let's do C for children."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "Let's define some variables that represent those things. So let's let, well, let's do A for adults. Let's let A equal the number of cupcakes that on average each adult eats. And let's do C for children. So C is the number of cupcakes for a child on average. So given that information, let's see how we can represent what the king has told us algebraically. So let's think about this orange information first."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "And let's do C for children. So C is the number of cupcakes for a child on average. So given that information, let's see how we can represent what the king has told us algebraically. So let's think about this orange information first. How could we represent this algebraically? Well, let's think about how many cupcakes the adults ate at that party. You had 500 adults, and on average, each of them ate exactly A cupcakes, so the total number of cupcakes that the adults ate were 500 times A."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "So let's think about this orange information first. How could we represent this algebraically? Well, let's think about how many cupcakes the adults ate at that party. You had 500 adults, and on average, each of them ate exactly A cupcakes, so the total number of cupcakes that the adults ate were 500 times A. How many did the children eat? Well, same logic. You had 200 children, and they each ate C cupcakes, so 200 times C is the total number of cupcakes that the children ate."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "You had 500 adults, and on average, each of them ate exactly A cupcakes, so the total number of cupcakes that the adults ate were 500 times A. How many did the children eat? Well, same logic. You had 200 children, and they each ate C cupcakes, so 200 times C is the total number of cupcakes that the children ate. Well, how much did they eat in total? Well, it's the total number that the adults ate plus the total number that the children ate, which is 2,900 cupcakes. 2,900."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "You had 200 children, and they each ate C cupcakes, so 200 times C is the total number of cupcakes that the children ate. Well, how much did they eat in total? Well, it's the total number that the adults ate plus the total number that the children ate, which is 2,900 cupcakes. 2,900. So let's do that, apply that same logic to the blue party right over here, this blue information. How can we represent this algebraically? Well, once again, how many total cupcakes did the adults eat?"}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "2,900. So let's do that, apply that same logic to the blue party right over here, this blue information. How can we represent this algebraically? Well, once again, how many total cupcakes did the adults eat? Well, you had 500 adults, and they each ate A cupcakes, which is an unknown right now. And then what about the children, where you had 300 children, and they each ate C cupcakes? And so if you add up all the cupcakes that the adults ate plus all the cupcakes that the children ate, you get to 3,100 cupcakes."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "Well, once again, how many total cupcakes did the adults eat? Well, you had 500 adults, and they each ate A cupcakes, which is an unknown right now. And then what about the children, where you had 300 children, and they each ate C cupcakes? And so if you add up all the cupcakes that the adults ate plus all the cupcakes that the children ate, you get to 3,100 cupcakes. So this is starting to look interesting. I have two equations. I have a system of two equations with two unknowns, and you know from your experience with the troll that you should be able to solve this."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "And so if you add up all the cupcakes that the adults ate plus all the cupcakes that the children ate, you get to 3,100 cupcakes. So this is starting to look interesting. I have two equations. I have a system of two equations with two unknowns, and you know from your experience with the troll that you should be able to solve this. You could solve it graphically like you did in the past, but now you feel that there could be another tool in your toolkit, which is really just an application of the algebra that you already know. So think a little bit about how you might do this. So let's rewrite this first equation right over here."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "I have a system of two equations with two unknowns, and you know from your experience with the troll that you should be able to solve this. You could solve it graphically like you did in the past, but now you feel that there could be another tool in your toolkit, which is really just an application of the algebra that you already know. So think a little bit about how you might do this. So let's rewrite this first equation right over here. So we have 500A plus 200C is equal to 2,900 cupcakes. Now, it would be good if we could get rid of this 500A somehow. Well, you might say, well, let me just subtract 500A."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "So let's rewrite this first equation right over here. So we have 500A plus 200C is equal to 2,900 cupcakes. Now, it would be good if we could get rid of this 500A somehow. Well, you might say, well, let me just subtract 500A. So you might say, well, I just wanna subtract 500A, but if you subtracted 500A from the left-hand side, you'd also have to subtract 500A from the right-hand side, and so the A wouldn't just disappear. It would just end up on the right-hand side, and you'd still have one equation with two unknowns, which isn't too helpful. But you see something interesting."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "Well, you might say, well, let me just subtract 500A. So you might say, well, I just wanna subtract 500A, but if you subtracted 500A from the left-hand side, you'd also have to subtract 500A from the right-hand side, and so the A wouldn't just disappear. It would just end up on the right-hand side, and you'd still have one equation with two unknowns, which isn't too helpful. But you see something interesting. You're like, well, this is a 500A here. What if I subtracted a 500A and this 300C? So if I subtracted the 500A and the 300C from the left-hand side, and you're like, well, why is that useful?"}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "But you see something interesting. You're like, well, this is a 500A here. What if I subtracted a 500A and this 300C? So if I subtracted the 500A and the 300C from the left-hand side, and you're like, well, why is that useful? You're gonna have to do the same thing on the right-hand side, and then you're gonna have an A and a C on the right-hand. And you just say, hold on, hold on one second here. Hold on, I guess you're talking to yourself."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "So if I subtracted the 500A and the 300C from the left-hand side, and you're like, well, why is that useful? You're gonna have to do the same thing on the right-hand side, and then you're gonna have an A and a C on the right-hand. And you just say, hold on, hold on one second here. Hold on, I guess you're talking to yourself. Hold on one second. I'm subtracting the left-hand side of this equation, but this left-hand side is the exact same thing as this right-hand side. So here I could subtract 500A and 300C, and I could do 500A and subtract 300C over here, but we know that subtracting 500A and 300C, that's the exact same thing as subtracting 3100."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "Hold on, I guess you're talking to yourself. Hold on one second. I'm subtracting the left-hand side of this equation, but this left-hand side is the exact same thing as this right-hand side. So here I could subtract 500A and 300C, and I could do 500A and subtract 300C over here, but we know that subtracting 500A and 300C, that's the exact same thing as subtracting 3100. 500, let me make it clear. This is 500A, 500A minus 300C is the exact same thing as subtracting 500A plus 300C, and we know that 500A plus 300C is exactly 3100. This is 3100."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "So here I could subtract 500A and 300C, and I could do 500A and subtract 300C over here, but we know that subtracting 500A and 300C, that's the exact same thing as subtracting 3100. 500, let me make it clear. This is 500A, 500A minus 300C is the exact same thing as subtracting 500A plus 300C, and we know that 500A plus 300C is exactly 3100. This is 3100. This is what the second information gave us. So instead of subtracting 500A minus 300C, we can just subtract 3100. So let me do that."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "This is 3100. This is what the second information gave us. So instead of subtracting 500A minus 300C, we can just subtract 3100. So let me do that. This is exciting. So let me clear that out. So let's clear that out."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "So let me do that. This is exciting. So let me clear that out. So let's clear that out. And so here, instead of doing this, I can subtract the exact same value, which we know is 3100. Subtract 3100. So look at it this way."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "So let's clear that out. And so here, instead of doing this, I can subtract the exact same value, which we know is 3100. Subtract 3100. So look at it this way. It looks like we're subtracting this bottom equation from the top equation, but we're really just subtracting the same thing from both sides. This is just very basic algebra here. But if we do that, let's see what happens."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "So look at it this way. It looks like we're subtracting this bottom equation from the top equation, but we're really just subtracting the same thing from both sides. This is just very basic algebra here. But if we do that, let's see what happens. So on the left-hand side, 500A minus 500A, those cancel out. 200C minus 300C, that gives us negative 100C. And on the right-hand side, 2900 minus 3100 is negative 200."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "But if we do that, let's see what happens. So on the left-hand side, 500A minus 500A, those cancel out. 200C minus 300C, that gives us negative 100C. And on the right-hand side, 2900 minus 3100 is negative 200. Well, now we have one equation with one unknown, and we know how to solve this. We can divide both sides by negative 100. Divide both sides by negative 100."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "And on the right-hand side, 2900 minus 3100 is negative 200. Well, now we have one equation with one unknown, and we know how to solve this. We can divide both sides by negative 100. Divide both sides by negative 100. These cancel out. And then over here, you end up with a positive two. So C is equal to positive two."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "Divide both sides by negative 100. These cancel out. And then over here, you end up with a positive two. So C is equal to positive two. So we've solved one of the unknowns. Each child, on average, drinks two cups. So C is equal to two."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "So C is equal to positive two. So we've solved one of the unknowns. Each child, on average, drinks two cups. So C is equal to two. So how can we figure out what A is? Well, now we can take this information and go back into either one of these and figure out what A has to be. So let's go back into the orange one right over here and figure out what A has to be."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "So C is equal to two. So how can we figure out what A is? Well, now we can take this information and go back into either one of these and figure out what A has to be. So let's go back into the orange one right over here and figure out what A has to be. So we had 500A, 500A, plus 200C, but we know what C is. C is two. So 200 times two is equal to 2,900, is equal to 2,900."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "So let's go back into the orange one right over here and figure out what A has to be. So we had 500A, 500A, plus 200C, but we know what C is. C is two. So 200 times two is equal to 2,900, is equal to 2,900. And now we just have to solve for A, one equation with one unknown. So we have 500A, 500A. 200 times two is 400, plus 400 is equal to 2,900."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "So 200 times two is equal to 2,900, is equal to 2,900. And now we just have to solve for A, one equation with one unknown. So we have 500A, 500A. 200 times two is 400, plus 400 is equal to 2,900. We can subtract 400 from both sides of this equation. Let me do that. Subtracting 400."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "200 times two is 400, plus 400 is equal to 2,900. We can subtract 400 from both sides of this equation. Let me do that. Subtracting 400. And we are left with, this cancels out. And on the left-hand side, we have 500A. This is very exciting."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "Subtracting 400. And we are left with, this cancels out. And on the left-hand side, we have 500A. This is very exciting. We're in the home stretch. On the right-hand side, you have 2,500. 500A equals 2,500."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "This is very exciting. We're in the home stretch. On the right-hand side, you have 2,500. 500A equals 2,500. We can divide both sides by 500. And we are left with 2,500 divided by 500 is just five. So you have A is equal to five and you're done."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "500A equals 2,500. We can divide both sides by 500. And we are left with 2,500 divided by 500 is just five. So you have A is equal to five and you're done. You have solved the King's Conundrum. Each child on average drinks two cups of water. Or sorry, not cups of water."}, {"video_title": "King's cupcakes Solving systems by elimination   Algebra II   Khan Academy.mp3", "Sentence": "So you have A is equal to five and you're done. You have solved the King's Conundrum. Each child on average drinks two cups of water. Or sorry, not cups of water. I don't know where I got that from. Each child eats two cupcakes and each adult will eat five cupcakes. A is equal to five."}, {"video_title": "What is a variable   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "So this entire expression, you can view this, is how much I might make in a given hour. Now, you might also realize that the number of tips or the amount of tips I might make in an hour can change dramatically from hour to hour. It can vary. One hour, it might be lunchtime, get a lot of tips, people might get some big ticket items. The next hour, I might not have any customers, and then my tips might be really low. So the tips part right over here, we consider that, the entire word, we consider that to be a variable. From scenario to scenario, it can change."}, {"video_title": "What is a variable   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "One hour, it might be lunchtime, get a lot of tips, people might get some big ticket items. The next hour, I might not have any customers, and then my tips might be really low. So the tips part right over here, we consider that, the entire word, we consider that to be a variable. From scenario to scenario, it can change. So for example, in one scenario, maybe it's lunchtime, I'm getting really big tips. So tips is equal to, let's say it's equal to $30. And so the total amount I might make in that hour is going, we can go back to this expression right over here, it's going to be 10 plus, instead of writing tips here, I'll write 30, because that's what my tips are in that hour."}, {"video_title": "What is a variable   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "From scenario to scenario, it can change. So for example, in one scenario, maybe it's lunchtime, I'm getting really big tips. So tips is equal to, let's say it's equal to $30. And so the total amount I might make in that hour is going, we can go back to this expression right over here, it's going to be 10 plus, instead of writing tips here, I'll write 30, because that's what my tips are in that hour. And so that is going to be equal to, it's going to be equal to 40. Let me do that, let me do that in yellow color. It's going to be equal to $40."}, {"video_title": "What is a variable   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And so the total amount I might make in that hour is going, we can go back to this expression right over here, it's going to be 10 plus, instead of writing tips here, I'll write 30, because that's what my tips are in that hour. And so that is going to be equal to, it's going to be equal to 40. Let me do that, let me do that in yellow color. It's going to be equal to $40. But let's say right after that, the restaurant slows down, we're out of the lunch hour for whatever reason, maybe the restaurant next door has a big sale or something. And so the next hour, my tips go down dramatically. My tips go down to $5 for that hour."}, {"video_title": "What is a variable   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "It's going to be equal to $40. But let's say right after that, the restaurant slows down, we're out of the lunch hour for whatever reason, maybe the restaurant next door has a big sale or something. And so the next hour, my tips go down dramatically. My tips go down to $5 for that hour. Now I go back to this expression. The total I make is my hourly wage plus the $5 in tips, plus the $5 in tips, which is equal to $15. As you see, this entire expression, the 10 plus tips, it changed depending on what the value of the variable tips is."}, {"video_title": "What is a variable   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "My tips go down to $5 for that hour. Now I go back to this expression. The total I make is my hourly wage plus the $5 in tips, plus the $5 in tips, which is equal to $15. As you see, this entire expression, the 10 plus tips, it changed depending on what the value of the variable tips is. Now, you won't see whole words typically used in algebra as variables. We get lazy, and so instead, we tend to use just easier to write symbols. And so in this context, instead of writing tips, maybe we could have just written 10 plus t, where t represents the tips that we get in an hour."}, {"video_title": "What is a variable   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "As you see, this entire expression, the 10 plus tips, it changed depending on what the value of the variable tips is. Now, you won't see whole words typically used in algebra as variables. We get lazy, and so instead, we tend to use just easier to write symbols. And so in this context, instead of writing tips, maybe we could have just written 10 plus t, where t represents the tips that we get in an hour. And so then we would say, okay, what happens when t is equal to 30? Well, if t is equal to 30, then we'd have, let me write, so what happens when t is equal to 30? Well, then we have a situation, t is equal to 30."}, {"video_title": "What is a variable   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And so in this context, instead of writing tips, maybe we could have just written 10 plus t, where t represents the tips that we get in an hour. And so then we would say, okay, what happens when t is equal to 30? Well, if t is equal to 30, then we'd have, let me write, so what happens when t is equal to 30? Well, then we have a situation, t is equal to 30. This evaluates to 10 plus 30, which would be 40. What would happen if t is equal to five? Well, then this would evaluate to 10 plus five, which is equal to 15."}, {"video_title": "What is a variable   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Well, then we have a situation, t is equal to 30. This evaluates to 10 plus 30, which would be 40. What would happen if t is equal to five? Well, then this would evaluate to 10 plus five, which is equal to 15. Now I wanna be clear, we didn't even have to use t. We didn't even really have to use a letter, although in traditional algebra, you almost do use a letter. We could have written it as 10 plus x, where x is your tips per hour. X might not be as natural."}, {"video_title": "What is a variable   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Well, then this would evaluate to 10 plus five, which is equal to 15. Now I wanna be clear, we didn't even have to use t. We didn't even really have to use a letter, although in traditional algebra, you almost do use a letter. We could have written it as 10 plus x, where x is your tips per hour. X might not be as natural. It's not the first letter in the word tips. Or you could have even written 10 plus, you could have even written 10 plus star, where you could say star represents the number of tips in an hour, but it just might have not made as much intuitive sense. But hopefully this gives you a general idea of just what a variable is."}, {"video_title": "How to evaluate an expression with variables   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "A local hospital is holding a raffle as a fundraiser. The individual cost of participating in the raffle is given by the following expression. 5t plus 3, or 5 times t plus 3, where t represents the number of tickets someone purchases. Evaluate the expression when t is equal to 1, t is equal to 8, and t is equal to 10. So let's first take the situation where t is equal to 1. Then this expression right over here becomes, and I'll use that same color, becomes 5 times 1 plus 3. And we know from order of operations, you do the multiplication before you do the addition."}, {"video_title": "How to evaluate an expression with variables   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Evaluate the expression when t is equal to 1, t is equal to 8, and t is equal to 10. So let's first take the situation where t is equal to 1. Then this expression right over here becomes, and I'll use that same color, becomes 5 times 1 plus 3. And we know from order of operations, you do the multiplication before you do the addition. So this will be 5 times 1 is 5 plus 3. And then this is clearly equal to 8. Now let's do it when t is equal to 8."}, {"video_title": "How to evaluate an expression with variables   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And we know from order of operations, you do the multiplication before you do the addition. So this will be 5 times 1 is 5 plus 3. And then this is clearly equal to 8. Now let's do it when t is equal to 8. So when t is equal to 8, this expression becomes, and I'll do the same colors again, 5 times 8 plus 3. And once again, 5 times 8 is 40. And then we have the plus 3 there."}, {"video_title": "How to evaluate an expression with variables   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Now let's do it when t is equal to 8. So when t is equal to 8, this expression becomes, and I'll do the same colors again, 5 times 8 plus 3. And once again, 5 times 8 is 40. And then we have the plus 3 there. So this is equal to 43. And so we have the last situation. With t is equal to 10, I'll do that in blue."}, {"video_title": "How to evaluate an expression with variables   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And then we have the plus 3 there. So this is equal to 43. And so we have the last situation. With t is equal to 10, I'll do that in blue. So we have 5 times 10. So 5t is 5 times 10. Instead of a t, we put a 10 there."}, {"video_title": "How to evaluate an expression with variables   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "With t is equal to 10, I'll do that in blue. So we have 5 times 10. So 5t is 5 times 10. Instead of a t, we put a 10 there. 5 times 10 plus 3. That's a slightly differentiated green, but I think you get the idea. 5 times 10 is 50."}, {"video_title": "How to evaluate an expression with variables   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Instead of a t, we put a 10 there. 5 times 10 plus 3. That's a slightly differentiated green, but I think you get the idea. 5 times 10 is 50. And then we're going to have to add 3 to that. And that is equal to 53. And we're done."}, {"video_title": "Why a negative times a negative is a positive   Pre-Algebra   Khan Academy.mp3", "Sentence": "And you know how to add and subtract negative numbers. But now you are faced with a conundrum. What happens when you multiply negative numbers? Either when you multiply a positive number times a negative number, or when you multiply two negative numbers. So for example, you aren't quite sure what should happen if you were to multiply, and I'm just picking two numbers where one is positive and one is negative. What would happen if you were to multiply five times negative three? You're not quite sure about this just yet."}, {"video_title": "Why a negative times a negative is a positive   Pre-Algebra   Khan Academy.mp3", "Sentence": "Either when you multiply a positive number times a negative number, or when you multiply two negative numbers. So for example, you aren't quite sure what should happen if you were to multiply, and I'm just picking two numbers where one is positive and one is negative. What would happen if you were to multiply five times negative three? You're not quite sure about this just yet. You're also not quite sure about what would happen if you multiply two negative numbers. So let's say negative two times negative six. This is also unclear to you."}, {"video_title": "Why a negative times a negative is a positive   Pre-Algebra   Khan Academy.mp3", "Sentence": "You're not quite sure about this just yet. You're also not quite sure about what would happen if you multiply two negative numbers. So let's say negative two times negative six. This is also unclear to you. What you do know, because you are a mathematician, is however you define this, or whatever this should be, it should hopefully be consistent with all of the other properties of mathematics that you already know, and preferably all of the other properties of multiplication. That would make you feel comfortable that you are getting this right. And later we can think about other ways to get the intuition for what these might be and why it actually makes sense."}, {"video_title": "Why a negative times a negative is a positive   Pre-Algebra   Khan Academy.mp3", "Sentence": "This is also unclear to you. What you do know, because you are a mathematician, is however you define this, or whatever this should be, it should hopefully be consistent with all of the other properties of mathematics that you already know, and preferably all of the other properties of multiplication. That would make you feel comfortable that you are getting this right. And later we can think about other ways to get the intuition for what these might be and why it actually makes sense. But to make this consistent with the rest of the mathematics that you know, you go into a little bit of a thought experiment. You say, well what should five times three plus negative three be? Well you already have a philosophy of adding negative numbers or adding positive to negative numbers."}, {"video_title": "Why a negative times a negative is a positive   Pre-Algebra   Khan Academy.mp3", "Sentence": "And later we can think about other ways to get the intuition for what these might be and why it actually makes sense. But to make this consistent with the rest of the mathematics that you know, you go into a little bit of a thought experiment. You say, well what should five times three plus negative three be? Well you already have a philosophy of adding negative numbers or adding positive to negative numbers. You know that negative three is the opposite of three. If you add three to negative three, you're going to get zero. So this is going to be equal to five times zero, based on how you already thought about adding a negative to a positive."}, {"video_title": "Why a negative times a negative is a positive   Pre-Algebra   Khan Academy.mp3", "Sentence": "Well you already have a philosophy of adding negative numbers or adding positive to negative numbers. You know that negative three is the opposite of three. If you add three to negative three, you're going to get zero. So this is going to be equal to five times zero, based on how you already thought about adding a negative to a positive. And anything times zero is going to be zero. So this expression right over here should be zero. But on the other hand you say, well I want multiplying positive and negative numbers to be consistent with the distributive property."}, {"video_title": "Why a negative times a negative is a positive   Pre-Algebra   Khan Academy.mp3", "Sentence": "So this is going to be equal to five times zero, based on how you already thought about adding a negative to a positive. And anything times zero is going to be zero. So this expression right over here should be zero. But on the other hand you say, well I want multiplying positive and negative numbers to be consistent with the distributive property. So I should be able to distribute this five. I should be able to distribute this five, and for math to be consistent, and math should be consistent, I should get the exact same answer. So let's distribute this five."}, {"video_title": "Why a negative times a negative is a positive   Pre-Algebra   Khan Academy.mp3", "Sentence": "But on the other hand you say, well I want multiplying positive and negative numbers to be consistent with the distributive property. So I should be able to distribute this five. I should be able to distribute this five, and for math to be consistent, and math should be consistent, I should get the exact same answer. So let's distribute this five. So if we get five times three, so five times three is, let me write it out. This is going to be five times three, and let me write it with a multiplication, not the dot. I'll write the x sign for multiplying."}, {"video_title": "Why a negative times a negative is a positive   Pre-Algebra   Khan Academy.mp3", "Sentence": "So let's distribute this five. So if we get five times three, so five times three is, let me write it out. This is going to be five times three, and let me write it with a multiplication, not the dot. I'll write the x sign for multiplying. Five times three, so I distribute it there, plus five times negative three. And I'll do that in yellow. Five times negative three."}, {"video_title": "Why a negative times a negative is a positive   Pre-Algebra   Khan Academy.mp3", "Sentence": "I'll write the x sign for multiplying. Five times three, so I distribute it there, plus five times negative three. And I'll do that in yellow. Five times negative three. Five times negative three. And this whole thing we just said should be equal to zero. This should be equal to zero."}, {"video_title": "Why a negative times a negative is a positive   Pre-Algebra   Khan Academy.mp3", "Sentence": "Five times negative three. Five times negative three. And this whole thing we just said should be equal to zero. This should be equal to zero. Well five times three, those are two positive numbers, we know what that should be. That is going to be 15. So now we get this thing."}, {"video_title": "Why a negative times a negative is a positive   Pre-Algebra   Khan Academy.mp3", "Sentence": "This should be equal to zero. Well five times three, those are two positive numbers, we know what that should be. That is going to be 15. So now we get this thing. 15 plus whatever five times negative three is, plus whatever five times negative three is, needs to be equal to zero in order to be consistent with all of the other mathematics that we know. Well what plus 15 is going to be equal to zero? Well the opposite of 15."}, {"video_title": "Why a negative times a negative is a positive   Pre-Algebra   Khan Academy.mp3", "Sentence": "So now we get this thing. 15 plus whatever five times negative three is, plus whatever five times negative three is, needs to be equal to zero in order to be consistent with all of the other mathematics that we know. Well what plus 15 is going to be equal to zero? Well the opposite of 15. In order for this to be true, in order for this to be consistent with all of the other mathematics we know, this right over here needs to be equal to negative 15. And so you say five times negative three, in order to be consistent with all the other mathematics we know, needs to be equal to negative 15. And that's also consistent with the intuition of adding negative three repeatedly five times."}, {"video_title": "Why a negative times a negative is a positive   Pre-Algebra   Khan Academy.mp3", "Sentence": "Well the opposite of 15. In order for this to be true, in order for this to be consistent with all of the other mathematics we know, this right over here needs to be equal to negative 15. And so you say five times negative three, in order to be consistent with all the other mathematics we know, needs to be equal to negative 15. And that's also consistent with the intuition of adding negative three repeatedly five times. Now a slightly harder to conceive idea is multiplying two negatives. But we can do the exact same thought experiment. We want whatever this answer to be to be consistent with the rest of mathematics that we know."}, {"video_title": "Why a negative times a negative is a positive   Pre-Algebra   Khan Academy.mp3", "Sentence": "And that's also consistent with the intuition of adding negative three repeatedly five times. Now a slightly harder to conceive idea is multiplying two negatives. But we can do the exact same thought experiment. We want whatever this answer to be to be consistent with the rest of mathematics that we know. So we can say, so we can do the same thought experiment. What would negative two times six plus negative six be equal to? Well six plus negative six is going to be zero."}, {"video_title": "Why a negative times a negative is a positive   Pre-Algebra   Khan Academy.mp3", "Sentence": "We want whatever this answer to be to be consistent with the rest of mathematics that we know. So we can say, so we can do the same thought experiment. What would negative two times six plus negative six be equal to? Well six plus negative six is going to be zero. Negative two times zero, anything times zero needs to be equal to zero. But then once again we can distribute. We can distribute negative two times six."}, {"video_title": "Why a negative times a negative is a positive   Pre-Algebra   Khan Academy.mp3", "Sentence": "Well six plus negative six is going to be zero. Negative two times zero, anything times zero needs to be equal to zero. But then once again we can distribute. We can distribute negative two times six. So we get negative two, negative two times six plus negative two times negative six. Plus negative two times negative six. And once again all of that's going to need to be equal to zero."}, {"video_title": "Why a negative times a negative is a positive   Pre-Algebra   Khan Academy.mp3", "Sentence": "We can distribute negative two times six. So we get negative two, negative two times six plus negative two times negative six. Plus negative two times negative six. And once again all of that's going to need to be equal to zero. Now based on the thought experiment we just did, we said well this needs to be equal to negative 12. Or we could view this as going to the six twice in the left direction on the number line which would get us to negative 12. Or you could say repeatedly adding negative two six times."}, {"video_title": "Why a negative times a negative is a positive   Pre-Algebra   Khan Academy.mp3", "Sentence": "And once again all of that's going to need to be equal to zero. Now based on the thought experiment we just did, we said well this needs to be equal to negative 12. Or we could view this as going to the six twice in the left direction on the number line which would get us to negative 12. Or you could say repeatedly adding negative two six times. That would also get you to negative 12. And now we also saw it over here that we multiplied a positive times a negative. We got the negative."}, {"video_title": "Why a negative times a negative is a positive   Pre-Algebra   Khan Academy.mp3", "Sentence": "Or you could say repeatedly adding negative two six times. That would also get you to negative 12. And now we also saw it over here that we multiplied a positive times a negative. We got the negative. So this could be, or we know that this is going to be equal to negative 12. And so we had negative 12 plus whatever this business is, whatever this business is going to have to be equal to zero. Is going to have to be equal to zero in order to be consistent with all of the other mathematics that we know."}, {"video_title": "Why a negative times a negative is a positive   Pre-Algebra   Khan Academy.mp3", "Sentence": "We got the negative. So this could be, or we know that this is going to be equal to negative 12. And so we had negative 12 plus whatever this business is, whatever this business is going to have to be equal to zero. Is going to have to be equal to zero in order to be consistent with all of the other mathematics that we know. And so what plus negative 12 is going to be equal to zero? Well positive 12 plus negative 12 is going to be equal to zero. So this needs to be equal to positive 12 in order to be consistent with all of the other mathematics we know."}, {"video_title": "Example 3 Factoring quadratics as a perfect square of a sum (a+b)^2   Algebra I   Khan Academy.mp3", "Sentence": "We need to factor negative 4t squared minus 12t minus 9. And a good place to start is to say, well, are there any common factors for all of these terms? And when you look at them, well, these first two are divisible by 4, these last two are divisible by 3, but not all of them are divisible by any one number. Well, but you could factor out a negative 1. But even if you factor out a negative 1, so you say this is the same thing as negative 1, times positive 4t squared plus 12t plus 9, you still end up with a non-1 coefficient out here and on the second degree term, on the t squared term. So you might want to immediately start grouping this. And if you did factor it by grouping, it would work."}, {"video_title": "Example 3 Factoring quadratics as a perfect square of a sum (a+b)^2   Algebra I   Khan Academy.mp3", "Sentence": "Well, but you could factor out a negative 1. But even if you factor out a negative 1, so you say this is the same thing as negative 1, times positive 4t squared plus 12t plus 9, you still end up with a non-1 coefficient out here and on the second degree term, on the t squared term. So you might want to immediately start grouping this. And if you did factor it by grouping, it would work. You would get the right answer. But there is something you might be able to see, or there is something about this equation that might pop out at you that might make it a little bit simpler to solve. And to understand that, let's take a little bit of a break here on the right-hand side and just think about what happens if you take a plus b times a plus b, if you just have a binomial squared."}, {"video_title": "Example 3 Factoring quadratics as a perfect square of a sum (a+b)^2   Algebra I   Khan Academy.mp3", "Sentence": "And if you did factor it by grouping, it would work. You would get the right answer. But there is something you might be able to see, or there is something about this equation that might pop out at you that might make it a little bit simpler to solve. And to understand that, let's take a little bit of a break here on the right-hand side and just think about what happens if you take a plus b times a plus b, if you just have a binomial squared. Well, you have a times a, which is a squared. Then you have a times that b, which is plus ab. Then you have b times a, which is the same thing as ab."}, {"video_title": "Example 3 Factoring quadratics as a perfect square of a sum (a+b)^2   Algebra I   Khan Academy.mp3", "Sentence": "And to understand that, let's take a little bit of a break here on the right-hand side and just think about what happens if you take a plus b times a plus b, if you just have a binomial squared. Well, you have a times a, which is a squared. Then you have a times that b, which is plus ab. Then you have b times a, which is the same thing as ab. And then you have b times b, or you have b squared. And so if you add these middle two terms right here, you're left with a squared plus 2ab plus b squared. This is the square of a binomial."}, {"video_title": "Example 3 Factoring quadratics as a perfect square of a sum (a+b)^2   Algebra I   Khan Academy.mp3", "Sentence": "Then you have b times a, which is the same thing as ab. And then you have b times b, or you have b squared. And so if you add these middle two terms right here, you're left with a squared plus 2ab plus b squared. This is the square of a binomial. Now, does this right here, does 4t squared plus 12t plus 9 fit this pattern? Well, if 4t squared is a squared, so if this right here is a squared, if that is a squared right there, then what does a have to be? If this is a squared, then a would be equal to the square root of this."}, {"video_title": "Example 3 Factoring quadratics as a perfect square of a sum (a+b)^2   Algebra I   Khan Academy.mp3", "Sentence": "This is the square of a binomial. Now, does this right here, does 4t squared plus 12t plus 9 fit this pattern? Well, if 4t squared is a squared, so if this right here is a squared, if that is a squared right there, then what does a have to be? If this is a squared, then a would be equal to the square root of this. It would be 2t. And if this is b squared, let me do that in a different color. If this right here is b squared, if the 9 is b squared right there, then that means that b is equal to 3, equal to the positive square root of the 9."}, {"video_title": "Example 3 Factoring quadratics as a perfect square of a sum (a+b)^2   Algebra I   Khan Academy.mp3", "Sentence": "If this is a squared, then a would be equal to the square root of this. It would be 2t. And if this is b squared, let me do that in a different color. If this right here is b squared, if the 9 is b squared right there, then that means that b is equal to 3, equal to the positive square root of the 9. Now, this number right here, and actually it doesn't have to just be equal to 3. It might have been negative 3 as well. It could be plus or minus 3."}, {"video_title": "Example 3 Factoring quadratics as a perfect square of a sum (a+b)^2   Algebra I   Khan Academy.mp3", "Sentence": "If this right here is b squared, if the 9 is b squared right there, then that means that b is equal to 3, equal to the positive square root of the 9. Now, this number right here, and actually it doesn't have to just be equal to 3. It might have been negative 3 as well. It could be plus or minus 3. But this number here, is it 2 times ab? That's the middle term that we care about. Is it 2 times ab?"}, {"video_title": "Example 3 Factoring quadratics as a perfect square of a sum (a+b)^2   Algebra I   Khan Academy.mp3", "Sentence": "It could be plus or minus 3. But this number here, is it 2 times ab? That's the middle term that we care about. Is it 2 times ab? Well, if we multiply 2t times 3, we get 6t. And then we multiply that times 2, you get 12t. This right here, 12t, is equal to 2 times 2t times 3."}, {"video_title": "Example 3 Factoring quadratics as a perfect square of a sum (a+b)^2   Algebra I   Khan Academy.mp3", "Sentence": "Is it 2 times ab? Well, if we multiply 2t times 3, we get 6t. And then we multiply that times 2, you get 12t. This right here, 12t, is equal to 2 times 2t times 3. It is 2 times ab. And if this was a negative 3, we would look to see if this was a negative 12, but this does work for a positive 3. So this does fit the pattern of a perfect square."}, {"video_title": "Example 3 Factoring quadratics as a perfect square of a sum (a+b)^2   Algebra I   Khan Academy.mp3", "Sentence": "This right here, 12t, is equal to 2 times 2t times 3. It is 2 times ab. And if this was a negative 3, we would look to see if this was a negative 12, but this does work for a positive 3. So this does fit the pattern of a perfect square. This is a special type of, or you could view this as a square of a binomial. So if you wanted to factor this, the stuff on the inside, you still have that negative 1 out there, the 4t squared plus 12t plus 9, you could immediately say, well that's going to be a plus b times a plus b. Or 2t plus 3 times 2t plus 3."}, {"video_title": "Example 3 Factoring quadratics as a perfect square of a sum (a+b)^2   Algebra I   Khan Academy.mp3", "Sentence": "So this does fit the pattern of a perfect square. This is a special type of, or you could view this as a square of a binomial. So if you wanted to factor this, the stuff on the inside, you still have that negative 1 out there, the 4t squared plus 12t plus 9, you could immediately say, well that's going to be a plus b times a plus b. Or 2t plus 3 times 2t plus 3. Or you could just say it's 2t plus 3 squared. It fits this pattern. And of course, you can't forget about this negative 1 out here."}, {"video_title": "Example 3 Factoring quadratics as a perfect square of a sum (a+b)^2   Algebra I   Khan Academy.mp3", "Sentence": "Or 2t plus 3 times 2t plus 3. Or you could just say it's 2t plus 3 squared. It fits this pattern. And of course, you can't forget about this negative 1 out here. You could have also solved it by grouping, but this might be a quicker thing to recognize. This is a number squared. That's another number squared."}, {"video_title": "Solving a more complicated equation   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Let's try some slightly more complicated equations. Let's say we have 3 times x plus 5 is equal to 17. So what's different about this than what we saw in the last video is all of a sudden now we have this plus 5. If it was just 3x is equal to 17, you could divide both sides by 3 and you'd get your answer. But now this 5 seems to mess things up a little bit. Now before we even solve it, let's think about what it's saying. Let's solve it kind of in a tangible way, and then we'll solve it using operations that hopefully will make sense after that."}, {"video_title": "Solving a more complicated equation   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "If it was just 3x is equal to 17, you could divide both sides by 3 and you'd get your answer. But now this 5 seems to mess things up a little bit. Now before we even solve it, let's think about what it's saying. Let's solve it kind of in a tangible way, and then we'll solve it using operations that hopefully will make sense after that. So 3 times x literally means, so let me write it over here. So we have 3 times x. So you literally have an x plus an x plus an x."}, {"video_title": "Solving a more complicated equation   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Let's solve it kind of in a tangible way, and then we'll solve it using operations that hopefully will make sense after that. So 3 times x literally means, so let me write it over here. So we have 3 times x. So you literally have an x plus an x plus an x. That right there is a 3x, and then that's plus 5. And I'm actually going to write it out as 5 objects. So plus 1, 2, 3, 4, 5."}, {"video_title": "Solving a more complicated equation   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So you literally have an x plus an x plus an x. That right there is a 3x, and then that's plus 5. And I'm actually going to write it out as 5 objects. So plus 1, 2, 3, 4, 5. That, this right here is 3x plus 5 is equal to 17. So let me write the equal sign is equal to 17. Now let me draw 17 objects here."}, {"video_title": "Solving a more complicated equation   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So plus 1, 2, 3, 4, 5. That, this right here is 3x plus 5 is equal to 17. So let me write the equal sign is equal to 17. Now let me draw 17 objects here. So 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17. Now these two things are equal. So anything you do to this side, you'd have to do to that side."}, {"video_title": "Solving a more complicated equation   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Now let me draw 17 objects here. So 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17. Now these two things are equal. So anything you do to this side, you'd have to do to that side. If we were to get rid of one object here, you'd want to get rid of one object there in order for the equality to still be true. Now what can we do to both sides of this equation so we can get it in the form that we're used to, where we only have a 3x on the left-hand side, where we don't have this 5? Well, ideally we would just get rid of these 5 objects here."}, {"video_title": "Solving a more complicated equation   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So anything you do to this side, you'd have to do to that side. If we were to get rid of one object here, you'd want to get rid of one object there in order for the equality to still be true. Now what can we do to both sides of this equation so we can get it in the form that we're used to, where we only have a 3x on the left-hand side, where we don't have this 5? Well, ideally we would just get rid of these 5 objects here. You would literally get rid of these 5 objects, 1, 2, 3, 4, 5, but like I said, if the original thing was equal to the original thing on the right, if we get rid of 5 objects from the left-hand side, we have to get rid of 5 objects from the right-hand side. So we have to do it here too, 1, 2, 3, 4, 5. Now what is a symbolic way of representing taking away 5 things?"}, {"video_title": "Solving a more complicated equation   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Well, ideally we would just get rid of these 5 objects here. You would literally get rid of these 5 objects, 1, 2, 3, 4, 5, but like I said, if the original thing was equal to the original thing on the right, if we get rid of 5 objects from the left-hand side, we have to get rid of 5 objects from the right-hand side. So we have to do it here too, 1, 2, 3, 4, 5. Now what is a symbolic way of representing taking away 5 things? Well, you're subtracting 5 from both sides of this equation, so that's what we're doing here when we took away 5 from the left and from the right. So we're subtracting 5 from the left. That's what we did here, and we're also subtracting 5 from the right."}, {"video_title": "Solving a more complicated equation   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Now what is a symbolic way of representing taking away 5 things? Well, you're subtracting 5 from both sides of this equation, so that's what we're doing here when we took away 5 from the left and from the right. So we're subtracting 5 from the left. That's what we did here, and we're also subtracting 5 from the right. Do that right over there. Now what does the left-hand side of the equation now become? The left-hand side, you have 5 minus 5."}, {"video_title": "Solving a more complicated equation   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "That's what we did here, and we're also subtracting 5 from the right. Do that right over there. Now what does the left-hand side of the equation now become? The left-hand side, you have 5 minus 5. These cancel out. You're just left with the 3x. You are just left with, it's a different shade of green, the 5 and the negative 5 canceled out."}, {"video_title": "Solving a more complicated equation   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "The left-hand side, you have 5 minus 5. These cancel out. You're just left with the 3x. You are just left with, it's a different shade of green, the 5 and the negative 5 canceled out. And you see that here. When you got rid of these 5 objects, we were just left with the 3x's. This right here is the 3x."}, {"video_title": "Solving a more complicated equation   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "You are just left with, it's a different shade of green, the 5 and the negative 5 canceled out. And you see that here. When you got rid of these 5 objects, we were just left with the 3x's. This right here is the 3x. And the whole reason why we subtracted 5 is because we wanted this 5 to go away. Now what does the right-hand side of the equation look like? So it's 3x is going to be, let me write the equality sign right under it, is equal to, well you can either just do it mathematically, say OK, 17 minus 5 is 12."}, {"video_title": "Solving a more complicated equation   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "This right here is the 3x. And the whole reason why we subtracted 5 is because we wanted this 5 to go away. Now what does the right-hand side of the equation look like? So it's 3x is going to be, let me write the equality sign right under it, is equal to, well you can either just do it mathematically, say OK, 17 minus 5 is 12. Or you could just count over here. I had 17 things. I took away 5."}, {"video_title": "Solving a more complicated equation   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So it's 3x is going to be, let me write the equality sign right under it, is equal to, well you can either just do it mathematically, say OK, 17 minus 5 is 12. Or you could just count over here. I had 17 things. I took away 5. I have 12 left. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. That's what subtraction is."}, {"video_title": "Solving a more complicated equation   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "I took away 5. I have 12 left. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. That's what subtraction is. It's just taking away 5 things. So now we have it in a pretty straightforward form. 3x is equal to 12."}, {"video_title": "Solving a more complicated equation   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "That's what subtraction is. It's just taking away 5 things. So now we have it in a pretty straightforward form. 3x is equal to 12. All we have to do is divide both sides of this equation by 3. So we're just left with an x on the left-hand side. So we divide by, let me pick a nicer color than that."}, {"video_title": "Solving a more complicated equation   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "3x is equal to 12. All we have to do is divide both sides of this equation by 3. So we're just left with an x on the left-hand side. So we divide by, let me pick a nicer color than that. Let me do this pink color. So you divide the left-hand side by 3, the right-hand side by 3. And remember what that's equivalent to."}, {"video_title": "Solving a more complicated equation   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So we divide by, let me pick a nicer color than that. Let me do this pink color. So you divide the left-hand side by 3, the right-hand side by 3. And remember what that's equivalent to. The left-hand side, none of this stuff exists anymore. So we should ignore it. None of this stuff exists anymore."}, {"video_title": "Solving a more complicated equation   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "And remember what that's equivalent to. The left-hand side, none of this stuff exists anymore. So we should ignore it. None of this stuff exists anymore. In fact, let me clear it out just so that we don't even have to look at it. We subtracted it. So let me clear it over here."}, {"video_title": "Solving a more complicated equation   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "None of this stuff exists anymore. In fact, let me clear it out just so that we don't even have to look at it. We subtracted it. So let me clear it over here. And so now we are dividing both sides by 3. Divide the left-hand side by 3. It's 1, 2, 3."}, {"video_title": "Solving a more complicated equation   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So let me clear it over here. And so now we are dividing both sides by 3. Divide the left-hand side by 3. It's 1, 2, 3. So 3 groups, each of them have an x in it. If you divide this right-hand side by 3, let's see, you have 1, 2, and 3. So it's 3 groups of 4."}, {"video_title": "Solving a more complicated equation   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "It's 1, 2, 3. So 3 groups, each of them have an x in it. If you divide this right-hand side by 3, let's see, you have 1, 2, and 3. So it's 3 groups of 4. So when you do it mathematically here, the 3's cancel out. 3 times something divided by 3 is just a something. So you're left with x is equal to, and then 12 divided by 3 is 4."}, {"video_title": "Solving a more complicated equation   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So it's 3 groups of 4. So when you do it mathematically here, the 3's cancel out. 3 times something divided by 3 is just a something. So you're left with x is equal to, and then 12 divided by 3 is 4. You get x is equal to 4. And you get that exact same thing over here. When you divided 3x into groups of 3, each of the groups had an x in it."}, {"video_title": "Solving a more complicated equation   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So you're left with x is equal to, and then 12 divided by 3 is 4. You get x is equal to 4. And you get that exact same thing over here. When you divided 3x into groups of 3, each of the groups had an x in it. And when you divided 12 into groups of 3, each of the groups have a 4 in it. So x must be equal to 4. x is equal to 4. Let's do another one."}, {"video_title": "Solving a more complicated equation   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "When you divided 3x into groups of 3, each of the groups had an x in it. And when you divided 12 into groups of 3, each of the groups have a 4 in it. So x must be equal to 4. x is equal to 4. Let's do another one. And this time I won't draw it all out like this, but hopefully you'll see that the same type of processes are involved. Let's say I have 7x. I'll do a slightly more complicated one this time."}, {"video_title": "Solving a more complicated equation   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Let's do another one. And this time I won't draw it all out like this, but hopefully you'll see that the same type of processes are involved. Let's say I have 7x. I'll do a slightly more complicated one this time. 7x minus 2 is equal to negative 10. Now this all of a sudden becomes a lot more, you have a negative sign, we have a negative over here, but we're going to do the exact same thing. The first thing we want to do if we want to get the left hand side simplified to just 7x is we want to get rid of this negative 2."}, {"video_title": "Solving a more complicated equation   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "I'll do a slightly more complicated one this time. 7x minus 2 is equal to negative 10. Now this all of a sudden becomes a lot more, you have a negative sign, we have a negative over here, but we're going to do the exact same thing. The first thing we want to do if we want to get the left hand side simplified to just 7x is we want to get rid of this negative 2. And what can we add or subtract to both sides of the equation to get rid of this negative 2? Well if we add 2 to the left hand side, these two guys will cancel out. But remember, this is equal to that."}, {"video_title": "Solving a more complicated equation   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "The first thing we want to do if we want to get the left hand side simplified to just 7x is we want to get rid of this negative 2. And what can we add or subtract to both sides of the equation to get rid of this negative 2? Well if we add 2 to the left hand side, these two guys will cancel out. But remember, this is equal to that. If we want the equality to still hold, if we add 2 to the left hand side, we also have to do it to the right hand side. So what is the new left hand side going to be equal to? So we have 7x, negative 2 plus 2 is just 0."}, {"video_title": "Solving a more complicated equation   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "But remember, this is equal to that. If we want the equality to still hold, if we add 2 to the left hand side, we also have to do it to the right hand side. So what is the new left hand side going to be equal to? So we have 7x, negative 2 plus 2 is just 0. I could write plus 0, I could just write nothing there, and I'll just write nothing. So we get 7x is equal to, now what's negative 10 plus 2? And this is a little bit of a review of adding and subtracting negative numbers."}, {"video_title": "Solving a more complicated equation   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So we have 7x, negative 2 plus 2 is just 0. I could write plus 0, I could just write nothing there, and I'll just write nothing. So we get 7x is equal to, now what's negative 10 plus 2? And this is a little bit of a review of adding and subtracting negative numbers. Remember, I'll draw the number line here for you. If I draw the number line, so if this is 0, this is 1, we could keep going in the positive direction. Negative 10 is out here."}, {"video_title": "Solving a more complicated equation   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "And this is a little bit of a review of adding and subtracting negative numbers. Remember, I'll draw the number line here for you. If I draw the number line, so if this is 0, this is 1, we could keep going in the positive direction. Negative 10 is out here. Negative 10, negative 9, negative 8, negative 7, there's a bunch of numbers here, dot, dot, dot. I don't have space to draw them all, but we're starting at negative 10. And we're adding 2 to it, so we're moving in the positive direction on the number line."}, {"video_title": "Solving a more complicated equation   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Negative 10 is out here. Negative 10, negative 9, negative 8, negative 7, there's a bunch of numbers here, dot, dot, dot. I don't have space to draw them all, but we're starting at negative 10. And we're adding 2 to it, so we're moving in the positive direction on the number line. So we're going 1, 2. So it's negative 8. Don't get confused."}, {"video_title": "Solving a more complicated equation   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "And we're adding 2 to it, so we're moving in the positive direction on the number line. So we're going 1, 2. So it's negative 8. Don't get confused. Don't say, OK, 10 plus 2 is 12, so negative 10 plus 2 is negative 12. No. Negative 10 minus 2 would be negative 12, because you'd be going more negative."}, {"video_title": "Solving a more complicated equation   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Don't get confused. Don't say, OK, 10 plus 2 is 12, so negative 10 plus 2 is negative 12. No. Negative 10 minus 2 would be negative 12, because you'd be going more negative. Here, we have a negative number, but we're going to the right. We're going in the positive direction. So this is negative 8."}, {"video_title": "Solving a more complicated equation   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Negative 10 minus 2 would be negative 12, because you'd be going more negative. Here, we have a negative number, but we're going to the right. We're going in the positive direction. So this is negative 8. So we have 7x is equal to negative 8. So now you might be saying, well, how do I do this type of a problem? I have a negative number here."}, {"video_title": "Solving a more complicated equation   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So this is negative 8. So we have 7x is equal to negative 8. So now you might be saying, well, how do I do this type of a problem? I have a negative number here. You do it the exact same way. If we want to just have an x on the left-hand side, we have to divide the left-hand side by 7, so that the 7x divided by 7, just the 7's cancel out. You're left with x."}, {"video_title": "Solving a more complicated equation   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "I have a negative number here. You do it the exact same way. If we want to just have an x on the left-hand side, we have to divide the left-hand side by 7, so that the 7x divided by 7, just the 7's cancel out. You're left with x. So let's do that. If you divide by 7, those cancel out. But you can't just do it to the left-hand side."}, {"video_title": "Solving a more complicated equation   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "You're left with x. So let's do that. If you divide by 7, those cancel out. But you can't just do it to the left-hand side. Anything you do to the left, you have to do to the right in order for the equality to still hold true. So let's divide the right by 7 as well. And we are left with just an x is equal to negative 8 divided by 7."}, {"video_title": "Solving a more complicated equation   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "But you can't just do it to the left-hand side. Anything you do to the left, you have to do to the right in order for the equality to still hold true. So let's divide the right by 7 as well. And we are left with just an x is equal to negative 8 divided by 7. We could work it out. It'll be some type of a decimal if you were to use a calculator, or you could just leave it in fraction form. Negative 8 divided by 7 is negative 8 sevenths."}, {"video_title": "Solving a more complicated equation   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "And we are left with just an x is equal to negative 8 divided by 7. We could work it out. It'll be some type of a decimal if you were to use a calculator, or you could just leave it in fraction form. Negative 8 divided by 7 is negative 8 sevenths. Or if you want to write it as a mixed number, x is equal to. 7 goes into 8 one time and has a remainder of 1. So it's negative 1 and 1 seventh."}, {"video_title": "Example translating parabola.mp3", "Sentence": "Write the equation for g of x. Now pause this video and see if you can work this out on your own. Alright, so whenever I think about shifting a function, and in this case we're shifting a parabola, I like to look for a distinctive point. And on a parabola, the vertex is going to be our most distinctive point. And if I focus on the vertex of f, it looks like if I shift that to the right by three, and then if I were to shift that down by four, at least our vertices would overlap. I would be able to shift the vertex to where the vertex of g is. And it does look, and we'll validate this at least visually in a little bit, so I'm gonna go minus four in the vertical direction, that not only would it make the vertices overlap, but it would make the entire curve overlap."}, {"video_title": "Example translating parabola.mp3", "Sentence": "And on a parabola, the vertex is going to be our most distinctive point. And if I focus on the vertex of f, it looks like if I shift that to the right by three, and then if I were to shift that down by four, at least our vertices would overlap. I would be able to shift the vertex to where the vertex of g is. And it does look, and we'll validate this at least visually in a little bit, so I'm gonna go minus four in the vertical direction, that not only would it make the vertices overlap, but it would make the entire curve overlap. So we're gonna first shift to the right by three, and we're gonna think about how would we change our equation so it shifts f to the right by three, and then we're gonna shift down by four. Shift down by four. Now, some of you might already be familiar with this, and I go into the intuition in a lot more depth in other videos, but in general, when you shift to the right by some value, in this case we're shifting to the right by three, you would replace x with x minus three."}, {"video_title": "Example translating parabola.mp3", "Sentence": "And it does look, and we'll validate this at least visually in a little bit, so I'm gonna go minus four in the vertical direction, that not only would it make the vertices overlap, but it would make the entire curve overlap. So we're gonna first shift to the right by three, and we're gonna think about how would we change our equation so it shifts f to the right by three, and then we're gonna shift down by four. Shift down by four. Now, some of you might already be familiar with this, and I go into the intuition in a lot more depth in other videos, but in general, when you shift to the right by some value, in this case we're shifting to the right by three, you would replace x with x minus three. So one way to think about this would be y is equal to f of x minus three, or y is equal to, instead of it being x squared, you would replace x with x minus three. So it'd be x minus three squared. Now, when I first learned this, this was counterintuitive."}, {"video_title": "Example translating parabola.mp3", "Sentence": "Now, some of you might already be familiar with this, and I go into the intuition in a lot more depth in other videos, but in general, when you shift to the right by some value, in this case we're shifting to the right by three, you would replace x with x minus three. So one way to think about this would be y is equal to f of x minus three, or y is equal to, instead of it being x squared, you would replace x with x minus three. So it'd be x minus three squared. Now, when I first learned this, this was counterintuitive. I'm shifting to the right by three. The x coordinate of my vertex is increasing by three, but I'm replacing x with x minus three. Why does this make sense?"}, {"video_title": "Example translating parabola.mp3", "Sentence": "Now, when I first learned this, this was counterintuitive. I'm shifting to the right by three. The x coordinate of my vertex is increasing by three, but I'm replacing x with x minus three. Why does this make sense? Well, let's graph the shifted version just to get a little bit more intuition here. Once again, I go into much more depth in other videos here. This is more of a worked example."}, {"video_title": "Example translating parabola.mp3", "Sentence": "Why does this make sense? Well, let's graph the shifted version just to get a little bit more intuition here. Once again, I go into much more depth in other videos here. This is more of a worked example. So this is what the shifted curve is gonna look like. Think about the behavior that we want right over here at x equals three. We want the same value that we used to have when x equals zero."}, {"video_title": "Example translating parabola.mp3", "Sentence": "This is more of a worked example. So this is what the shifted curve is gonna look like. Think about the behavior that we want right over here at x equals three. We want the same value that we used to have when x equals zero. When x equals zero for the original f, zero squared was zero, y equals zero. We still want y equals zero. Well, the way that we can do that is if we are squaring zero, and the way that we're gonna square zero is if we subtract three from x."}, {"video_title": "Example translating parabola.mp3", "Sentence": "We want the same value that we used to have when x equals zero. When x equals zero for the original f, zero squared was zero, y equals zero. We still want y equals zero. Well, the way that we can do that is if we are squaring zero, and the way that we're gonna square zero is if we subtract three from x. And you can validate that at other points. Think about what happens now when x equals four. Four minus three is one squared."}, {"video_title": "Example translating parabola.mp3", "Sentence": "Well, the way that we can do that is if we are squaring zero, and the way that we're gonna square zero is if we subtract three from x. And you can validate that at other points. Think about what happens now when x equals four. Four minus three is one squared. It does indeed equal one. The same behavior that you used to get at x is equal to one. So it does look like we have indeed shifted to the right by three when we replace x with x minus three."}, {"video_title": "Example translating parabola.mp3", "Sentence": "Four minus three is one squared. It does indeed equal one. The same behavior that you used to get at x is equal to one. So it does look like we have indeed shifted to the right by three when we replace x with x minus three. If you replaced x with x plus three, it would have had the opposite effect. You would have shifted to the left by three. I encourage you to think about why that actually makes sense."}, {"video_title": "Example translating parabola.mp3", "Sentence": "So it does look like we have indeed shifted to the right by three when we replace x with x minus three. If you replaced x with x plus three, it would have had the opposite effect. You would have shifted to the left by three. I encourage you to think about why that actually makes sense. So now that we've shifted to the right by three, the next step is to shift down by four. And this one is a little bit more intuitive. So let's start with our shifted to the right."}, {"video_title": "Example translating parabola.mp3", "Sentence": "I encourage you to think about why that actually makes sense. So now that we've shifted to the right by three, the next step is to shift down by four. And this one is a little bit more intuitive. So let's start with our shifted to the right. So that's y is equal to x minus three squared. But now, whatever y value we were getting, we wanna get four less than that. So when x equals three, instead of getting y equals zero, we wanna get y equals four less, or negative four."}, {"video_title": "Example translating parabola.mp3", "Sentence": "So let's start with our shifted to the right. So that's y is equal to x minus three squared. But now, whatever y value we were getting, we wanna get four less than that. So when x equals three, instead of getting y equals zero, we wanna get y equals four less, or negative four. When x equals four, instead of getting one, we wanna get y is equal to negative three. So whatever y value we were getting, we wanna now get four less than that. So the shifting in the vertical direction is a little bit more intuitive."}, {"video_title": "Example translating parabola.mp3", "Sentence": "So when x equals three, instead of getting y equals zero, we wanna get y equals four less, or negative four. When x equals four, instead of getting one, we wanna get y is equal to negative three. So whatever y value we were getting, we wanna now get four less than that. So the shifting in the vertical direction is a little bit more intuitive. If we shift down, we subtract that amount. If we shift up, we add that amount. So this right over here is the equation for g of x. G of x is going to be equal to x minus three squared minus four."}, {"video_title": "Example translating parabola.mp3", "Sentence": "So the shifting in the vertical direction is a little bit more intuitive. If we shift down, we subtract that amount. If we shift up, we add that amount. So this right over here is the equation for g of x. G of x is going to be equal to x minus three squared minus four. And once again, just to review, replacing the x with x minus three on f of x, that's what shifted right by three, by three. And then subtracting the four, that shifted us down by four. Shifted down by four to give us this next graph."}, {"video_title": "Examples of simplifying expressions involving rational numbers  7th grade   Khan Academy.mp3", "Sentence": "And like always, try to pause this video and see if you can simplify this expression before I take a stab at it. Alright, I'm assuming you have attempted it. Now let's look at it. We have negative 5.55 minus 8.55c plus 4.35c. So the first thing I want to do is, can I combine these c terms? And I definitely can. This is if you, we can add negative 8.55c to 4.35c first, and then that would be, let's see, that would be negative 8.55 plus 4.35, I'm just adding the coefficients, times c, and of course we still have that negative 5.55 out front."}, {"video_title": "Examples of simplifying expressions involving rational numbers  7th grade   Khan Academy.mp3", "Sentence": "We have negative 5.55 minus 8.55c plus 4.35c. So the first thing I want to do is, can I combine these c terms? And I definitely can. This is if you, we can add negative 8.55c to 4.35c first, and then that would be, let's see, that would be negative 8.55 plus 4.35, I'm just adding the coefficients, times c, and of course we still have that negative 5.55 out front. Negative 5.55, and I'll just put a plus there. Now how do we calculate negative 8.55 plus 4.35? Well there's a couple of ways to think about it or visualize it."}, {"video_title": "Examples of simplifying expressions involving rational numbers  7th grade   Khan Academy.mp3", "Sentence": "This is if you, we can add negative 8.55c to 4.35c first, and then that would be, let's see, that would be negative 8.55 plus 4.35, I'm just adding the coefficients, times c, and of course we still have that negative 5.55 out front. Negative 5.55, and I'll just put a plus there. Now how do we calculate negative 8.55 plus 4.35? Well there's a couple of ways to think about it or visualize it. One way is to say, well this is the same thing as the negative of 8.55 minus 4.35. And 8.55 minus 4.35, let's see, 8 minus 4 is gonna be the negative, 8 minus 4 is 4, 55 hundredths minus 35 hundredths is 20 hundredths, so I can write 4.20, which is really just the same thing as 4.2. So all of this, all of this can be replaced with a negative 4.2."}, {"video_title": "Examples of simplifying expressions involving rational numbers  7th grade   Khan Academy.mp3", "Sentence": "Well there's a couple of ways to think about it or visualize it. One way is to say, well this is the same thing as the negative of 8.55 minus 4.35. And 8.55 minus 4.35, let's see, 8 minus 4 is gonna be the negative, 8 minus 4 is 4, 55 hundredths minus 35 hundredths is 20 hundredths, so I can write 4.20, which is really just the same thing as 4.2. So all of this, all of this can be replaced with a negative 4.2. So my entire expression has simplified to negative 5.55, and instead of saying plus negative 4.2c, I can just write it as minus 4.2, 4.2c, and we're done. We can't simplify this anymore. We can't add this term that doesn't involve the variable to this term that does involve the variable."}, {"video_title": "Examples of simplifying expressions involving rational numbers  7th grade   Khan Academy.mp3", "Sentence": "So all of this, all of this can be replaced with a negative 4.2. So my entire expression has simplified to negative 5.55, and instead of saying plus negative 4.2c, I can just write it as minus 4.2, 4.2c, and we're done. We can't simplify this anymore. We can't add this term that doesn't involve the variable to this term that does involve the variable. So this is about as simple as we're gonna get. So let's do another example. So here I have these, I have some more hairy numbers involved, these are all expressed as fractions."}, {"video_title": "Examples of simplifying expressions involving rational numbers  7th grade   Khan Academy.mp3", "Sentence": "We can't add this term that doesn't involve the variable to this term that does involve the variable. So this is about as simple as we're gonna get. So let's do another example. So here I have these, I have some more hairy numbers involved, these are all expressed as fractions. And so let's see, I have 2 fifths m, minus 4 fifths, minus 3 fifths m. So how can I simplify? Well I could add all the m terms together. So let me just change the order, I could rewrite this as 2 fifths m, minus 3 fifths m, minus 4 fifths."}, {"video_title": "Examples of simplifying expressions involving rational numbers  7th grade   Khan Academy.mp3", "Sentence": "So here I have these, I have some more hairy numbers involved, these are all expressed as fractions. And so let's see, I have 2 fifths m, minus 4 fifths, minus 3 fifths m. So how can I simplify? Well I could add all the m terms together. So let me just change the order, I could rewrite this as 2 fifths m, minus 3 fifths m, minus 4 fifths. All I did is I changed the order, and we can see that I have these 2 m terms, I can add those two together. So this is going to be 2 fifths, minus 3 fifths, times m, and then I have the minus 4 fifths still on the right hand side. Now what's 2 fifths minus 3 fifths?"}, {"video_title": "Examples of simplifying expressions involving rational numbers  7th grade   Khan Academy.mp3", "Sentence": "So let me just change the order, I could rewrite this as 2 fifths m, minus 3 fifths m, minus 4 fifths. All I did is I changed the order, and we can see that I have these 2 m terms, I can add those two together. So this is going to be 2 fifths, minus 3 fifths, times m, and then I have the minus 4 fifths still on the right hand side. Now what's 2 fifths minus 3 fifths? Well that's gonna be negative 1 fifth. It's gonna be negative 1 fifth. So I have negative 1 fifth m, minus 4 fifths."}, {"video_title": "Examples of simplifying expressions involving rational numbers  7th grade   Khan Academy.mp3", "Sentence": "Now what's 2 fifths minus 3 fifths? Well that's gonna be negative 1 fifth. It's gonna be negative 1 fifth. So I have negative 1 fifth m, minus 4 fifths. Minus 4 fifths. And once again I'm done, I can't simplify it anymore, I can't add this term that involves m somehow to this negative 4 fifths. So we are done here."}, {"video_title": "Examples of simplifying expressions involving rational numbers  7th grade   Khan Academy.mp3", "Sentence": "So I have negative 1 fifth m, minus 4 fifths. Minus 4 fifths. And once again I'm done, I can't simplify it anymore, I can't add this term that involves m somehow to this negative 4 fifths. So we are done here. Let's do one more example. So here, this is interesting, I have a parentheses and all the rest. And like always, pause the video, see if you can simplify this."}, {"video_title": "Examples of simplifying expressions involving rational numbers  7th grade   Khan Academy.mp3", "Sentence": "So we are done here. Let's do one more example. So here, this is interesting, I have a parentheses and all the rest. And like always, pause the video, see if you can simplify this. Alright, let's work through it together. Now the first thing that I want to do is, let's distribute this 2 so that we just have three terms that are just being added and subtracted. So if we distribute this 2, we're gonna get 2 times 1 fifth m is 2 fifths m. Let me make sure you see that m, m is right here."}, {"video_title": "Examples of simplifying expressions involving rational numbers  7th grade   Khan Academy.mp3", "Sentence": "And like always, pause the video, see if you can simplify this. Alright, let's work through it together. Now the first thing that I want to do is, let's distribute this 2 so that we just have three terms that are just being added and subtracted. So if we distribute this 2, we're gonna get 2 times 1 fifth m is 2 fifths m. Let me make sure you see that m, m is right here. 2 times negative 2 fifths is negative 4 fifths, and then I have plus 3 fifths. Now how can we simplify this more? Well I have these two terms here that don't involve the variable, those are just numbers, I can add them to each other."}, {"video_title": "Examples of simplifying expressions involving rational numbers  7th grade   Khan Academy.mp3", "Sentence": "So if we distribute this 2, we're gonna get 2 times 1 fifth m is 2 fifths m. Let me make sure you see that m, m is right here. 2 times negative 2 fifths is negative 4 fifths, and then I have plus 3 fifths. Now how can we simplify this more? Well I have these two terms here that don't involve the variable, those are just numbers, I can add them to each other. So I have negative 4 fifths plus 3 fifths. So what's negative 4 plus 3? That's gonna be negative 1."}, {"video_title": "Examples of simplifying expressions involving rational numbers  7th grade   Khan Academy.mp3", "Sentence": "Well I have these two terms here that don't involve the variable, those are just numbers, I can add them to each other. So I have negative 4 fifths plus 3 fifths. So what's negative 4 plus 3? That's gonna be negative 1. So this is gonna be negative 1 fifth. What we have in yellow here, and I still have the 2 over 5 m. 2 fifths m minus 1 fifth. And we're done, we've simplified that as much as we can."}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "One are these x masses, and we know that they have the same identical mass, and we'll call that identical. Each of them have a mass of x. But then we have this other, this blue thing, and that has a mass of y, which isn't necessarily going to be the same as a mass of x. But when I have two of these x's and a y, it seems like their total mass, y is the case, that their total mass balances out these 8 kilograms right over here. Each of these are 1-kilogram blocks. It balances them out. So the first question I'm going to ask you is, can you express this mathematically?"}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "But when I have two of these x's and a y, it seems like their total mass, y is the case, that their total mass balances out these 8 kilograms right over here. Each of these are 1-kilogram blocks. It balances them out. So the first question I'm going to ask you is, can you express this mathematically? Can you express what we're seeing here, the fact that this total mass balances out with this total mass? Can you express that mathematically? Well, let's just think about our total mass on this side."}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So the first question I'm going to ask you is, can you express this mathematically? Can you express what we're seeing here, the fact that this total mass balances out with this total mass? Can you express that mathematically? Well, let's just think about our total mass on this side. We have two masses of mass x, so those two are going to total at 2x. And then you have a mass of y, so then you're going to have another y. So the total mass on the left-hand side, and let me write it a little bit closer to the center, just so it doesn't get too spread out."}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Well, let's just think about our total mass on this side. We have two masses of mass x, so those two are going to total at 2x. And then you have a mass of y, so then you're going to have another y. So the total mass on the left-hand side, and let me write it a little bit closer to the center, just so it doesn't get too spread out. On the left-hand side, I have 2x plus a mass of y. That's the total mass. The total mass on the left-hand side is 2x plus y."}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So the total mass on the left-hand side, and let me write it a little bit closer to the center, just so it doesn't get too spread out. On the left-hand side, I have 2x plus a mass of y. That's the total mass. The total mass on the left-hand side is 2x plus y. The total mass on the right-hand side is just 8. 1, 2, 3, 4, 5, 6, 7, 8. It is equal to 8."}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "The total mass on the left-hand side is 2x plus y. The total mass on the right-hand side is just 8. 1, 2, 3, 4, 5, 6, 7, 8. It is equal to 8. And since we see that the scale is balanced, this total mass must be equal to this total mass, so we can write an equal sign there. Now my question to you. Is there anything we can do, just based on the information that we have here, to solve for either the mass x or for the mass y?"}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "It is equal to 8. And since we see that the scale is balanced, this total mass must be equal to this total mass, so we can write an equal sign there. Now my question to you. Is there anything we can do, just based on the information that we have here, to solve for either the mass x or for the mass y? Is there anything that we can do? Well, the simple answer is, just with this information here, there is actually very little. You might say, well, let me take a y from both sides."}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Is there anything we can do, just based on the information that we have here, to solve for either the mass x or for the mass y? Is there anything that we can do? Well, the simple answer is, just with this information here, there is actually very little. You might say, well, let me take a y from both sides. You might take this y block up. But if you take this y block up, you have to take away y from this side, and you don't know what y is, and if you think about it algebraically, you might get rid of the y here, subtracting y, but then you're going to have to subtract y from this side too. So you're not going to get rid of the y."}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "You might say, well, let me take a y from both sides. You might take this y block up. But if you take this y block up, you have to take away y from this side, and you don't know what y is, and if you think about it algebraically, you might get rid of the y here, subtracting y, but then you're going to have to subtract y from this side too. So you're not going to get rid of the y. Same thing with the x's. You actually don't have enough information. y depends on what x is, and x depends on what y is."}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So you're not going to get rid of the y. Same thing with the x's. You actually don't have enough information. y depends on what x is, and x depends on what y is. Lucky for us, however, we do have some more of these blocks laying around. And what we do is, we take one of these x blocks, and I stick it over here, and I also take one of the y blocks, and I stick them over here, and I stick it right over there. And then I keep adding these ones until I balance this thing out."}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "y depends on what x is, and x depends on what y is. Lucky for us, however, we do have some more of these blocks laying around. And what we do is, we take one of these x blocks, and I stick it over here, and I also take one of the y blocks, and I stick them over here, and I stick it right over there. And then I keep adding these ones until I balance this thing out. So I keep adding these ones. So obviously, if I just place this, this will go down, because there's nothing on that side. But I keep adding these blocks until it all balances out."}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "And then I keep adding these ones until I balance this thing out. So I keep adding these ones. So obviously, if I just place this, this will go down, because there's nothing on that side. But I keep adding these blocks until it all balances out. And I find that my scale balances once I have 5 kilograms on the right-hand side. So once again, let me ask you this information, that now having an x and a y on the left-hand side, and a 5 kilograms on the right-hand side, and the fact that they're balanced, how can we represent that mathematically? Well, our total mass on the left-hand side is x plus y, and our total mass on the right-hand side is x plus y, and on the right-hand side I have 5 kilograms."}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "But I keep adding these blocks until it all balances out. And I find that my scale balances once I have 5 kilograms on the right-hand side. So once again, let me ask you this information, that now having an x and a y on the left-hand side, and a 5 kilograms on the right-hand side, and the fact that they're balanced, how can we represent that mathematically? Well, our total mass on the left-hand side is x plus y, and our total mass on the right-hand side is x plus y, and on the right-hand side I have 5 kilograms. And we know this is what actually balanced the scale. So these total masses must be equal to each other. And this information by itself, once again, there's nothing I can do with it."}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Well, our total mass on the left-hand side is x plus y, and our total mass on the right-hand side is x plus y, and on the right-hand side I have 5 kilograms. And we know this is what actually balanced the scale. So these total masses must be equal to each other. And this information by itself, once again, there's nothing I can do with it. I don't know what x and y are. If y is 4, maybe x is 1. Maybe x is 4, y is 1. Who knows what these are?"}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "And this information by itself, once again, there's nothing I can do with it. I don't know what x and y are. If y is 4, maybe x is 1. Maybe x is 4, y is 1. Who knows what these are? The interesting thing is we can actually use both of this information to figure out what x and y actually is. And I'll give you a few seconds to think about how we can approach this situation. Well, think about it this way."}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Maybe x is 4, y is 1. Who knows what these are? The interesting thing is we can actually use both of this information to figure out what x and y actually is. And I'll give you a few seconds to think about how we can approach this situation. Well, think about it this way. We know that x plus y is equal to 5. So if we were to get rid of an x and a y on this side, on the left-hand side of the equation, what would we have to get rid of on the right-hand side of the\u2014 If we get rid of x and the y on the left-hand side of the scale, what would we have to get rid of on the right-hand side of the scale to take away the same mass? Well, if we take away an x and a y on the left-hand side, we know that an x plus y is 5 kilograms."}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Well, think about it this way. We know that x plus y is equal to 5. So if we were to get rid of an x and a y on this side, on the left-hand side of the equation, what would we have to get rid of on the right-hand side of the\u2014 If we get rid of x and the y on the left-hand side of the scale, what would we have to get rid of on the right-hand side of the scale to take away the same mass? Well, if we take away an x and a y on the left-hand side, we know that an x plus y is 5 kilograms. So we would just have to take 5 kilograms from the right-hand side. So let's think about what that would do. Well, then I would just have an x left over here, and I would just have some of these masses left over there, and I would know what x is."}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Well, if we take away an x and a y on the left-hand side, we know that an x plus y is 5 kilograms. So we would just have to take 5 kilograms from the right-hand side. So let's think about what that would do. Well, then I would just have an x left over here, and I would just have some of these masses left over there, and I would know what x is. Let's think about how we can represent that algebraically. Essentially, if we're taking an x and a y from the left-hand side, if I'm taking an x and a y from the left-hand side, I'm subtracting an x, I'm subtracting an x, and I am subtracting, actually, let me think of it this way. I am subtracting, I am subtracting an x, I am subtracting an x plus y. I'm subtracting an x and a y from the left-hand side, but then what am I gonna have to do on the right-hand side?"}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Well, then I would just have an x left over here, and I would just have some of these masses left over there, and I would know what x is. Let's think about how we can represent that algebraically. Essentially, if we're taking an x and a y from the left-hand side, if I'm taking an x and a y from the left-hand side, I'm subtracting an x, I'm subtracting an x, and I am subtracting, actually, let me think of it this way. I am subtracting, I am subtracting an x, I am subtracting an x plus y. I'm subtracting an x and a y from the left-hand side, but then what am I gonna have to do on the right-hand side? Well, an x and a y we know has a mass of five, so we can subtract five from the right-hand side. And the only way I'm able to do this is because of the information that we got from the second scale. And so I can take away five, so this is going to be equal to taking away five."}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "I am subtracting, I am subtracting an x, I am subtracting an x plus y. I'm subtracting an x and a y from the left-hand side, but then what am I gonna have to do on the right-hand side? Well, an x and a y we know has a mass of five, so we can subtract five from the right-hand side. And the only way I'm able to do this is because of the information that we got from the second scale. And so I can take away five, so this is going to be equal to taking away five. Taking away an x and a y is the same thing as taking away five, and we know that because an x and a y is equal to five kilograms. And so if we take away an x and a y on the left-hand side, what are we left with? Well, this is going to be the same thing."}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "And so I can take away five, so this is going to be equal to taking away five. Taking away an x and a y is the same thing as taking away five, and we know that because an x and a y is equal to five kilograms. And so if we take away an x and a y on the left-hand side, what are we left with? Well, this is going to be the same thing. Let me rewrite, let me rewrite this part. This taking away an x and a y is the same thing, is the same thing if you distribute the negative sign as taking away an x, taking away an x and taking away a y, and taking away a y. And so on the left-hand side, we are left with, on the left-hand side, we are left with just two x, and we take away one of the x's."}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Well, this is going to be the same thing. Let me rewrite, let me rewrite this part. This taking away an x and a y is the same thing, is the same thing if you distribute the negative sign as taking away an x, taking away an x and taking away a y, and taking away a y. And so on the left-hand side, we are left with, on the left-hand side, we are left with just two x, and we take away one of the x's. We're left with just an x. And we had a y and we took away one of the y's, so we're left with no y's left. And we see that here visually."}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "And so on the left-hand side, we are left with, on the left-hand side, we are left with just two x, and we take away one of the x's. We're left with just an x. And we had a y and we took away one of the y's, so we're left with no y's left. And we see that here visually. We have just an x here. And what do we have on the right-hand side? We had eight, and we knew that an x and a y is equal to five so we took away five, to keep the scale balanced."}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "And we see that here visually. We have just an x here. And what do we have on the right-hand side? We had eight, and we knew that an x and a y is equal to five so we took away five, to keep the scale balanced. And so eight minus five is going to be three. Eight minus five is equal to three. And just like that, using this extra information, we were able to figure out that the mass of x is equal to, the mass of x is equal to three."}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "We had eight, and we knew that an x and a y is equal to five so we took away five, to keep the scale balanced. And so eight minus five is going to be three. Eight minus five is equal to three. And just like that, using this extra information, we were able to figure out that the mass of x is equal to, the mass of x is equal to three. Now, one final question. We were able to figure out what the mass of x is, but can you now figure out what the mass of y is? Well, we can go back to either one of these scales, but it'll probably be simpler to go back to this one."}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "And just like that, using this extra information, we were able to figure out that the mass of x is equal to, the mass of x is equal to three. Now, one final question. We were able to figure out what the mass of x is, but can you now figure out what the mass of y is? Well, we can go back to either one of these scales, but it'll probably be simpler to go back to this one. We know that the mass of x plus the mass of y is equal to five. So you could say, so one thing we know, that x is now equal to three. We know that this is now a three kilogram mass."}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Well, we can go back to either one of these scales, but it'll probably be simpler to go back to this one. We know that the mass of x plus the mass of y is equal to five. So you could say, so one thing we know, that x is now equal to three. We know that this is now a three kilogram mass. We can rewrite this as three, three plus y, three plus y is equal to, is equal to five. Well now we say, well, we could take three away from both sides. If I take three away from the left-hand side, I just have to take three away from the right-hand side to keep my scale balanced, and I'll be left with that the mass of y is balanced with a mass of two, or y is equal to two."}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "We know that this is now a three kilogram mass. We can rewrite this as three, three plus y, three plus y is equal to, is equal to five. Well now we say, well, we could take three away from both sides. If I take three away from the left-hand side, I just have to take three away from the right-hand side to keep my scale balanced, and I'll be left with that the mass of y is balanced with a mass of two, or y is equal to two. That's analogous to taking three from both sides of this equation. And on the left-hand side, I'm just left with a y. And on the right-hand side, I am just left with a two."}, {"video_title": "Why we do the same thing to both sides basic systems   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "If I take three away from the left-hand side, I just have to take three away from the right-hand side to keep my scale balanced, and I'll be left with that the mass of y is balanced with a mass of two, or y is equal to two. That's analogous to taking three from both sides of this equation. And on the left-hand side, I'm just left with a y. And on the right-hand side, I am just left with a two. So x is equal to three kilograms, and y is equal to two kilograms. And what I encourage you to do is verify that it made sense right up here. Figure out what the total mass on the left-hand and the right-hand, or verify what the total mass right over here really was eight to begin with."}, {"video_title": "Explicit and recursive definitions of sequences   Precalculus   Khan Academy.mp3", "Sentence": "And all a sequence is is an ordered list of numbers. So for example, I could have a finite sequence. That means I don't have an infinite number of numbers in it. Or let's say I start at 1 and I keep adding 3. So 1 plus 3 is 4. 4 plus 3 is 7. 7 plus 3 is 10."}, {"video_title": "Explicit and recursive definitions of sequences   Precalculus   Khan Academy.mp3", "Sentence": "Or let's say I start at 1 and I keep adding 3. So 1 plus 3 is 4. 4 plus 3 is 7. 7 plus 3 is 10. And let's say I only have these four terms right over here. So this one we would call a finite sequence. I could also have an infinite sequence."}, {"video_title": "Explicit and recursive definitions of sequences   Precalculus   Khan Academy.mp3", "Sentence": "7 plus 3 is 10. And let's say I only have these four terms right over here. So this one we would call a finite sequence. I could also have an infinite sequence. So an example of an infinite sequence, let's say we start at 3 and we keep adding 4. So go to 3, to 7, to 11, 15. And you don't always have to add the same thing."}, {"video_title": "Explicit and recursive definitions of sequences   Precalculus   Khan Academy.mp3", "Sentence": "I could also have an infinite sequence. So an example of an infinite sequence, let's say we start at 3 and we keep adding 4. So go to 3, to 7, to 11, 15. And you don't always have to add the same thing. We'll explore fancier sequences. The sequences where you keep adding the same amount we call these arithmetic sequences, which we will also explore in more detail. But to show that this is infinite, to show that we keep this pattern going on and on and on, I'll put three dots."}, {"video_title": "Explicit and recursive definitions of sequences   Precalculus   Khan Academy.mp3", "Sentence": "And you don't always have to add the same thing. We'll explore fancier sequences. The sequences where you keep adding the same amount we call these arithmetic sequences, which we will also explore in more detail. But to show that this is infinite, to show that we keep this pattern going on and on and on, I'll put three dots. This just means we're going to keep going on and on and on. So we could call this an infinite sequence. Now, there's a bunch of different notations that seem fancy for denoting sequences, but this is all they refer to."}, {"video_title": "Explicit and recursive definitions of sequences   Precalculus   Khan Academy.mp3", "Sentence": "But to show that this is infinite, to show that we keep this pattern going on and on and on, I'll put three dots. This just means we're going to keep going on and on and on. So we could call this an infinite sequence. Now, there's a bunch of different notations that seem fancy for denoting sequences, but this is all they refer to. But I want to make us comfortable with how we can denote sequences and also how we can define them. So we could say that this right over here is the sequence a sub k for k is going from 1 to 4 is equal to this right over here. So when we look at it this way, we can look at each of these as the terms in the sequence."}, {"video_title": "Explicit and recursive definitions of sequences   Precalculus   Khan Academy.mp3", "Sentence": "Now, there's a bunch of different notations that seem fancy for denoting sequences, but this is all they refer to. But I want to make us comfortable with how we can denote sequences and also how we can define them. So we could say that this right over here is the sequence a sub k for k is going from 1 to 4 is equal to this right over here. So when we look at it this way, we can look at each of these as the terms in the sequence. And this right over here would be the first term. We would call that a sub 1. This right over here would be the second term."}, {"video_title": "Explicit and recursive definitions of sequences   Precalculus   Khan Academy.mp3", "Sentence": "So when we look at it this way, we can look at each of these as the terms in the sequence. And this right over here would be the first term. We would call that a sub 1. This right over here would be the second term. We'd call it a sub 2. I think you get the picture. a sub 3."}, {"video_title": "Explicit and recursive definitions of sequences   Precalculus   Khan Academy.mp3", "Sentence": "This right over here would be the second term. We'd call it a sub 2. I think you get the picture. a sub 3. This right over here is a sub 4. So this just says all of the a sub k's from k equals 1 from our first term all the way to the fourth term. Now, I could also define it by not explicitly writing the sequence like this."}, {"video_title": "Explicit and recursive definitions of sequences   Precalculus   Khan Academy.mp3", "Sentence": "a sub 3. This right over here is a sub 4. So this just says all of the a sub k's from k equals 1 from our first term all the way to the fourth term. Now, I could also define it by not explicitly writing the sequence like this. I could essentially do it using defining our sequence as explicitly using a function notation or something close to function notation. So the same exact sequence, I could define it as a sub k from k equals 1 to 4 with, instead of explicitly writing the numbers here, I could say a sub k is equal to some function of k. So let's see what happens. When k is 1, we get 1."}, {"video_title": "Explicit and recursive definitions of sequences   Precalculus   Khan Academy.mp3", "Sentence": "Now, I could also define it by not explicitly writing the sequence like this. I could essentially do it using defining our sequence as explicitly using a function notation or something close to function notation. So the same exact sequence, I could define it as a sub k from k equals 1 to 4 with, instead of explicitly writing the numbers here, I could say a sub k is equal to some function of k. So let's see what happens. When k is 1, we get 1. When k is 2, we get 4. When k is 3, we get 7. So let's see, when k is 3, we added 3 twice."}, {"video_title": "Explicit and recursive definitions of sequences   Precalculus   Khan Academy.mp3", "Sentence": "When k is 1, we get 1. When k is 2, we get 4. When k is 3, we get 7. So let's see, when k is 3, we added 3 twice. Let me make it clear. So this was a plus 3. This right over here was a plus 3."}, {"video_title": "Explicit and recursive definitions of sequences   Precalculus   Khan Academy.mp3", "Sentence": "So let's see, when k is 3, we added 3 twice. Let me make it clear. So this was a plus 3. This right over here was a plus 3. This right over here is a plus 3. So whatever k is, we started at 1, and we added 3, 1 less than the k term times. So we could say that this is going to be equal to 1 plus k minus 1 times 3."}, {"video_title": "Explicit and recursive definitions of sequences   Precalculus   Khan Academy.mp3", "Sentence": "This right over here was a plus 3. This right over here is a plus 3. So whatever k is, we started at 1, and we added 3, 1 less than the k term times. So we could say that this is going to be equal to 1 plus k minus 1 times 3. Or maybe I should write 3 times k minus 1, same thing. 3 times k minus 1. And you can verify that this works."}, {"video_title": "Explicit and recursive definitions of sequences   Precalculus   Khan Academy.mp3", "Sentence": "So we could say that this is going to be equal to 1 plus k minus 1 times 3. Or maybe I should write 3 times k minus 1, same thing. 3 times k minus 1. And you can verify that this works. So you can verify that this works. If k is equal to 1, you're going to get 1 minus 1 is 0. And so a sub 1 is going to be 1."}, {"video_title": "Explicit and recursive definitions of sequences   Precalculus   Khan Academy.mp3", "Sentence": "And you can verify that this works. So you can verify that this works. If k is equal to 1, you're going to get 1 minus 1 is 0. And so a sub 1 is going to be 1. If k is equal to 2, you're going to have 1 plus 3, which is 4. If k is equal to 3, you get 3 times 2 plus 1 is 7. So it works out."}, {"video_title": "Explicit and recursive definitions of sequences   Precalculus   Khan Academy.mp3", "Sentence": "And so a sub 1 is going to be 1. If k is equal to 2, you're going to have 1 plus 3, which is 4. If k is equal to 3, you get 3 times 2 plus 1 is 7. So it works out. So this is one way to explicitly define our sequence with this function notation. I want to make it clear. I've essentially defined a function here."}, {"video_title": "Explicit and recursive definitions of sequences   Precalculus   Khan Academy.mp3", "Sentence": "So it works out. So this is one way to explicitly define our sequence with this function notation. I want to make it clear. I've essentially defined a function here. If I wanted a more traditional function notation, I could have written a of k, where k is the term that I care about, a of k is equal to 1 plus 3 times k minus 1. This is essentially a function where in allowable input, the domain is restricted to positive integers. Now, how would I denote this business right over here?"}, {"video_title": "Explicit and recursive definitions of sequences   Precalculus   Khan Academy.mp3", "Sentence": "I've essentially defined a function here. If I wanted a more traditional function notation, I could have written a of k, where k is the term that I care about, a of k is equal to 1 plus 3 times k minus 1. This is essentially a function where in allowable input, the domain is restricted to positive integers. Now, how would I denote this business right over here? Well, I could say that this is equal to, and people tend to use a, but I could use the notation b sub k or anything else. But I'll do a again. a sub k. And here we're going from our first term."}, {"video_title": "Explicit and recursive definitions of sequences   Precalculus   Khan Academy.mp3", "Sentence": "Now, how would I denote this business right over here? Well, I could say that this is equal to, and people tend to use a, but I could use the notation b sub k or anything else. But I'll do a again. a sub k. And here we're going from our first term. So this is a sub 1. This is a sub 2, all the way to infinity. Or we could define it, if we wanted to define it explicitly as a function, we could write this sequence as a sub k, where k starts at the first term and goes to infinity with a sub k is equaling."}, {"video_title": "Explicit and recursive definitions of sequences   Precalculus   Khan Academy.mp3", "Sentence": "a sub k. And here we're going from our first term. So this is a sub 1. This is a sub 2, all the way to infinity. Or we could define it, if we wanted to define it explicitly as a function, we could write this sequence as a sub k, where k starts at the first term and goes to infinity with a sub k is equaling. So we're starting at 3. And we are adding 4 one less time. For the second term, we added 4 once."}, {"video_title": "Explicit and recursive definitions of sequences   Precalculus   Khan Academy.mp3", "Sentence": "Or we could define it, if we wanted to define it explicitly as a function, we could write this sequence as a sub k, where k starts at the first term and goes to infinity with a sub k is equaling. So we're starting at 3. And we are adding 4 one less time. For the second term, we added 4 once. For the third term, we add 4 twice. For the fourth term, we add 4 three times. So we're adding 4 one less than the term that we're at."}, {"video_title": "Explicit and recursive definitions of sequences   Precalculus   Khan Academy.mp3", "Sentence": "For the second term, we added 4 once. For the third term, we add 4 twice. For the fourth term, we add 4 three times. So we're adding 4 one less than the term that we're at. So it's going to be plus 4 times k minus 1. 4 times k minus 1. So this is another way of defining this infinite sequence."}, {"video_title": "Explicit and recursive definitions of sequences   Precalculus   Khan Academy.mp3", "Sentence": "So we're adding 4 one less than the term that we're at. So it's going to be plus 4 times k minus 1. 4 times k minus 1. So this is another way of defining this infinite sequence. Now, in both of these cases, I defined it as an explicit function. So this right over here is explicit. That's not an attractive color."}, {"video_title": "Explicit and recursive definitions of sequences   Precalculus   Khan Academy.mp3", "Sentence": "So this is another way of defining this infinite sequence. Now, in both of these cases, I defined it as an explicit function. So this right over here is explicit. That's not an attractive color. Let me write this in. This is an explicit function. And so you might say, well, what's another way of defining these functions?"}, {"video_title": "Explicit and recursive definitions of sequences   Precalculus   Khan Academy.mp3", "Sentence": "That's not an attractive color. Let me write this in. This is an explicit function. And so you might say, well, what's another way of defining these functions? Well, we can also define it, especially something like this. Like an arithmetic sequence, we can also define it recursively. And I want to be clear."}, {"video_title": "Explicit and recursive definitions of sequences   Precalculus   Khan Academy.mp3", "Sentence": "And so you might say, well, what's another way of defining these functions? Well, we can also define it, especially something like this. Like an arithmetic sequence, we can also define it recursively. And I want to be clear. Not every sequence can be defined as either an explicit function like this or as a recursive function. But many can, including this, which is an arithmetic sequence where we keep adding the same quantity over and over again. So how would we do that?"}, {"video_title": "Explicit and recursive definitions of sequences   Precalculus   Khan Academy.mp3", "Sentence": "And I want to be clear. Not every sequence can be defined as either an explicit function like this or as a recursive function. But many can, including this, which is an arithmetic sequence where we keep adding the same quantity over and over again. So how would we do that? Well, we could also, another way of defining this first sequence, we could say a sub k, starting at k equals 1 and going to 4. And when you define things recursively, when you define a sequence recursively, you want to define what your first term is with a sub 1, equaling 1. And then you can define every other term in terms of the term before it."}, {"video_title": "Explicit and recursive definitions of sequences   Precalculus   Khan Academy.mp3", "Sentence": "So how would we do that? Well, we could also, another way of defining this first sequence, we could say a sub k, starting at k equals 1 and going to 4. And when you define things recursively, when you define a sequence recursively, you want to define what your first term is with a sub 1, equaling 1. And then you can define every other term in terms of the term before it. And so then we could write a of k plus, or let me write it this way, a sub k is equal to the previous term. So this is a sub k minus 1. So a given term is equal to the previous term."}, {"video_title": "Explicit and recursive definitions of sequences   Precalculus   Khan Academy.mp3", "Sentence": "And then you can define every other term in terms of the term before it. And so then we could write a of k plus, or let me write it this way, a sub k is equal to the previous term. So this is a sub k minus 1. So a given term is equal to the previous term. Let me make it clear. This is the previous term plus, in this case, we're adding 3 every time. Now, how does this make sense?"}, {"video_title": "Explicit and recursive definitions of sequences   Precalculus   Khan Academy.mp3", "Sentence": "So a given term is equal to the previous term. Let me make it clear. This is the previous term plus, in this case, we're adding 3 every time. Now, how does this make sense? Well, we're defining what a sub 1 is. And if someone says, well, what happens when k equals 2? Well, they're saying, well, it's going to be a sub 2 minus 1."}, {"video_title": "Explicit and recursive definitions of sequences   Precalculus   Khan Academy.mp3", "Sentence": "Now, how does this make sense? Well, we're defining what a sub 1 is. And if someone says, well, what happens when k equals 2? Well, they're saying, well, it's going to be a sub 2 minus 1. So it's going to be a sub 1 plus 3. Well, we know a sub 1 is 1. So it's going to be 1 plus 3, which is 4."}, {"video_title": "Explicit and recursive definitions of sequences   Precalculus   Khan Academy.mp3", "Sentence": "Well, they're saying, well, it's going to be a sub 2 minus 1. So it's going to be a sub 1 plus 3. Well, we know a sub 1 is 1. So it's going to be 1 plus 3, which is 4. Well, what about a sub 3? Well, it's going to be a sub 2 plus 3. a sub 2, we just calculated as 4. You add 3, it's going to be 7."}, {"video_title": "Explicit and recursive definitions of sequences   Precalculus   Khan Academy.mp3", "Sentence": "So it's going to be 1 plus 3, which is 4. Well, what about a sub 3? Well, it's going to be a sub 2 plus 3. a sub 2, we just calculated as 4. You add 3, it's going to be 7. This is essentially what we mentally did when I first wrote out the sequence, when I said, hey, I'm just going to start with 1. I'm just going to start with 1. And I'm just going to add 3 for every successive term."}, {"video_title": "Explicit and recursive definitions of sequences   Precalculus   Khan Academy.mp3", "Sentence": "You add 3, it's going to be 7. This is essentially what we mentally did when I first wrote out the sequence, when I said, hey, I'm just going to start with 1. I'm just going to start with 1. And I'm just going to add 3 for every successive term. So how would we do this one? Well, once again, we could write this as a sub k, starting at k, the first term, going to infinity, with our first term, a sub 1, is going to be 3 now. And every successive term, a sub k, is going to be the previous term, a sub k minus 1 plus 4."}, {"video_title": "Explicit and recursive definitions of sequences   Precalculus   Khan Academy.mp3", "Sentence": "And I'm just going to add 3 for every successive term. So how would we do this one? Well, once again, we could write this as a sub k, starting at k, the first term, going to infinity, with our first term, a sub 1, is going to be 3 now. And every successive term, a sub k, is going to be the previous term, a sub k minus 1 plus 4. And once again, you start at 3. And then if you want the second term, it's going to be the first term plus 4. It's going to be 3 plus 4."}, {"video_title": "Explicit and recursive definitions of sequences   Precalculus   Khan Academy.mp3", "Sentence": "And every successive term, a sub k, is going to be the previous term, a sub k minus 1 plus 4. And once again, you start at 3. And then if you want the second term, it's going to be the first term plus 4. It's going to be 3 plus 4. You get to 7, and you keep adding 4. So both of these, this right over here, is a recursive definition. We started with kind of a base case."}, {"video_title": "Introduction to exponential decay.mp3", "Sentence": "So let's review exponential growth. Let's say we have something that, and I'll do this on a table here. Oops, let me make that straight. So let's say this is our x and this is our y. Now let's say when x is zero, y is equal to three. And every time we increase x by one, we double y. So y is gonna go from three to six."}, {"video_title": "Introduction to exponential decay.mp3", "Sentence": "So let's say this is our x and this is our y. Now let's say when x is zero, y is equal to three. And every time we increase x by one, we double y. So y is gonna go from three to six. If x increases by one again, so we go to two, then we're gonna double y again. And so six times two is 12. This right over here is exponential growth."}, {"video_title": "Introduction to exponential decay.mp3", "Sentence": "So y is gonna go from three to six. If x increases by one again, so we go to two, then we're gonna double y again. And so six times two is 12. This right over here is exponential growth. And you could even go for negative x's. When x is negative one, well if we're going back one in x, we would divide by two. So this is going to be 3 1\u20442, 3 1\u20442."}, {"video_title": "Introduction to exponential decay.mp3", "Sentence": "This right over here is exponential growth. And you could even go for negative x's. When x is negative one, well if we're going back one in x, we would divide by two. So this is going to be 3 1\u20442, 3 1\u20442. And notice, if you go from negative one to zero, you once again, you keep multiplying by two and this will keep on happening. And you can describe this with an equation. You could say that y is equal to, and sometimes people might call this your y-intercept or your initial value, is equal to three."}, {"video_title": "Introduction to exponential decay.mp3", "Sentence": "So this is going to be 3 1\u20442, 3 1\u20442. And notice, if you go from negative one to zero, you once again, you keep multiplying by two and this will keep on happening. And you can describe this with an equation. You could say that y is equal to, and sometimes people might call this your y-intercept or your initial value, is equal to three. Essentially what happens when x equals zero is equal to three times our common ratio. And our common ratio is, well what are we multiplying by every time we increase x by one? So three times our common ratio to the x power."}, {"video_title": "Introduction to exponential decay.mp3", "Sentence": "You could say that y is equal to, and sometimes people might call this your y-intercept or your initial value, is equal to three. Essentially what happens when x equals zero is equal to three times our common ratio. And our common ratio is, well what are we multiplying by every time we increase x by one? So three times our common ratio to the x power. And you can verify that, pick any of these. When x is equal to two, it's gonna be three times two squared, which is three times four, which is indeed equal to 12. And we can see that on a graph."}, {"video_title": "Introduction to exponential decay.mp3", "Sentence": "So three times our common ratio to the x power. And you can verify that, pick any of these. When x is equal to two, it's gonna be three times two squared, which is three times four, which is indeed equal to 12. And we can see that on a graph. So let me draw a quick graph right over here. So I'm having trouble drawing a straight line. All right, there we go."}, {"video_title": "Introduction to exponential decay.mp3", "Sentence": "And we can see that on a graph. So let me draw a quick graph right over here. So I'm having trouble drawing a straight line. All right, there we go. And let's see, we could go, and they're gonna be on a slightly different scale, my x and y-axes. So this is x-axis, y-axis, and we go from negative one to one to two. And so we're going all the way up to 12."}, {"video_title": "Introduction to exponential decay.mp3", "Sentence": "All right, there we go. And let's see, we could go, and they're gonna be on a slightly different scale, my x and y-axes. So this is x-axis, y-axis, and we go from negative one to one to two. And so we're going all the way up to 12. So let's say that this is three, six, nine, and let's say this is 12. And we could just plot these points here. When x is negative one, y is 3 1\u20442."}, {"video_title": "Introduction to exponential decay.mp3", "Sentence": "And so we're going all the way up to 12. So let's say that this is three, six, nine, and let's say this is 12. And we could just plot these points here. When x is negative one, y is 3 1\u20442. So it looks like that. Then at y equals zero, when x is zero, y is three. When x equals one, y has doubled, it's now at six."}, {"video_title": "Introduction to exponential decay.mp3", "Sentence": "When x is negative one, y is 3 1\u20442. So it looks like that. Then at y equals zero, when x is zero, y is three. When x equals one, y has doubled, it's now at six. When x is equal to two, y is 12. And you will see this telltale curve. And so there's a couple of key features that we've, well, we've already talked about several of them."}, {"video_title": "Introduction to exponential decay.mp3", "Sentence": "When x equals one, y has doubled, it's now at six. When x is equal to two, y is 12. And you will see this telltale curve. And so there's a couple of key features that we've, well, we've already talked about several of them. But if you go to increasingly negative x values, you will asymptote towards the x-axis. It'll never quite get to zero as you get to more and more negative values, but it'll definitely approach it. And as you get to more and more positive values, it just kind of skyrockets up."}, {"video_title": "Introduction to exponential decay.mp3", "Sentence": "And so there's a couple of key features that we've, well, we've already talked about several of them. But if you go to increasingly negative x values, you will asymptote towards the x-axis. It'll never quite get to zero as you get to more and more negative values, but it'll definitely approach it. And as you get to more and more positive values, it just kind of skyrockets up. We always, we've talked about in previous videos how this will pass up any linear function or any linear graph eventually. Now, let's compare that to exponential decay. Exponential, exponential decay."}, {"video_title": "Introduction to exponential decay.mp3", "Sentence": "And as you get to more and more positive values, it just kind of skyrockets up. We always, we've talked about in previous videos how this will pass up any linear function or any linear graph eventually. Now, let's compare that to exponential decay. Exponential, exponential decay. And an easy way to think about it, instead of growing every time you're increasing x, you're going to shrink by about, by a certain amount. You are going to decay. So let's set up another table here with x and y values."}, {"video_title": "Introduction to exponential decay.mp3", "Sentence": "Exponential, exponential decay. And an easy way to think about it, instead of growing every time you're increasing x, you're going to shrink by about, by a certain amount. You are going to decay. So let's set up another table here with x and y values. That was really a very, I'm supposed to, this is supposed to, when I press shift, it should create a straight line, but my computer, I've been eating next to my computer. Maybe there's crumbs in the keyboard or something. All right, so here we go."}, {"video_title": "Introduction to exponential decay.mp3", "Sentence": "So let's set up another table here with x and y values. That was really a very, I'm supposed to, this is supposed to, when I press shift, it should create a straight line, but my computer, I've been eating next to my computer. Maybe there's crumbs in the keyboard or something. All right, so here we go. We have x and we have y. And so let's start with, let's say we start in the same place. So when x is zero, y is three."}, {"video_title": "Introduction to exponential decay.mp3", "Sentence": "All right, so here we go. We have x and we have y. And so let's start with, let's say we start in the same place. So when x is zero, y is three. But instead of doubling every time we increase x by one, let's go by half every time we increase x by one. So when x is equal to one, we're gonna multiply by 1 1.5, and so we're going to get to 3 1.5. And then when x is equal to two, we'll multiply by 1 1.5 again, and so we're going to get to three, we're going to get to 3 4."}, {"video_title": "Introduction to exponential decay.mp3", "Sentence": "So when x is zero, y is three. But instead of doubling every time we increase x by one, let's go by half every time we increase x by one. So when x is equal to one, we're gonna multiply by 1 1.5, and so we're going to get to 3 1.5. And then when x is equal to two, we'll multiply by 1 1.5 again, and so we're going to get to three, we're going to get to 3 4. And so on and so forth. And if we were to go to negative values, when x is equal to negative one, well, to go, if we're going backwards in x by one, we would divide by 1 1.5, and so we would get to six. Or going from negative one to zero, as we increase x by one, once again, we're multiplying by 1 1.5."}, {"video_title": "Introduction to exponential decay.mp3", "Sentence": "And then when x is equal to two, we'll multiply by 1 1.5 again, and so we're going to get to three, we're going to get to 3 4. And so on and so forth. And if we were to go to negative values, when x is equal to negative one, well, to go, if we're going backwards in x by one, we would divide by 1 1.5, and so we would get to six. Or going from negative one to zero, as we increase x by one, once again, we're multiplying by 1 1.5. And so how would we write this as an equation? I encourage you to pause the video and see if you can write it in a similar way. Well, it's gonna look something like this."}, {"video_title": "Introduction to exponential decay.mp3", "Sentence": "Or going from negative one to zero, as we increase x by one, once again, we're multiplying by 1 1.5. And so how would we write this as an equation? I encourage you to pause the video and see if you can write it in a similar way. Well, it's gonna look something like this. It's gonna be y is equal to, you have your, you could have your y-intercept here, the value of y when x is equal to zero, so it's three, times, what's our common ratio now? Well, every time we increase x by one, we're multiplying by 1 1.5. So 1 1.5, and we're gonna raise that to the x power."}, {"video_title": "Introduction to exponential decay.mp3", "Sentence": "Well, it's gonna look something like this. It's gonna be y is equal to, you have your, you could have your y-intercept here, the value of y when x is equal to zero, so it's three, times, what's our common ratio now? Well, every time we increase x by one, we're multiplying by 1 1.5. So 1 1.5, and we're gonna raise that to the x power. And so notice, these are both exponentials. We have some, you could say y-intercept or initial value, and it's being multiplied by some common ratio to the power x, some common ratio to the power x. But notice, when you're growing, our common ratio, and it actually turns out to be a general idea, when you're growing, your common ratio, the absolute value of your common ratio is going to be greater than one."}, {"video_title": "Introduction to exponential decay.mp3", "Sentence": "So 1 1.5, and we're gonna raise that to the x power. And so notice, these are both exponentials. We have some, you could say y-intercept or initial value, and it's being multiplied by some common ratio to the power x, some common ratio to the power x. But notice, when you're growing, our common ratio, and it actually turns out to be a general idea, when you're growing, your common ratio, the absolute value of your common ratio is going to be greater than one. Let me write it then. So the absolute value of two in this case is greater than one but when you're shrinking, the absolute value of it is less than one. And that makes sense because if you have something where the absolute value is less than one, like 1 1.5 or 3 4ths or 0.9, every time you multiply it, you're gonna get a lower and lower and lower value."}, {"video_title": "Introduction to exponential decay.mp3", "Sentence": "But notice, when you're growing, our common ratio, and it actually turns out to be a general idea, when you're growing, your common ratio, the absolute value of your common ratio is going to be greater than one. Let me write it then. So the absolute value of two in this case is greater than one but when you're shrinking, the absolute value of it is less than one. And that makes sense because if you have something where the absolute value is less than one, like 1 1.5 or 3 4ths or 0.9, every time you multiply it, you're gonna get a lower and lower and lower value. And you could actually see that in a graph. Let's graph the same information right over here. And let me do it in a different color."}, {"video_title": "Introduction to exponential decay.mp3", "Sentence": "And that makes sense because if you have something where the absolute value is less than one, like 1 1.5 or 3 4ths or 0.9, every time you multiply it, you're gonna get a lower and lower and lower value. And you could actually see that in a graph. Let's graph the same information right over here. And let me do it in a different color. I'll do it in a blue color. So when x is equal to negative one, y is equal to six. When x is equal to zero, y is equal to three."}, {"video_title": "Introduction to exponential decay.mp3", "Sentence": "And let me do it in a different color. I'll do it in a blue color. So when x is equal to negative one, y is equal to six. When x is equal to zero, y is equal to three. When x is equal to one, y is equal to 3 1.5. When x is equal to two, y is equal to 3 4ths, and so on and so forth. And notice, because our common ratios are the reciprocal of each other, that these two graphs look like they've been flipped over, they look like they've been flipped horizontally or flipped over the y-axis."}, {"video_title": "Introduction to exponential decay.mp3", "Sentence": "When x is equal to zero, y is equal to three. When x is equal to one, y is equal to 3 1.5. When x is equal to two, y is equal to 3 4ths, and so on and so forth. And notice, because our common ratios are the reciprocal of each other, that these two graphs look like they've been flipped over, they look like they've been flipped horizontally or flipped over the y-axis. They're symmetric around that y-axis. And what you will see in exponential decay is that things will get smaller and smaller and smaller, but they'll never quite exactly get to zero. It'll approach zero."}, {"video_title": "Introduction to exponential decay.mp3", "Sentence": "And notice, because our common ratios are the reciprocal of each other, that these two graphs look like they've been flipped over, they look like they've been flipped horizontally or flipped over the y-axis. They're symmetric around that y-axis. And what you will see in exponential decay is that things will get smaller and smaller and smaller, but they'll never quite exactly get to zero. It'll approach zero. It'll asymptote towards the x-axis as x becomes more and more positive. Just as for exponential growth, if x becomes more and more and more negative, we asymptote towards the x-axis. So that's the introduction."}, {"video_title": "Introduction to exponential decay.mp3", "Sentence": "It'll approach zero. It'll asymptote towards the x-axis as x becomes more and more positive. Just as for exponential growth, if x becomes more and more and more negative, we asymptote towards the x-axis. So that's the introduction. I'd use a very specific example, but in general, if you have an equation of the form y is equal to a times some common ratio to the x power, and we could write it like that just to make it a little bit clearer. Well, there's a bunch of different ways that we could write it. This is going to be exponential growth."}, {"video_title": "Introduction to exponential decay.mp3", "Sentence": "So that's the introduction. I'd use a very specific example, but in general, if you have an equation of the form y is equal to a times some common ratio to the x power, and we could write it like that just to make it a little bit clearer. Well, there's a bunch of different ways that we could write it. This is going to be exponential growth. So if the absolute value of r is greater than one, then we're dealing with growth because every time you multiply, every time you increase x, you're multiplying by more and more r's is one way to think about it. And if the absolute value of r is less than one, you're dealing with decay. You are shrinking as x increases."}, {"video_title": "Introduction to exponential decay.mp3", "Sentence": "This is going to be exponential growth. So if the absolute value of r is greater than one, then we're dealing with growth because every time you multiply, every time you increase x, you're multiplying by more and more r's is one way to think about it. And if the absolute value of r is less than one, you're dealing with decay. You are shrinking as x increases. And I'll let you think about what happens when r is equal to one. What are we dealing with in that situation? And it's a bit of a trick question because it's actually quite straight."}, {"video_title": "Introduction to exponential decay.mp3", "Sentence": "You are shrinking as x increases. And I'll let you think about what happens when r is equal to one. What are we dealing with in that situation? And it's a bit of a trick question because it's actually quite straight. Oh, I'll just tell you. If r is equal to one, well, then this thing right over here is always going to be equal to one, and you boil down to just the constant equation y is equal to a. So this would just be a horizontal line."}, {"video_title": "Talking bird solves systems with substitution   Algebra II   Khan Academy.mp3", "Sentence": "And the king says, well, the bird says that he thinks that there's another way to do the problem. And you're not used to taking advice from birds, and so being a little bit defensive, you say, well, if the bird thinks he knows so much, let him do this problem. And so the bird whispers a little bit more in the king's ear and says, okay, well, I'll have to do the writing because the bird does not have any hands, or at least can't manipulate chalk. And so the king, the bird continues to whisper in the king's ear, and the king translates and says, well, the bird says, let's use one of these equations to solve for a variable. So let's say, let's use this blue equation right over here to solve for a variable. And that's essentially going to be a constraint of one variable in terms of another. So let's see if we can do that."}, {"video_title": "Talking bird solves systems with substitution   Algebra II   Khan Academy.mp3", "Sentence": "And so the king, the bird continues to whisper in the king's ear, and the king translates and says, well, the bird says, let's use one of these equations to solve for a variable. So let's say, let's use this blue equation right over here to solve for a variable. And that's essentially going to be a constraint of one variable in terms of another. So let's see if we can do that. So here, if we want to solve for m, we could subtract 400w from both sides, and we would have 100m is, if we subtract 400w from the left, this 400w goes away. If we subtract 400w from the right, we have is equal to negative 400w plus 1,100. So what got us from here to here is just subtracting 400w from both sides."}, {"video_title": "Talking bird solves systems with substitution   Algebra II   Khan Academy.mp3", "Sentence": "So let's see if we can do that. So here, if we want to solve for m, we could subtract 400w from both sides, and we would have 100m is, if we subtract 400w from the left, this 400w goes away. If we subtract 400w from the right, we have is equal to negative 400w plus 1,100. So what got us from here to here is just subtracting 400w from both sides. And if we want to solve for m, we just divide both sides by 100. So we just divide all of the terms by 100, and then we get m is equal to, m is equal to negative 400 divided by 100 is negative 4w. 1,100 divided by 100 is 11, plus 11."}, {"video_title": "Talking bird solves systems with substitution   Algebra II   Khan Academy.mp3", "Sentence": "So what got us from here to here is just subtracting 400w from both sides. And if we want to solve for m, we just divide both sides by 100. So we just divide all of the terms by 100, and then we get m is equal to, m is equal to negative 400 divided by 100 is negative 4w. 1,100 divided by 100 is 11, plus 11. So now we've constrained m in terms of w, and this is what the bird is saying, using the king as his translator. Why don't we take this constraint and substitute it back for m in the first equation, and then we will have one equation with one unknown. And so the king starts to write at the bird's direction, 200, 200, so he's looking at that first equation now."}, {"video_title": "Talking bird solves systems with substitution   Algebra II   Khan Academy.mp3", "Sentence": "1,100 divided by 100 is 11, plus 11. So now we've constrained m in terms of w, and this is what the bird is saying, using the king as his translator. Why don't we take this constraint and substitute it back for m in the first equation, and then we will have one equation with one unknown. And so the king starts to write at the bird's direction, 200, 200, so he's looking at that first equation now. He says 200, 200, and instead of putting an m there, the bird says, well, m, by the second constraint, m is equal to negative 4w plus 11. So instead of writing an m, we substitute the value for m, we substitute for m the expression negative 4w plus 11. So negative 4w plus 11, and then we have the rest of it, plus 300w, plus 300w is equal to 1,200."}, {"video_title": "Talking bird solves systems with substitution   Algebra II   Khan Academy.mp3", "Sentence": "And so the king starts to write at the bird's direction, 200, 200, so he's looking at that first equation now. He says 200, 200, and instead of putting an m there, the bird says, well, m, by the second constraint, m is equal to negative 4w plus 11. So instead of writing an m, we substitute the value for m, we substitute for m the expression negative 4w plus 11. So negative 4w plus 11, and then we have the rest of it, plus 300w, plus 300w is equal to 1,200. So just to be clear, everywhere we saw an m, we replaced it with this right over here in that first equation. So the first thing, you start to scratch your head, and say, is this a legitimate thing to do? Will I get the same answer as I got when I solved the same problem with elimination?"}, {"video_title": "Talking bird solves systems with substitution   Algebra II   Khan Academy.mp3", "Sentence": "So negative 4w plus 11, and then we have the rest of it, plus 300w, plus 300w is equal to 1,200. So just to be clear, everywhere we saw an m, we replaced it with this right over here in that first equation. So the first thing, you start to scratch your head, and say, is this a legitimate thing to do? Will I get the same answer as I got when I solved the same problem with elimination? And I want you to sit and think about that for a second. But then the bird starts whispering in the king's ear, and the king just progresses to just work through the algebra. This is one equation with one unknown now."}, {"video_title": "Talking bird solves systems with substitution   Algebra II   Khan Academy.mp3", "Sentence": "Will I get the same answer as I got when I solved the same problem with elimination? And I want you to sit and think about that for a second. But then the bird starts whispering in the king's ear, and the king just progresses to just work through the algebra. This is one equation with one unknown now. So the first step would be to distribute the 200. So 200 times negative 4w is negative 800w. 200 times 11 is 2,200, plus 2,200."}, {"video_title": "Talking bird solves systems with substitution   Algebra II   Khan Academy.mp3", "Sentence": "This is one equation with one unknown now. So the first step would be to distribute the 200. So 200 times negative 4w is negative 800w. 200 times 11 is 2,200, plus 2,200. And then we have the plus 300w, plus 300w is equal to positive 1,200. Now we just need to solve for w. We first might wanna group this negative 800w with this 300w. Negative 800 of something plus 300 of something is going to be negative 500w."}, {"video_title": "Talking bird solves systems with substitution   Algebra II   Khan Academy.mp3", "Sentence": "200 times 11 is 2,200, plus 2,200. And then we have the plus 300w, plus 300w is equal to positive 1,200. Now we just need to solve for w. We first might wanna group this negative 800w with this 300w. Negative 800 of something plus 300 of something is going to be negative 500w. And then we still have this plus 2,200. Plus 2,200 is equal to 1,200. Now to solve for w, we'd wanna subtract 2,200 from both sides so subtract 2,200, subtract 2,200."}, {"video_title": "Talking bird solves systems with substitution   Algebra II   Khan Academy.mp3", "Sentence": "Negative 800 of something plus 300 of something is going to be negative 500w. And then we still have this plus 2,200. Plus 2,200 is equal to 1,200. Now to solve for w, we'd wanna subtract 2,200 from both sides so subtract 2,200, subtract 2,200. On the left-hand side, you're left just with the negative 500w, negative 500w. And on the right-hand side, you are left with negative 1,000. And this is starting to look interesting because if we divide both sides by negative 500, we get w is equal to 2, which is the exact same answer that we got when we tried to solve, when we tried to figure out how many bags of chips each woman would eat on average."}, {"video_title": "Talking bird solves systems with substitution   Algebra II   Khan Academy.mp3", "Sentence": "Now to solve for w, we'd wanna subtract 2,200 from both sides so subtract 2,200, subtract 2,200. On the left-hand side, you're left just with the negative 500w, negative 500w. And on the right-hand side, you are left with negative 1,000. And this is starting to look interesting because if we divide both sides by negative 500, we get w is equal to 2, which is the exact same answer that we got when we tried to solve, when we tried to figure out how many bags of chips each woman would eat on average. When we tried to solve it with elimination, we got the exact right answer. So for at least for this example, it seems like the substitution method that this bird came up with worked just as well as the elimination method that you had originally done the first time you wanted to figure out the potato chip conundrum. And if now you actually wanted to figure out how many chips the men would eat, well, you could do exactly the same thing you did last time."}, {"video_title": "Solving a two-step inequality   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "We have the inequality 2 thirds is greater than negative 4y minus 8 and 1 third. Now the first thing I want to do here, just because mixed numbers bother me, they're actually hard to deal with mathematically. They're easy to think about, oh it's a little bit more than 8. Let's convert this to a improper fraction. So 8 and 1 third is equal to, the denominator is going to be 3. 3 times 8 is 24, plus 1 is 25. So this thing over here is the same thing as 25 over 3."}, {"video_title": "Solving a two-step inequality   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Let's convert this to a improper fraction. So 8 and 1 third is equal to, the denominator is going to be 3. 3 times 8 is 24, plus 1 is 25. So this thing over here is the same thing as 25 over 3. Let me just rewrite the whole thing. So it's 2 thirds is greater than negative 4y minus 25 over 3. Now the next thing I want to do, just because dealing with fractions are a bit of a pain, is multiply both sides of this inequality by some quantity that will eliminate the fractions."}, {"video_title": "Solving a two-step inequality   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So this thing over here is the same thing as 25 over 3. Let me just rewrite the whole thing. So it's 2 thirds is greater than negative 4y minus 25 over 3. Now the next thing I want to do, just because dealing with fractions are a bit of a pain, is multiply both sides of this inequality by some quantity that will eliminate the fractions. And the easiest one I can think of is multiply both sides by 3. That will get rid of the 3's in the denominator. So let's multiply both sides of this equation by 3."}, {"video_title": "Solving a two-step inequality   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Now the next thing I want to do, just because dealing with fractions are a bit of a pain, is multiply both sides of this inequality by some quantity that will eliminate the fractions. And the easiest one I can think of is multiply both sides by 3. That will get rid of the 3's in the denominator. So let's multiply both sides of this equation by 3. That's the left hand side. And then I'm going to multiply the right hand side. 3, I'll put in parentheses like that."}, {"video_title": "Solving a two-step inequality   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So let's multiply both sides of this equation by 3. That's the left hand side. And then I'm going to multiply the right hand side. 3, I'll put in parentheses like that. And, well one point that I want to point out is that I did not have to swap the inequality sign because I multiplied both sides by a positive number. If the 3 was a negative number, if I multiplied both sides by negative 3 or negative 1 or negative whatever, I would have had to swap the inequality sign. Anyway, let's simplify this."}, {"video_title": "Solving a two-step inequality   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "3, I'll put in parentheses like that. And, well one point that I want to point out is that I did not have to swap the inequality sign because I multiplied both sides by a positive number. If the 3 was a negative number, if I multiplied both sides by negative 3 or negative 1 or negative whatever, I would have had to swap the inequality sign. Anyway, let's simplify this. So the left hand side, we have 3 times 2 thirds, which is just 2. 2 is greater than, and then we can distribute this 3. 3 times negative 4y is negative 12y."}, {"video_title": "Solving a two-step inequality   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Anyway, let's simplify this. So the left hand side, we have 3 times 2 thirds, which is just 2. 2 is greater than, and then we can distribute this 3. 3 times negative 4y is negative 12y. And then 3 times negative 25 over 3 is just negative 25. Now, we want to get all of our constant terms on one side of the inequality and all of our variable terms. The only variable here is y on the other side."}, {"video_title": "Solving a two-step inequality   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "3 times negative 4y is negative 12y. And then 3 times negative 25 over 3 is just negative 25. Now, we want to get all of our constant terms on one side of the inequality and all of our variable terms. The only variable here is y on the other side. The y is already sitting here, so let's guess it's 25 on the other side of the inequality. And we can do that by adding 25 to both sides of this equation. So let's add 25 to both sides of this equation."}, {"video_title": "Solving a two-step inequality   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "The only variable here is y on the other side. The y is already sitting here, so let's guess it's 25 on the other side of the inequality. And we can do that by adding 25 to both sides of this equation. So let's add 25 to both sides of this equation. Adding 25. And the left hand side, 2 plus 25 is 27. And we're going to get 27 is greater than."}, {"video_title": "Solving a two-step inequality   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So let's add 25 to both sides of this equation. Adding 25. And the left hand side, 2 plus 25 is 27. And we're going to get 27 is greater than. The right hand side of the inequality is negative 12y. And the negative 25 plus 25, those cancel out. That was the whole point."}, {"video_title": "Solving a two-step inequality   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "And we're going to get 27 is greater than. The right hand side of the inequality is negative 12y. And the negative 25 plus 25, those cancel out. That was the whole point. So we're left with 27 is greater than negative 12y. Now, to isolate the y, we can multiply, or you can either multiply both sides by negative 1 twelfth, or you could say let's just divide both sides by negative 12. Now, because I'm multiplying or dividing by a negative number here, I'm going to need to swap the inequality."}, {"video_title": "Solving a two-step inequality   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "That was the whole point. So we're left with 27 is greater than negative 12y. Now, to isolate the y, we can multiply, or you can either multiply both sides by negative 1 twelfth, or you could say let's just divide both sides by negative 12. Now, because I'm multiplying or dividing by a negative number here, I'm going to need to swap the inequality. So let me write this. If I divide both sides of this equation by negative 12, then it becomes 27 over negative 12 is less than. I'm swapping the inequality."}, {"video_title": "Solving a two-step inequality   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Now, because I'm multiplying or dividing by a negative number here, I'm going to need to swap the inequality. So let me write this. If I divide both sides of this equation by negative 12, then it becomes 27 over negative 12 is less than. I'm swapping the inequality. Let me do this in a different color. Is less than negative 12y over negative 12. Notice, when I divide both sides of the inequality by a negative number, I swap the inequality."}, {"video_title": "Solving a two-step inequality   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "I'm swapping the inequality. Let me do this in a different color. Is less than negative 12y over negative 12. Notice, when I divide both sides of the inequality by a negative number, I swap the inequality. I swap the greater than becomes a less than. When it was positive, I didn't have to swap it. So 27 divided by negative 12, well, they're both divisible by 3."}, {"video_title": "Solving a two-step inequality   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Notice, when I divide both sides of the inequality by a negative number, I swap the inequality. I swap the greater than becomes a less than. When it was positive, I didn't have to swap it. So 27 divided by negative 12, well, they're both divisible by 3. So we're going to get, if we do the numerator and the denominator by 3, we get negative 9 over 4 is less than. These cancel out. y."}, {"video_title": "Solving a two-step inequality   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So 27 divided by negative 12, well, they're both divisible by 3. So we're going to get, if we do the numerator and the denominator by 3, we get negative 9 over 4 is less than. These cancel out. y. So y is greater than negative 9 fourths, or negative 9 fourths is less than y. And if you wanted to write that, let me write this. So our answer is y is greater than negative 9 fourths."}, {"video_title": "Solving a two-step inequality   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "y. So y is greater than negative 9 fourths, or negative 9 fourths is less than y. And if you wanted to write that, let me write this. So our answer is y is greater than negative 9 fourths. I just swapped the order. You can say negative 9 fourths is less than y. Or if you want to visualize that a little bit better, 9 fourths is 2 and 1 fourth."}, {"video_title": "Solving a two-step inequality   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So our answer is y is greater than negative 9 fourths. I just swapped the order. You can say negative 9 fourths is less than y. Or if you want to visualize that a little bit better, 9 fourths is 2 and 1 fourth. So we could also say y is greater than negative 2 and 1 fourth if we want to put it as a mixed number. And if we wanted to graph it on the number line, let me draw a number line right here, a real simple one. Maybe this is 0."}, {"video_title": "Solving a two-step inequality   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Or if you want to visualize that a little bit better, 9 fourths is 2 and 1 fourth. So we could also say y is greater than negative 2 and 1 fourth if we want to put it as a mixed number. And if we wanted to graph it on the number line, let me draw a number line right here, a real simple one. Maybe this is 0. Negative 2 is right over, let's say negative 1. Negative 2. Then say negative 3 is right there."}, {"video_title": "Solving a two-step inequality   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Maybe this is 0. Negative 2 is right over, let's say negative 1. Negative 2. Then say negative 3 is right there. Negative 2 and 1 fourth is going to be right here. And it's greater than. So we're not going to include that in the solution set."}, {"video_title": "Why a negative times a negative makes intuitive sense   Pre-Algebra   Khan Academy.mp3", "Sentence": "So you, as the ancient philosopher mathematician, have concluded that in order for the multiplication of positive and negative numbers to be consistent with everything that you've been constructing so far, all of the other properties of multiplication that you know so far, that you need a negative number times a positive number, or a positive times a negative, to give you a negative number, and a negative times a negative to give you a positive number. And so you accept it, so it's all consistent, but it still does not make complete, concrete sense to you. You want to have a slightly deeper intuition than just having to accept it so that it's consistent with the distributive property and whatever else. And so you try another thought experiment. You say, well, what is just basic multiplication doing? So if I say two times, if I say two times three, one way to conceptualize this basic multiplication here is really repeated addition. So you could view this as two threes, so two threes would literally be three plus three, and notice there are two of them, there are two of these, or you could view this as three twos."}, {"video_title": "Why a negative times a negative makes intuitive sense   Pre-Algebra   Khan Academy.mp3", "Sentence": "And so you try another thought experiment. You say, well, what is just basic multiplication doing? So if I say two times, if I say two times three, one way to conceptualize this basic multiplication here is really repeated addition. So you could view this as two threes, so two threes would literally be three plus three, and notice there are two of them, there are two of these, or you could view this as three twos. And so this is the same thing as two plus two plus two, and there are three of them. And either way you conceptualize them, you get the same exact answer. This is going to be equal to six."}, {"video_title": "Why a negative times a negative makes intuitive sense   Pre-Algebra   Khan Academy.mp3", "Sentence": "So you could view this as two threes, so two threes would literally be three plus three, and notice there are two of them, there are two of these, or you could view this as three twos. And so this is the same thing as two plus two plus two, and there are three of them. And either way you conceptualize them, you get the same exact answer. This is going to be equal to six. Fair enough, you knew that before you even tried to tackle negative numbers. Now let's try to make one of these negatives and see what, one of these negative and see what happens. Let's do two, two times, two times negative three."}, {"video_title": "Why a negative times a negative makes intuitive sense   Pre-Algebra   Khan Academy.mp3", "Sentence": "This is going to be equal to six. Fair enough, you knew that before you even tried to tackle negative numbers. Now let's try to make one of these negatives and see what, one of these negative and see what happens. Let's do two, two times, two times negative three. And I'm going to make the negative in a different color. Two times negative, two times negative three. Well, one way you could view this is just the same analogy here."}, {"video_title": "Why a negative times a negative makes intuitive sense   Pre-Algebra   Khan Academy.mp3", "Sentence": "Let's do two, two times, two times negative three. And I'm going to make the negative in a different color. Two times negative, two times negative three. Well, one way you could view this is just the same analogy here. It's negative three twice. So it would be, it would be negative three, and I'll try to color code it, negative three, and then another negative, another negative three, or you could say negative three minus three. Or, and now this is the interesting thing, instead of over here, since you had two times positive three, you added two three times, but since here it's two times negative three, you could also imagine that you're going to subtract two three times."}, {"video_title": "Why a negative times a negative makes intuitive sense   Pre-Algebra   Khan Academy.mp3", "Sentence": "Well, one way you could view this is just the same analogy here. It's negative three twice. So it would be, it would be negative three, and I'll try to color code it, negative three, and then another negative, another negative three, or you could say negative three minus three. Or, and now this is the interesting thing, instead of over here, since you had two times positive three, you added two three times, but since here it's two times negative three, you could also imagine that you're going to subtract two three times. So instead of, and up here I could have literally written plus two, plus two, plus two, because this is a positive three right over here, all of a sudden since we're dealing with a negative three, we could imagine that this is subtracting two three times. So this would be subtracting two, subtracting two, mind another, subtract another two right over here, subtract another two, and then you subtract another two, another two right over there. Let me color code it, make sure I don't get the colors messed up."}, {"video_title": "Why a negative times a negative makes intuitive sense   Pre-Algebra   Khan Academy.mp3", "Sentence": "Or, and now this is the interesting thing, instead of over here, since you had two times positive three, you added two three times, but since here it's two times negative three, you could also imagine that you're going to subtract two three times. So instead of, and up here I could have literally written plus two, plus two, plus two, because this is a positive three right over here, all of a sudden since we're dealing with a negative three, we could imagine that this is subtracting two three times. So this would be subtracting two, subtracting two, mind another, subtract another two right over here, subtract another two, and then you subtract another two, another two right over there. Let me color code it, make sure I don't get the colors messed up. And you have another two right over there. And notice, you did it, once again, you did it three times. So this was a negative three, essentially you are subtracting two three times."}, {"video_title": "Why a negative times a negative makes intuitive sense   Pre-Algebra   Khan Academy.mp3", "Sentence": "Let me color code it, make sure I don't get the colors messed up. And you have another two right over there. And notice, you did it, once again, you did it three times. So this was a negative three, essentially you are subtracting two three times. And either way you conceptualize it right over here, you are going to get negative six. You are going to get negative six as the answer. Now, so you're already starting to feel good about this part right over here, negative times a positive, negative times a positive, or a positive times a negative, give you a negative."}, {"video_title": "Why a negative times a negative makes intuitive sense   Pre-Algebra   Khan Academy.mp3", "Sentence": "So this was a negative three, essentially you are subtracting two three times. And either way you conceptualize it right over here, you are going to get negative six. You are going to get negative six as the answer. Now, so you're already starting to feel good about this part right over here, negative times a positive, negative times a positive, or a positive times a negative, give you a negative. Now let's take to the really unintuitive one. A negative times a negative, all of a sudden the negatives kind of cancel out and give you a positive. Why is that the case?"}, {"video_title": "Why a negative times a negative makes intuitive sense   Pre-Algebra   Khan Academy.mp3", "Sentence": "Now, so you're already starting to feel good about this part right over here, negative times a positive, negative times a positive, or a positive times a negative, give you a negative. Now let's take to the really unintuitive one. A negative times a negative, all of a sudden the negatives kind of cancel out and give you a positive. Why is that the case? Well, we can just build from this example right over here. Let's say that we had negative two, let's say that we had negative two, let me do it in a different color. Let's say that we had negative two, I already used that color, negative two times negative three, times negative three."}, {"video_title": "Why a negative times a negative makes intuitive sense   Pre-Algebra   Khan Academy.mp3", "Sentence": "Why is that the case? Well, we can just build from this example right over here. Let's say that we had negative two, let's say that we had negative two, let me do it in a different color. Let's say that we had negative two, I already used that color, negative two times negative three, times negative three. So now we can do it a couple of ways. Actually, I'll do this one first. We're still multiplying something by negative three, so we're going to repeatedly subtract that thing three times, whatever that thing is."}, {"video_title": "Why a negative times a negative makes intuitive sense   Pre-Algebra   Khan Academy.mp3", "Sentence": "Let's say that we had negative two, I already used that color, negative two times negative three, times negative three. So now we can do it a couple of ways. Actually, I'll do this one first. We're still multiplying something by negative three, so we're going to repeatedly subtract that thing three times, whatever that thing is. But now that thing isn't a positive two, that thing that we're going to subtract three times is a negative two. So let me make it clear. This says we're going to subtract something three times."}, {"video_title": "Why a negative times a negative makes intuitive sense   Pre-Algebra   Khan Academy.mp3", "Sentence": "We're still multiplying something by negative three, so we're going to repeatedly subtract that thing three times, whatever that thing is. But now that thing isn't a positive two, that thing that we're going to subtract three times is a negative two. So let me make it clear. This says we're going to subtract something three times. We're going to subtract something three times. So subtracting something, subtracting something, subtracting something three times. That's what this part right over here tells us."}, {"video_title": "Why a negative times a negative makes intuitive sense   Pre-Algebra   Khan Academy.mp3", "Sentence": "This says we're going to subtract something three times. We're going to subtract something three times. So subtracting something, subtracting something, subtracting something three times. That's what this part right over here tells us. And we're going to do this, we're going to do it exactly three times. Over here was a positive two that we subtracted three times. Now we're going to do a negative two."}, {"video_title": "Why a negative times a negative makes intuitive sense   Pre-Algebra   Khan Academy.mp3", "Sentence": "That's what this part right over here tells us. And we're going to do this, we're going to do it exactly three times. Over here was a positive two that we subtracted three times. Now we're going to do a negative two. Now we're going to do a negative two. And we know from subtracting negative numbers, we've already built this intuition that subtracting a negative is the same thing, it's like taking away someone's debt. It's the same thing as adding a positive."}, {"video_title": "Why a negative times a negative makes intuitive sense   Pre-Algebra   Khan Academy.mp3", "Sentence": "Now we're going to do a negative two. Now we're going to do a negative two. And we know from subtracting negative numbers, we've already built this intuition that subtracting a negative is the same thing, it's like taking away someone's debt. It's the same thing as adding a positive. And so this is going to be the same thing as two plus two plus two, which will once again give you positive six. You can use the same logic over here. Now instead of adding negative three twice, and really I could have written this as negative three in this example, negative three, let me write it this way, negative three, negative three, and we added it, we added it, and I'll write a plus out here to make it clear."}, {"video_title": "Why a negative times a negative makes intuitive sense   Pre-Algebra   Khan Academy.mp3", "Sentence": "It's the same thing as adding a positive. And so this is going to be the same thing as two plus two plus two, which will once again give you positive six. You can use the same logic over here. Now instead of adding negative three twice, and really I could have written this as negative three in this example, negative three, let me write it this way, negative three, negative three, and we added it, we added it, and I'll write a plus out here to make it clear. Over here we added it twice. We added negative three two times. Over here, since we now have a negative two, we're going to subtract negative three twice."}, {"video_title": "Why a negative times a negative makes intuitive sense   Pre-Algebra   Khan Academy.mp3", "Sentence": "Now instead of adding negative three twice, and really I could have written this as negative three in this example, negative three, let me write it this way, negative three, negative three, and we added it, we added it, and I'll write a plus out here to make it clear. Over here we added it twice. We added negative three two times. Over here, since we now have a negative two, we're going to subtract negative three twice. So we're going to subtract something, and we're going to subtract something again, and that something is going to be our negative three. It's going to be our negative three. So negative, negative, and put our three right over here."}, {"video_title": "Why a negative times a negative makes intuitive sense   Pre-Algebra   Khan Academy.mp3", "Sentence": "Over here, since we now have a negative two, we're going to subtract negative three twice. So we're going to subtract something, and we're going to subtract something again, and that something is going to be our negative three. It's going to be our negative three. So negative, negative, and put our three right over here. And once again, subtracting a negative, it's like taking away someone's debt, which is essentially giving them money, this is the same thing as adding three plus three, which is once again six. So now you as the ancient philosopher feel pretty good. Not only is this all consistent with all of the mathematics you know, the distributive property, the associative property, multiplying something times zero, all of these things that you already know, it now actually makes conceptual sense to you, that this actually is very consistent with your notion, your original notion, or one of the possible notions of multiplication, which is as repeated addition."}, {"video_title": "Powers of products & quotients (integer exponents)   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "And I want you to pause this video and see if you could simplify this on your own. So the key realization here, there's a couple of ways that you can tackle it, but the key thing to realize is if you have the product of two things and then you're raising that to some type of a exponent, that is going to be the same thing as raising each of these things to that exponent and then taking the product. So this is going to be the same thing as three to the negative eight and then that to the negative two times seven to the third to the negative two. So I'll do seven to the third right over here. And if I wanna simplify this, three to the negative eight to the negative two, we have the other exponent property that if you're raising to an exponent and then raising that whole thing to another exponent, then you can just multiply the exponents. So this is going to be three to the negative eight times negative two power. Well, negative eight times negative two is positive 16."}, {"video_title": "Powers of products & quotients (integer exponents)   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "So I'll do seven to the third right over here. And if I wanna simplify this, three to the negative eight to the negative two, we have the other exponent property that if you're raising to an exponent and then raising that whole thing to another exponent, then you can just multiply the exponents. So this is going to be three to the negative eight times negative two power. Well, negative eight times negative two is positive 16. So this is gonna be three to the 16th power right over there. And then this part right over here, seven to the third to the negative two, that's gonna be seven to the three times negative two, which is seven to the negative sixth power. So that is seven to the negative sixth."}, {"video_title": "Powers of products & quotients (integer exponents)   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "Well, negative eight times negative two is positive 16. So this is gonna be three to the 16th power right over there. And then this part right over here, seven to the third to the negative two, that's gonna be seven to the three times negative two, which is seven to the negative sixth power. So that is seven to the negative sixth. And this would be about as much as you could simplify it. You could rewrite it different ways. Seven to the negative sixth, the same thing as one over seven to the sixth."}, {"video_title": "Powers of products & quotients (integer exponents)   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "So that is seven to the negative sixth. And this would be about as much as you could simplify it. You could rewrite it different ways. Seven to the negative sixth, the same thing as one over seven to the sixth. So you could write it like three to the 16th. Let me use that same shade of blue. Three to the 16th over seven to the sixth."}, {"video_title": "Powers of products & quotients (integer exponents)   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "Seven to the negative sixth, the same thing as one over seven to the sixth. So you could write it like three to the 16th. Let me use that same shade of blue. Three to the 16th over seven to the sixth. But these two are equivalent. And there's other ways that you could have tackled this. You could have said that this original thing right over here this is the same thing as three to the negative eight is the same thing as one over three to the eighth."}, {"video_title": "Powers of products & quotients (integer exponents)   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "Three to the 16th over seven to the sixth. But these two are equivalent. And there's other ways that you could have tackled this. You could have said that this original thing right over here this is the same thing as three to the negative eight is the same thing as one over three to the eighth. So you could have said this is the same thing as seven to the third over three to the eighth. And then you're raising that to the negative two, in which case you would raise this numerator to the negative two and the denominator to negative two. But you would have gotten to the exact same place."}, {"video_title": "Powers of products & quotients (integer exponents)   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "You could have said that this original thing right over here this is the same thing as three to the negative eight is the same thing as one over three to the eighth. So you could have said this is the same thing as seven to the third over three to the eighth. And then you're raising that to the negative two, in which case you would raise this numerator to the negative two and the denominator to negative two. But you would have gotten to the exact same place. Let's do another one of these. So let's say, let me, so let's say that we have, we have got a to the negative two times eight to the seventh power. We wanna raise all of that to the second power."}, {"video_title": "Powers of products & quotients (integer exponents)   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "But you would have gotten to the exact same place. Let's do another one of these. So let's say, let me, so let's say that we have, we have got a to the negative two times eight to the seventh power. We wanna raise all of that to the second power. Well, like before, I can raise each of these things to the second power. So this is the same thing as a to the negative two to the second power times this thing to the second power, eight to the seventh to the second power. And then here, negative two times two is negative four."}, {"video_title": "Powers of products & quotients (integer exponents)   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "We wanna raise all of that to the second power. Well, like before, I can raise each of these things to the second power. So this is the same thing as a to the negative two to the second power times this thing to the second power, eight to the seventh to the second power. And then here, negative two times two is negative four. So that's a to the negative four times eight to the seven times two is 14, eight to the 14th power. In other videos, we go into more depth about why this should hopefully make intuitive sense. Here you have eight to the seventh times eight to the seventh where you would then add the two exponents and you would get to eight to the 14th."}, {"video_title": "Powers of products & quotients (integer exponents)   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "And then here, negative two times two is negative four. So that's a to the negative four times eight to the seven times two is 14, eight to the 14th power. In other videos, we go into more depth about why this should hopefully make intuitive sense. Here you have eight to the seventh times eight to the seventh where you would then add the two exponents and you would get to eight to the 14th. So however many times you have eight to the seventh, you would just keep adding the exponents or you would multiply by seven that many times. Hopefully that didn't sound too confusing. But the general idea is if you raise something to an exponent and then another exponent, you can multiply those exponents."}, {"video_title": "Powers of products & quotients (integer exponents)   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "Here you have eight to the seventh times eight to the seventh where you would then add the two exponents and you would get to eight to the 14th. So however many times you have eight to the seventh, you would just keep adding the exponents or you would multiply by seven that many times. Hopefully that didn't sound too confusing. But the general idea is if you raise something to an exponent and then another exponent, you can multiply those exponents. Let's do one more example where we are dealing with quotients, which that first example could have even been perceived as. So let's say we have, let's say we have two to the negative 10 divided by four squared, and we're gonna raise all of that to the seventh power. Well, this is equivalent to two to the negative 10 raised to the seventh power over four squared raised to the seventh power."}, {"video_title": "Powers of products & quotients (integer exponents)   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "But the general idea is if you raise something to an exponent and then another exponent, you can multiply those exponents. Let's do one more example where we are dealing with quotients, which that first example could have even been perceived as. So let's say we have, let's say we have two to the negative 10 divided by four squared, and we're gonna raise all of that to the seventh power. Well, this is equivalent to two to the negative 10 raised to the seventh power over four squared raised to the seventh power. So if you have the difference of two things and you're raising it to some power, that's the same thing as the numerator raised to that power divided by the denominator raised to that power. Well, what's our numerator going to be? Well, we've done this drill before."}, {"video_title": "Powers of products & quotients (integer exponents)   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "Well, this is equivalent to two to the negative 10 raised to the seventh power over four squared raised to the seventh power. So if you have the difference of two things and you're raising it to some power, that's the same thing as the numerator raised to that power divided by the denominator raised to that power. Well, what's our numerator going to be? Well, we've done this drill before. It'd be two to the negative 10 times seventh power. So this would be equal to two to the negative 70th power. And then in the denominator, four to the second power, then that raised to the seventh power, well, two times seven is 14."}, {"video_title": "Powers of products & quotients (integer exponents)   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "Well, we've done this drill before. It'd be two to the negative 10 times seventh power. So this would be equal to two to the negative 70th power. And then in the denominator, four to the second power, then that raised to the seventh power, well, two times seven is 14. So that's going to be four to the 17th, four to the 17th power. Now, we actually could think about simplifying this even more. There's multiple ways that you could rewrite this."}, {"video_title": "Powers of products & quotients (integer exponents)   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "And then in the denominator, four to the second power, then that raised to the seventh power, well, two times seven is 14. So that's going to be four to the 17th, four to the 17th power. Now, we actually could think about simplifying this even more. There's multiple ways that you could rewrite this. But one thing you could do is say, hey, look, four is a power of two. So you could rewrite this as, this is equal to two to the negative 70th power over, instead of writing four, instead of writing four to the 17th power, why did I write 17th power? It should be four to the 14th power."}, {"video_title": "Powers of products & quotients (integer exponents)   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "There's multiple ways that you could rewrite this. But one thing you could do is say, hey, look, four is a power of two. So you could rewrite this as, this is equal to two to the negative 70th power over, instead of writing four, instead of writing four to the 17th power, why did I write 17th power? It should be four to the 14th power. Let me correct that. Instead of writing four to the 14th power, I instead could write, so this is two, let me get the colors right. This is two to the negative 70th over, over, instead of writing four, I could write two squared to the 14th power."}, {"video_title": "Powers of products & quotients (integer exponents)   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "It should be four to the 14th power. Let me correct that. Instead of writing four to the 14th power, I instead could write, so this is two, let me get the colors right. This is two to the negative 70th over, over, instead of writing four, I could write two squared to the 14th power. Four is the same thing as two squared. And so now I can rewrite this whole thing as two to the negative 70th power over, well, two to the second, and then that to the 14th, well, that's two to the 28th power. Two to the 28th power."}, {"video_title": "Powers of products & quotients (integer exponents)   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "This is two to the negative 70th over, over, instead of writing four, I could write two squared to the 14th power. Four is the same thing as two squared. And so now I can rewrite this whole thing as two to the negative 70th power over, well, two to the second, and then that to the 14th, well, that's two to the 28th power. Two to the 28th power. And so can I simplify this even more? Well, this is going to be equal to two to the, I can just, if I'm taking a quotient with the same base, I can subtract the exponents. So it's gonna be negative 70."}, {"video_title": "Powers of products & quotients (integer exponents)   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "Two to the 28th power. And so can I simplify this even more? Well, this is going to be equal to two to the, I can just, if I'm taking a quotient with the same base, I can subtract the exponents. So it's gonna be negative 70. It's going to be negative 70 minus 28th power. Minus 28. And so this is going to simplify two to the negative 98th power."}, {"video_title": "Find measure of vertical angles   Angles and intersecting lines   Geometry   Khan Academy.mp3", "Sentence": "So that's one of the lines right over there. And then I have another line right over here. So those are my two intersecting lines. And let's say we know that the measure of this angle right over here is equal to 7x plus 182. And this is being given in degrees. So it's 7x plus 182 degrees. And we know that the measure of this angle right over here is 9x plus 194 degrees."}, {"video_title": "Find measure of vertical angles   Angles and intersecting lines   Geometry   Khan Academy.mp3", "Sentence": "And let's say we know that the measure of this angle right over here is equal to 7x plus 182. And this is being given in degrees. So it's 7x plus 182 degrees. And we know that the measure of this angle right over here is 9x plus 194 degrees. So my question to you is, what is the measure of each of these angles? And I encourage you to pause the video and to think about it. Well, the thing that might jump out at you is that these two things are vertical angles."}, {"video_title": "Find measure of vertical angles   Angles and intersecting lines   Geometry   Khan Academy.mp3", "Sentence": "And we know that the measure of this angle right over here is 9x plus 194 degrees. So my question to you is, what is the measure of each of these angles? And I encourage you to pause the video and to think about it. Well, the thing that might jump out at you is that these two things are vertical angles. They're the opposite angles when we have these intersecting lines right over here. And vertical angles are equal to each other. So we know, because these are vertical angles, that 9x plus 194 degrees must be equal to 7x plus 182 degrees."}, {"video_title": "Find measure of vertical angles   Angles and intersecting lines   Geometry   Khan Academy.mp3", "Sentence": "Well, the thing that might jump out at you is that these two things are vertical angles. They're the opposite angles when we have these intersecting lines right over here. And vertical angles are equal to each other. So we know, because these are vertical angles, that 9x plus 194 degrees must be equal to 7x plus 182 degrees. And now we just have to solve for x. So if we want all the x terms on the left-hand side, we could subtract 7x from here. We've got to do it to both sides, of course, in order to maintain the equality."}, {"video_title": "Find measure of vertical angles   Angles and intersecting lines   Geometry   Khan Academy.mp3", "Sentence": "So we know, because these are vertical angles, that 9x plus 194 degrees must be equal to 7x plus 182 degrees. And now we just have to solve for x. So if we want all the x terms on the left-hand side, we could subtract 7x from here. We've got to do it to both sides, of course, in order to maintain the equality. And then we could put all of our constant terms on the right-hand side. So we can subtract 194 from the left. We have to subtract 194 from the right in order to maintain the inequality."}, {"video_title": "Find measure of vertical angles   Angles and intersecting lines   Geometry   Khan Academy.mp3", "Sentence": "We've got to do it to both sides, of course, in order to maintain the equality. And then we could put all of our constant terms on the right-hand side. So we can subtract 194 from the left. We have to subtract 194 from the right in order to maintain the inequality. And on the left, what we're left with is just 2x. And on the right, what we're left with, let's see, 182 minus 194. So if it was 194 minus 182, it would be positive 12."}, {"video_title": "Find measure of vertical angles   Angles and intersecting lines   Geometry   Khan Academy.mp3", "Sentence": "We have to subtract 194 from the right in order to maintain the inequality. And on the left, what we're left with is just 2x. And on the right, what we're left with, let's see, 182 minus 194. So if it was 194 minus 182, it would be positive 12. But now it's going to be negative 12. We're subtracting the larger from the smaller. So it's equal to negative 12."}, {"video_title": "Find measure of vertical angles   Angles and intersecting lines   Geometry   Khan Academy.mp3", "Sentence": "So if it was 194 minus 182, it would be positive 12. But now it's going to be negative 12. We're subtracting the larger from the smaller. So it's equal to negative 12. And then divide both sides by 2. And we get x is equal to negative 6. And now we can use that information to find out the measure of either one of these angles, which is the same as the other one."}, {"video_title": "Find measure of vertical angles   Angles and intersecting lines   Geometry   Khan Academy.mp3", "Sentence": "So it's equal to negative 12. And then divide both sides by 2. And we get x is equal to negative 6. And now we can use that information to find out the measure of either one of these angles, which is the same as the other one. So we can see here that if we take 7 times negative 6 plus 182, so 7 times negative 6 is negative 42, plus 182 is going to be equal to 140 degrees. And you'll see the same thing over here. If we say 9 times negative 6, which is negative 54 plus 194, this also equals 140 degrees."}, {"video_title": "How to write basic expressions with variables   6th grade   Khan Academy.mp3", "Sentence": "Well, you could just have A, but if you want 11 more than A, you would want to add 11. So you could write that as A plus 11. You could also write that as 11 plus A. Both of them would be 11 more than A. So let's check our answer here. We got it right. Let's do a few more of these."}, {"video_title": "How to write basic expressions with variables   6th grade   Khan Academy.mp3", "Sentence": "Both of them would be 11 more than A. So let's check our answer here. We got it right. Let's do a few more of these. Write an expression to represent the sum of D and nine. So the sum of D and nine. That means you're gonna add D and nine."}, {"video_title": "How to write basic expressions with variables   6th grade   Khan Academy.mp3", "Sentence": "Let's do a few more of these. Write an expression to represent the sum of D and nine. So the sum of D and nine. That means you're gonna add D and nine. So I could write that as D plus nine, or I could write that as nine plus nine plus D. Check our answer. Got that right. Let's do a few more of these."}, {"video_title": "How to write basic expressions with variables   6th grade   Khan Academy.mp3", "Sentence": "That means you're gonna add D and nine. So I could write that as D plus nine, or I could write that as nine plus nine plus D. Check our answer. Got that right. Let's do a few more of these. Write an expression to represent J minus 15. Well, I could just write it with math symbols instead of writing the word minus. Instead of writing M-I-N-U-S, I could write J minus 15."}, {"video_title": "How to write basic expressions with variables   6th grade   Khan Academy.mp3", "Sentence": "Let's do a few more of these. Write an expression to represent J minus 15. Well, I could just write it with math symbols instead of writing the word minus. Instead of writing M-I-N-U-S, I could write J minus 15. And then I check my answer. Got it right. Let's do a few more of these."}, {"video_title": "How to write basic expressions with variables   6th grade   Khan Academy.mp3", "Sentence": "Instead of writing M-I-N-U-S, I could write J minus 15. And then I check my answer. Got it right. Let's do a few more of these. This is a lot of fun. Write an expression to represent seven times R. Well, seven, there's a couple ways I could do it. I could use this little dot right over here."}, {"video_title": "How to write basic expressions with variables   6th grade   Khan Academy.mp3", "Sentence": "Let's do a few more of these. This is a lot of fun. Write an expression to represent seven times R. Well, seven, there's a couple ways I could do it. I could use this little dot right over here. I could do seven times R like that. That would be correct. I could literally just write seven R. If I just wrote seven R, that would also count."}, {"video_title": "How to write basic expressions with variables   6th grade   Khan Academy.mp3", "Sentence": "I could use this little dot right over here. I could do seven times R like that. That would be correct. I could literally just write seven R. If I just wrote seven R, that would also count. Let me check my answer. That's right. Let me do a couple of other of these just so you can see that I could have just done 10."}, {"video_title": "How to write basic expressions with variables   6th grade   Khan Academy.mp3", "Sentence": "I could literally just write seven R. If I just wrote seven R, that would also count. Let me check my answer. That's right. Let me do a couple of other of these just so you can see that I could have just done 10. And this is not a decimal. This is, you know, it's a little bit higher than a decimal. It's multiplication."}, {"video_title": "How to write basic expressions with variables   6th grade   Khan Academy.mp3", "Sentence": "Let me do a couple of other of these just so you can see that I could have just done 10. And this is not a decimal. This is, you know, it's a little bit higher than a decimal. It's multiplication. And the reason why once you start doing algebra, you use this symbol instead of that kind of cross for multiplication is that that X-looking thing gets confused with X when you're using X as a variable. So that's why this is a lot more useful. So we want to write 10 times U."}, {"video_title": "How to write basic expressions with variables   6th grade   Khan Academy.mp3", "Sentence": "It's multiplication. And the reason why once you start doing algebra, you use this symbol instead of that kind of cross for multiplication is that that X-looking thing gets confused with X when you're using X as a variable. So that's why this is a lot more useful. So we want to write 10 times U. 10 times U. Let's check our answer. Got it right."}, {"video_title": "How to write basic expressions with variables   6th grade   Khan Academy.mp3", "Sentence": "So we want to write 10 times U. 10 times U. Let's check our answer. Got it right. Let's do one more. Write an expression to represent eight divided by D. So you could write it as eight, and then I could write a slash like that. Eight divided by D. And there you go."}, {"video_title": "How to write basic expressions with variables   6th grade   Khan Academy.mp3", "Sentence": "Got it right. Let's do one more. Write an expression to represent eight divided by D. So you could write it as eight, and then I could write a slash like that. Eight divided by D. And there you go. This is eight divided by D. Let me check the answer. I'll do one more of these. Well, six divided by B."}, {"video_title": "How to write basic expressions with variables   6th grade   Khan Academy.mp3", "Sentence": "Eight divided by D. And there you go. This is eight divided by D. Let me check the answer. I'll do one more of these. Well, six divided by B. All right, same thing. So six, I could use this tool right over here. It does the same thing as if I were to press the backslash."}, {"video_title": "How to write basic expressions with variables   6th grade   Khan Academy.mp3", "Sentence": "Well, six divided by B. All right, same thing. So six, I could use this tool right over here. It does the same thing as if I were to press the backslash. So six divided by B. Check my answer. Got it right."}, {"video_title": "Example 2 Variables on both sides   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So the way I like to do these is we just like to separate the constant terms, which are the 20 and the negative 6 on one side of the equation. I'll put them on the right hand side. And then we'll put all the x terms, the negative 7x and the 6x, we'll put it all on the left hand side. So to get the 20 out of the way from the left hand side, let's subtract it from the left hand side. But this is an equation. Anything you do to the left hand side, you also have to do to the right hand side. If that is equal to that, in order for them to still be equal, anything I do to the left hand side, I have to do to the right hand side."}, {"video_title": "Example 2 Variables on both sides   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So to get the 20 out of the way from the left hand side, let's subtract it from the left hand side. But this is an equation. Anything you do to the left hand side, you also have to do to the right hand side. If that is equal to that, in order for them to still be equal, anything I do to the left hand side, I have to do to the right hand side. So I subtracted 20 from the left. Let me also subtract 20 from the right. And so the left hand side of the equation, 20 minus 20 is just 0."}, {"video_title": "Example 2 Variables on both sides   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "If that is equal to that, in order for them to still be equal, anything I do to the left hand side, I have to do to the right hand side. So I subtracted 20 from the left. Let me also subtract 20 from the right. And so the left hand side of the equation, 20 minus 20 is just 0. That was the whole point. They cancel out. Don't have to write it down."}, {"video_title": "Example 2 Variables on both sides   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "And so the left hand side of the equation, 20 minus 20 is just 0. That was the whole point. They cancel out. Don't have to write it down. And then I have a negative 7x, just gets carried down. And then that is equal to the right hand side of the equation. I have a 6x."}, {"video_title": "Example 2 Variables on both sides   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Don't have to write it down. And then I have a negative 7x, just gets carried down. And then that is equal to the right hand side of the equation. I have a 6x. I'm not adding or subtracting anything to that. But then I have a negative 6 minus 20. So if I'm already 6 below 0 on the number line and I go another 20 below that, that's at negative 26."}, {"video_title": "Example 2 Variables on both sides   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "I have a 6x. I'm not adding or subtracting anything to that. But then I have a negative 6 minus 20. So if I'm already 6 below 0 on the number line and I go another 20 below that, that's at negative 26. Now the next thing we want to do is we want to get all of the x terms on the left hand side. So we don't want this 6x here. So maybe we subtract 6x from both sides."}, {"video_title": "Example 2 Variables on both sides   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So if I'm already 6 below 0 on the number line and I go another 20 below that, that's at negative 26. Now the next thing we want to do is we want to get all of the x terms on the left hand side. So we don't want this 6x here. So maybe we subtract 6x from both sides. So let's subtract 6x from the right. Subtract 6x from the left. And what do we get?"}, {"video_title": "Example 2 Variables on both sides   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So maybe we subtract 6x from both sides. So let's subtract 6x from the right. Subtract 6x from the left. And what do we get? The left hand side, negative 7x minus 6x, that's negative 13x. Negative 7 of something minus another 6 of that something is going to be negative 13 of that something. And that is going to be equal to 6x minus 6x."}, {"video_title": "Example 2 Variables on both sides   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "And what do we get? The left hand side, negative 7x minus 6x, that's negative 13x. Negative 7 of something minus another 6 of that something is going to be negative 13 of that something. And that is going to be equal to 6x minus 6x. That cancels out. That was the whole point behind subtracting negative 6x. And then we have just a negative 26 or minus 26, depending on how you want to view it."}, {"video_title": "Example 2 Variables on both sides   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "And that is going to be equal to 6x minus 6x. That cancels out. That was the whole point behind subtracting negative 6x. And then we have just a negative 26 or minus 26, depending on how you want to view it. So just negative 13x is equal to negative 26. Now our whole goal, just to remember, is to isolate the x. And we have a negative 13 times the x here."}, {"video_title": "Example 2 Variables on both sides   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "And then we have just a negative 26 or minus 26, depending on how you want to view it. So just negative 13x is equal to negative 26. Now our whole goal, just to remember, is to isolate the x. And we have a negative 13 times the x here. So the best way to isolate it is if we have something times x, if we divide by that something, we'll isolate the x. So let's divide by negative 13. Now, you know by now, anything you do to the left hand side of an equation, you have to do to the right hand side."}, {"video_title": "Example 2 Variables on both sides   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "And we have a negative 13 times the x here. So the best way to isolate it is if we have something times x, if we divide by that something, we'll isolate the x. So let's divide by negative 13. Now, you know by now, anything you do to the left hand side of an equation, you have to do to the right hand side. So we're going to divide both sides of the equation by negative 13. Now what does the left hand side become? Negative 13 times x divided by negative 13, that's just going to be x."}, {"video_title": "Example 2 Variables on both sides   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Now, you know by now, anything you do to the left hand side of an equation, you have to do to the right hand side. So we're going to divide both sides of the equation by negative 13. Now what does the left hand side become? Negative 13 times x divided by negative 13, that's just going to be x. You multiply something times x, divided by the something, it's just going to be left with an x. So the left hand side just becomes an x. x is equal to negative 26 divided by negative 13. Well, that's just positive 2."}, {"video_title": "Example 2 Variables on both sides   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Negative 13 times x divided by negative 13, that's just going to be x. You multiply something times x, divided by the something, it's just going to be left with an x. So the left hand side just becomes an x. x is equal to negative 26 divided by negative 13. Well, that's just positive 2. A negative divided by a negative is a positive. 26 divided by 13 is 2. And that is our answer."}, {"video_title": "Example 2 Variables on both sides   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Well, that's just positive 2. A negative divided by a negative is a positive. 26 divided by 13 is 2. And that is our answer. Now let's verify that it really works. That's the fun thing about algebra. You can always make sure that you got the right answer."}, {"video_title": "Example 2 Variables on both sides   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "And that is our answer. Now let's verify that it really works. That's the fun thing about algebra. You can always make sure that you got the right answer. So let's substitute it back into the original equation. So we have 20 minus 7 times x. x is 2. Minus 7 times 2 is equal to 6 times x."}, {"video_title": "Example 2 Variables on both sides   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "You can always make sure that you got the right answer. So let's substitute it back into the original equation. So we have 20 minus 7 times x. x is 2. Minus 7 times 2 is equal to 6 times x. We've solved for x. It is 2 minus 6. So let's verify that this left hand side really does equal this right hand side."}, {"video_title": "Example 2 Variables on both sides   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Minus 7 times 2 is equal to 6 times x. We've solved for x. It is 2 minus 6. So let's verify that this left hand side really does equal this right hand side. So the left hand side simplifies to 20 minus 7 times 2, which is 14. 20 minus 14 is 6. That's what the left hand side simplifies to."}, {"video_title": "Example 2 Variables on both sides   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So let's verify that this left hand side really does equal this right hand side. So the left hand side simplifies to 20 minus 7 times 2, which is 14. 20 minus 14 is 6. That's what the left hand side simplifies to. The right hand side, we have 6 times 2, which is 12 minus 6. 12 minus 6 is 6. So they are indeed equal."}, {"video_title": "Evaluating exponent expressions with variables.mp3", "Sentence": "So pause this video and see if you can figure out what is this expression equal when x equals two? All right, now let's work through this together. So what we want to do is everywhere we see an x, we want to replace it with a two. So this expression for x equals two would be five to the second power minus three to the second power. Well, what's that going to be equal to? Well, five to the second power, that's the same thing as five times five, and then from that, we are going to subtract three times three, three times three. And now order of operations would tell us to do the multiplication or do the exponents first, which is this multiplication, but just to make it clear, I'll put some parentheses here."}, {"video_title": "Evaluating exponent expressions with variables.mp3", "Sentence": "So this expression for x equals two would be five to the second power minus three to the second power. Well, what's that going to be equal to? Well, five to the second power, that's the same thing as five times five, and then from that, we are going to subtract three times three, three times three. And now order of operations would tell us to do the multiplication or do the exponents first, which is this multiplication, but just to make it clear, I'll put some parentheses here. And this is going to be equal to five times five is 25 minus nine, which is equal to, what's 25 minus nine? It is equal to 16. So that's what that expression equals for x equals two."}, {"video_title": "Evaluating exponent expressions with variables.mp3", "Sentence": "And now order of operations would tell us to do the multiplication or do the exponents first, which is this multiplication, but just to make it clear, I'll put some parentheses here. And this is going to be equal to five times five is 25 minus nine, which is equal to, what's 25 minus nine? It is equal to 16. So that's what that expression equals for x equals two. Let's do another example. So now we are asked, what is the value of y squared minus x to the fourth when y is equal to nine and x equals two? So once again, pause this video and see if you can evaluate that."}, {"video_title": "Evaluating exponent expressions with variables.mp3", "Sentence": "So that's what that expression equals for x equals two. Let's do another example. So now we are asked, what is the value of y squared minus x to the fourth when y is equal to nine and x equals two? So once again, pause this video and see if you can evaluate that. Alright, so here we are, we have variables as the bases as opposed to being the exponents, and we have two different variables, but all we have to do is wherever we see a y, we substitute it with a nine, and wherever we see an x, we substitute it with a two. So y squared is going to be the same thing as nine squared minus x, which is two. That minus looks a little bit funny, let me see."}, {"video_title": "Evaluating exponent expressions with variables.mp3", "Sentence": "So once again, pause this video and see if you can evaluate that. Alright, so here we are, we have variables as the bases as opposed to being the exponents, and we have two different variables, but all we have to do is wherever we see a y, we substitute it with a nine, and wherever we see an x, we substitute it with a two. So y squared is going to be the same thing as nine squared minus x, which is two. That minus looks a little bit funny, let me see. So this is gonna be nine squared minus x, which is two, two to the fourth power. Now what is this going to be equal to? Well, nine squared is nine times nine, so this whole thing is going to be equal to 81, this whole thing right over here is nine times nine, nine times nine is that right over there, and then from that, we're going to subtract two to the fourth power."}, {"video_title": "Evaluating exponent expressions with variables.mp3", "Sentence": "That minus looks a little bit funny, let me see. So this is gonna be nine squared minus x, which is two, two to the fourth power. Now what is this going to be equal to? Well, nine squared is nine times nine, so this whole thing is going to be equal to 81, this whole thing right over here is nine times nine, nine times nine is that right over there, and then from that, we're going to subtract two to the fourth power. Well, what's two to the fourth power? That is two times two times two times two, so this is going to be two times two is four, four times two is eight, eight times two is 16, so it's 81 minus 16. Now what is that going to be equal to?"}, {"video_title": "Evaluating exponent expressions with variables.mp3", "Sentence": "Well, nine squared is nine times nine, so this whole thing is going to be equal to 81, this whole thing right over here is nine times nine, nine times nine is that right over there, and then from that, we're going to subtract two to the fourth power. Well, what's two to the fourth power? That is two times two times two times two, so this is going to be two times two is four, four times two is eight, eight times two is 16, so it's 81 minus 16. Now what is that going to be equal to? Let's see, 81 minus six is 75, and then minus another 10 is going to be 65. So there you have it. Y squared minus x to the fourth, when y is equal to nine and x equals two, is equal to 65, and we're done."}, {"video_title": "Word problem solving equations   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So let's draw this garden here, Tina's garden. So it's a rectangle. They tell us that it's a rectangular garden. So it looks something like this. And let's say that this is the width. So if this is the width, then this is also going to be the width. And this is the length up here."}, {"video_title": "Word problem solving equations   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So it looks something like this. And let's say that this is the width. So if this is the width, then this is also going to be the width. And this is the length up here. And they tell us that the length of the garden is twice the width. So if this is w, then the length is going to be 2w. It's going to be twice the width."}, {"video_title": "Word problem solving equations   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "And this is the length up here. And they tell us that the length of the garden is twice the width. So if this is w, then the length is going to be 2w. It's going to be twice the width. This is also going to be 2w over here. Now what's the perimeter of this garden? Well, it's going to be w plus w plus 2w plus 2w."}, {"video_title": "Word problem solving equations   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "It's going to be twice the width. This is also going to be 2w over here. Now what's the perimeter of this garden? Well, it's going to be w plus w plus 2w plus 2w. Let me write this down. The perimeter of this garden is going to be equal to w plus 2w plus w plus 2w, which is equal to what? This is w plus 2w is 3w, 4w, 6w."}, {"video_title": "Word problem solving equations   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Well, it's going to be w plus w plus 2w plus 2w. Let me write this down. The perimeter of this garden is going to be equal to w plus 2w plus w plus 2w, which is equal to what? This is w plus 2w is 3w, 4w, 6w. So this is equal to 6w. That's the perimeter in terms of the width. They also tell us that the actual numerical value of the perimeter is 60 feet."}, {"video_title": "Word problem solving equations   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "This is w plus 2w is 3w, 4w, 6w. So this is equal to 6w. That's the perimeter in terms of the width. They also tell us that the actual numerical value of the perimeter is 60 feet. So this perimeter, 6w, must be equal to 60 if we assume that we're dealing with feet. So we just have the equation 6w is equal to 60. We can divide both sides of this equation by 6 so that we have just a w on the left-hand side."}, {"video_title": "Word problem solving equations   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "They also tell us that the actual numerical value of the perimeter is 60 feet. So this perimeter, 6w, must be equal to 60 if we assume that we're dealing with feet. So we just have the equation 6w is equal to 60. We can divide both sides of this equation by 6 so that we have just a w on the left-hand side. 6w divided by 6 is just w. And then 60 divided by 6 is 10. So we have w is equal to 10. So the width of the garden is 10."}, {"video_title": "Word problem solving equations   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "We can divide both sides of this equation by 6 so that we have just a w on the left-hand side. 6w divided by 6 is just w. And then 60 divided by 6 is 10. So we have w is equal to 10. So the width of the garden is 10. So this distance over here is 10. And then what is the length of the garden? Well, it's 2 times the width."}, {"video_title": "Word problem solving equations   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So the width of the garden is 10. So this distance over here is 10. And then what is the length of the garden? Well, it's 2 times the width. So this is equal to 20. The length is equal to 20. And so we're done."}, {"video_title": "Fraction to decimal   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "And so the main realization here is that 7 over 8 is the same thing as 7 divided by 8, which is the same thing as 7 divided by 8. These are all different ways of writing the same thing. So let's actually divide 8 into 7. And I'll do it down here just so I have some more real estate to work with. I'm going to divide 8 into 7. And I'm going to add a decimal point here just because we know that this value is going to be less than 1. 7 8ths is less than 1."}, {"video_title": "Fraction to decimal   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "And I'll do it down here just so I have some more real estate to work with. I'm going to divide 8 into 7. And I'm going to add a decimal point here just because we know that this value is going to be less than 1. 7 8ths is less than 1. We're going to have some digits beyond or to the right of the decimal point. And let me put the decimal point right up here, right above the decimal point in 7. And then we start dividing."}, {"video_title": "Fraction to decimal   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "7 8ths is less than 1. We're going to have some digits beyond or to the right of the decimal point. And let me put the decimal point right up here, right above the decimal point in 7. And then we start dividing. And now this really turns into a long division problem. And we just have to make sure we keep track of the decimal sign. So 8 doesn't go into 7 at all, but it does go into 7t."}, {"video_title": "Fraction to decimal   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "And then we start dividing. And now this really turns into a long division problem. And we just have to make sure we keep track of the decimal sign. So 8 doesn't go into 7 at all, but it does go into 7t. So 8 goes into 7t 8 times. 8 times 8 is 64. And then you subtract."}, {"video_title": "Fraction to decimal   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "So 8 doesn't go into 7 at all, but it does go into 7t. So 8 goes into 7t 8 times. 8 times 8 is 64. And then you subtract. 70 minus 64 is 6. And then bring down another 0 because we still have a remainder. We want to get to the point that we have no remainders, assuming that this thing doesn't repeat forever."}, {"video_title": "Fraction to decimal   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "And then you subtract. 70 minus 64 is 6. And then bring down another 0 because we still have a remainder. We want to get to the point that we have no remainders, assuming that this thing doesn't repeat forever. And then there's other ways we can deal with that. 8 goes into 60. Well, let's see, it doesn't go into it 8 times because that's 64."}, {"video_title": "Fraction to decimal   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "We want to get to the point that we have no remainders, assuming that this thing doesn't repeat forever. And then there's other ways we can deal with that. 8 goes into 60. Well, let's see, it doesn't go into it 8 times because that's 64. 8 goes into 60 7 times. 7 times 8 is 56. And then we subtract again."}, {"video_title": "Fraction to decimal   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "Well, let's see, it doesn't go into it 8 times because that's 64. 8 goes into 60 7 times. 7 times 8 is 56. And then we subtract again. 60 minus 56 is 4. And now we can bring down another 0 right over here. And 8 goes into 40."}, {"video_title": "Fraction to decimal   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "And then we subtract again. 60 minus 56 is 4. And now we can bring down another 0 right over here. And 8 goes into 40. Well, it goes into 40 exactly 5 times. 5 times 8 is 40. And we have nothing left over."}, {"video_title": "Fraction to decimal   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "And 8 goes into 40. Well, it goes into 40 exactly 5 times. 5 times 8 is 40. And we have nothing left over. And so we're done. 7 divided by 8, or 7 8s is equal to 7 divided by 8, which is equal to 0.875. But I'll put a leading 0 here just so it makes it clear that this is where the decimal is."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "The king's advisor, Arbegla, is watching all of this discourse between you, the king, and the bird. And he's starting to feel a little bit jealous, because he's supposed to be the wise man in the kingdom, the king's closest advisor. So he steps in and says, OK, if you and this bird are so smart, how about you tackle the riddle of the fruit prices? And the king says, yes, that is something that we haven't been able to figure out, the fruit prices. Arbegla tell them the riddle of the fruit prices. And so Arbegla says, well, we want to keep track of how much our fruit costs, but we forgot to actually log how much it costs when we went to the market. But we know how much in total we spent."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "And the king says, yes, that is something that we haven't been able to figure out, the fruit prices. Arbegla tell them the riddle of the fruit prices. And so Arbegla says, well, we want to keep track of how much our fruit costs, but we forgot to actually log how much it costs when we went to the market. But we know how much in total we spent. We know how much we got. We know that one week ago, when we went to the fruit market, we bought two pounds of apples and one pound of bananas. And the total cost that time was $3."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "But we know how much in total we spent. We know how much we got. We know that one week ago, when we went to the fruit market, we bought two pounds of apples and one pound of bananas. And the total cost that time was $3. So there was $3 in total cost. And then when we went the time before that, we bought six pounds of apples and three pounds of bananas. And the total cost at that point was $15."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "And the total cost that time was $3. So there was $3 in total cost. And then when we went the time before that, we bought six pounds of apples and three pounds of bananas. And the total cost at that point was $15. So what is the cost of apples and bananas? So you look at the bird. The bird looks at you."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "And the total cost at that point was $15. So what is the cost of apples and bananas? So you look at the bird. The bird looks at you. The bird whispers into the king's ear. And the king says, well, the bird says, well, just start defining some variables here so we can express this thing algebraically. So you go about doing that."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "The bird looks at you. The bird whispers into the king's ear. And the king says, well, the bird says, well, just start defining some variables here so we can express this thing algebraically. So you go about doing that. What we want to figure out is the cost of apples and the cost of bananas per pound. So we set some variables. So let's let A equal the cost of apples per pound."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "So you go about doing that. What we want to figure out is the cost of apples and the cost of bananas per pound. So we set some variables. So let's let A equal the cost of apples per pound. And let's let B equal the cost of bananas per pound. So how could we interpret this first information right over here? 2 pounds of apples and a pound of bananas cost $3."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "So let's let A equal the cost of apples per pound. And let's let B equal the cost of bananas per pound. So how could we interpret this first information right over here? 2 pounds of apples and a pound of bananas cost $3. Well, how much are the apples going to cost? Well, it's going to cost 2 pounds times the cost per pound times A. That's going to be the total cost of the apples in this scenario."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "2 pounds of apples and a pound of bananas cost $3. Well, how much are the apples going to cost? Well, it's going to cost 2 pounds times the cost per pound times A. That's going to be the total cost of the apples in this scenario. And what's the total cost of the bananas? Well, it's 1 pound times the cost per pound. So you're just going to have B."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "That's going to be the total cost of the apples in this scenario. And what's the total cost of the bananas? Well, it's 1 pound times the cost per pound. So you're just going to have B. That's going to be the total cost of the bananas, because we know we bought 1 pound. So the total cost of the apples and bananas are going to be 2A plus B. And we know what that total cost is."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "So you're just going to have B. That's going to be the total cost of the bananas, because we know we bought 1 pound. So the total cost of the apples and bananas are going to be 2A plus B. And we know what that total cost is. It is $3. Now let's do the same thing for the other time that we went to the market. 6 pounds of apples, the total cost is going to be 6 pounds times A dollars per pound."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "And we know what that total cost is. It is $3. Now let's do the same thing for the other time that we went to the market. 6 pounds of apples, the total cost is going to be 6 pounds times A dollars per pound. And the total cost of bananas is going to be, well, we bought 3 pounds of bananas. And the cost per pound is B. And so the total cost of the apples and bananas, this scenario, is going to be equal to $15."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "6 pounds of apples, the total cost is going to be 6 pounds times A dollars per pound. And the total cost of bananas is going to be, well, we bought 3 pounds of bananas. And the cost per pound is B. And so the total cost of the apples and bananas, this scenario, is going to be equal to $15. So let's think about how we might want to solve it. We could use elimination. We could use substitution."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "And so the total cost of the apples and bananas, this scenario, is going to be equal to $15. So let's think about how we might want to solve it. We could use elimination. We could use substitution. Whatever we want. We could even do it graphically. Let's try it first with elimination."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "We could use substitution. Whatever we want. We could even do it graphically. Let's try it first with elimination. So the first thing I might want to do is maybe I want to eliminate the A variable right over here. So I have 2A over here. I have 6A over here."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "Let's try it first with elimination. So the first thing I might want to do is maybe I want to eliminate the A variable right over here. So I have 2A over here. I have 6A over here. So if I multiply this entire white equation by negative 3, then this 2A would become a negative 6A. And then it might be able to cancel out with that. So let me do that."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "I have 6A over here. So if I multiply this entire white equation by negative 3, then this 2A would become a negative 6A. And then it might be able to cancel out with that. So let me do that. Let me multiply this entire equation, the entire equation, times negative 3. So negative 3 times 2A is negative 6A. Negative 3 times B is negative 3B."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "So let me do that. Let me multiply this entire equation, the entire equation, times negative 3. So negative 3 times 2A is negative 6A. Negative 3 times B is negative 3B. And then negative 3 times 3 is negative 9. And now we can essentially add the two equations, or add the left side of this equation to the left side of that, and the right side of this equation to the right side of that. We're essentially adding the same thing to both sides of this green equation, because we know that this is equal to that."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "Negative 3 times B is negative 3B. And then negative 3 times 3 is negative 9. And now we can essentially add the two equations, or add the left side of this equation to the left side of that, and the right side of this equation to the right side of that. We're essentially adding the same thing to both sides of this green equation, because we know that this is equal to that. So let's do that. Let's do it. So on the left-hand side, 6A and 6A cancel out."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "We're essentially adding the same thing to both sides of this green equation, because we know that this is equal to that. So let's do that. Let's do it. So on the left-hand side, 6A and 6A cancel out. But something else interesting happens. The 3B and the 3B cancels out as well. So we're just left with 0 on the left-hand side."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "So on the left-hand side, 6A and 6A cancel out. But something else interesting happens. The 3B and the 3B cancels out as well. So we're just left with 0 on the left-hand side. And on the right-hand side, what do we have? 15 minus 9 is equal to 6. So we get this bizarre statement."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "So we're just left with 0 on the left-hand side. And on the right-hand side, what do we have? 15 minus 9 is equal to 6. So we get this bizarre statement. All of our variables have gone away, and we're left with this bizarre, nonsensical statement that 0 is equal to 6, which we know is definitely not the case. So what's going on over here? What's going on?"}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "So we get this bizarre statement. All of our variables have gone away, and we're left with this bizarre, nonsensical statement that 0 is equal to 6, which we know is definitely not the case. So what's going on over here? What's going on? And then you say, what's going on? And you look at the bird, because the bird seems to be the most knowledgeable person in the room, or at least the most knowledgeable vertebrate in the room. And so the bird whispers into the king's ear."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "What's going on? And then you say, what's going on? And you look at the bird, because the bird seems to be the most knowledgeable person in the room, or at least the most knowledgeable vertebrate in the room. And so the bird whispers into the king's ear. And the king says, well, he says that there's no solution, and you should at least try to graph it to see why. And so you say, well, the bird seems to know what he's talking about. So let me attempt to graph these two equations and see what's going on."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "And so the bird whispers into the king's ear. And the king says, well, he says that there's no solution, and you should at least try to graph it to see why. And so you say, well, the bird seems to know what he's talking about. So let me attempt to graph these two equations and see what's going on. And so what you do is you take each of the equation. And when you graph it, you like to put it in kind of the y-intercept form or slope-intercept form. And so you do that."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "So let me attempt to graph these two equations and see what's going on. And so what you do is you take each of the equation. And when you graph it, you like to put it in kind of the y-intercept form or slope-intercept form. And so you do that. So you say, well, let me solve both of these for b. So if you want to solve this first equation for b, you just subtract 2a from both sides. If you subtract 2a from both sides of this first equation, you get b is equal to negative 2a plus 3."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "And so you do that. So you say, well, let me solve both of these for b. So if you want to solve this first equation for b, you just subtract 2a from both sides. If you subtract 2a from both sides of this first equation, you get b is equal to negative 2a plus 3. Now let's solve this second equation for b. So the first thing you might want to do is subtract 6a from both sides. So you would get 3b is equal to negative 6a plus 15."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "If you subtract 2a from both sides of this first equation, you get b is equal to negative 2a plus 3. Now let's solve this second equation for b. So the first thing you might want to do is subtract 6a from both sides. So you would get 3b is equal to negative 6a plus 15. And then you can divide both sides by 3. You get b is equal to negative 2a plus 5. So the second equation, let me revert back to that other shade of green, is b is equal to negative 2a plus 5."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "So you would get 3b is equal to negative 6a plus 15. And then you can divide both sides by 3. You get b is equal to negative 2a plus 5. So the second equation, let me revert back to that other shade of green, is b is equal to negative 2a plus 5. And we haven't even graphed it yet, but it looks like something interesting is going on. They both have the exact same slope when you solve for b, but they seem to have different, I guess you could call them b-intercepts. Let's graph it to actually see what's going on."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "So the second equation, let me revert back to that other shade of green, is b is equal to negative 2a plus 5. And we haven't even graphed it yet, but it looks like something interesting is going on. They both have the exact same slope when you solve for b, but they seem to have different, I guess you could call them b-intercepts. Let's graph it to actually see what's going on. So let me draw some axes over here. So let's call that my b-axis. And then this could be my a-axis."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "Let's graph it to actually see what's going on. So let me draw some axes over here. So let's call that my b-axis. And then this could be my a-axis. And this first equation has a b-intercept of positive 3. So let's see, 1, 2, 3, 4, 5. The first one has a b-intercept of positive 3, and then has a slope of negative 2."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "And then this could be my a-axis. And this first equation has a b-intercept of positive 3. So let's see, 1, 2, 3, 4, 5. The first one has a b-intercept of positive 3, and then has a slope of negative 2. So you go down, or you go to the right one, you go down 2. So the line looks something like this. I'm trying my best to draw it straight."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "The first one has a b-intercept of positive 3, and then has a slope of negative 2. So you go down, or you go to the right one, you go down 2. So the line looks something like this. I'm trying my best to draw it straight. So it looks something like that. And now let's draw this green one. This green one, our b-intercept is 5, so it's right over here."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "I'm trying my best to draw it straight. So it looks something like that. And now let's draw this green one. This green one, our b-intercept is 5, so it's right over here. But we have the exact same slope, a slope of negative 2. So it looks something like that right over there. And you immediately see now that the bird was right."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "This green one, our b-intercept is 5, so it's right over here. But we have the exact same slope, a slope of negative 2. So it looks something like that right over there. And you immediately see now that the bird was right. There is no solution, because these two constraints can be represented by lines that don't intersect. So the lines don't intersect. And so the bird is right."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "And you immediately see now that the bird was right. There is no solution, because these two constraints can be represented by lines that don't intersect. So the lines don't intersect. And so the bird is right. There's no solution. There's no x and y that can make this statement equal true, or that can make 0 equal 6. There is no possible."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "And so the bird is right. There's no solution. There's no x and y that can make this statement equal true, or that can make 0 equal 6. There is no possible. There is no overlap between these two things. And so something gets into your brain. You realize that our bagel is trying to stump you."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "There is no possible. There is no overlap between these two things. And so something gets into your brain. You realize that our bagel is trying to stump you. And you say, our bagel, you have given me inconsistent information. This is an inconsistent system of equations. Which happens to be the word that is sometimes used to refer to a system that has no solutions, where the lines do not intersect."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "You realize that our bagel is trying to stump you. And you say, our bagel, you have given me inconsistent information. This is an inconsistent system of equations. Which happens to be the word that is sometimes used to refer to a system that has no solutions, where the lines do not intersect. And therefore, this information is incorrect. We cannot assume that the apple or banana, either you are lying, which is possible, or you accounted for it wrong, or maybe the prices of apples and bananas actually changed between the two visits to the market. At which point, the bird whispered into the king's ear and says, oh, this character isn't so bad at this algebra stuff."}, {"video_title": "Setting up systems of linear equations example.mp3", "Sentence": "You can do that if you like, just for kicks. But really, we're going to just focus on setting it up. So here we're told Lauren uses a blend of dark roast beans and light roast beans to make coffee at her cafe. She needs 80 kilograms of beans in total for her next order. Dark roast beans cost $3 per kilogram. Light roast beans cost $2 per kilogram. And she wants to spend $220 total."}, {"video_title": "Setting up systems of linear equations example.mp3", "Sentence": "She needs 80 kilograms of beans in total for her next order. Dark roast beans cost $3 per kilogram. Light roast beans cost $2 per kilogram. And she wants to spend $220 total. Let D be the number of kilograms of dark roast beans she buys, and L be the number of kilograms of light roast beans she buys. All right, so based on this information that we've been given, see if you can pause this video and set up a system of equations. And it's going to have two equations with two unknowns, D and L, that in theory we could solve to figure out the right number of kilograms of dark roast beans and light roast beans that Karen should use."}, {"video_title": "Setting up systems of linear equations example.mp3", "Sentence": "And she wants to spend $220 total. Let D be the number of kilograms of dark roast beans she buys, and L be the number of kilograms of light roast beans she buys. All right, so based on this information that we've been given, see if you can pause this video and set up a system of equations. And it's going to have two equations with two unknowns, D and L, that in theory we could solve to figure out the right number of kilograms of dark roast beans and light roast beans that Karen should use. So pause this video and try to work that out. All right, now let's do it together. And what I'm going to do is I'm going to underline, so let's see, we know that D is dark roast beans and L is the number of kilograms of light roast beans."}, {"video_title": "Setting up systems of linear equations example.mp3", "Sentence": "And it's going to have two equations with two unknowns, D and L, that in theory we could solve to figure out the right number of kilograms of dark roast beans and light roast beans that Karen should use. So pause this video and try to work that out. All right, now let's do it together. And what I'm going to do is I'm going to underline, so let's see, we know that D is dark roast beans and L is the number of kilograms of light roast beans. And then they tell us here, they say she needs 80 kilograms of beans in total. So what we could say is, hey, the number of kilograms of dark roast beans plus the number of kilograms of light roast beans needs to be equal to 80 kilograms. So the number of kilograms of dark roast beans plus the number of kilograms of light roast beans, I'm having trouble saying light roast beans, well, this, what I just underlined here, it says needs to be 80 kilograms in total."}, {"video_title": "Setting up systems of linear equations example.mp3", "Sentence": "And what I'm going to do is I'm going to underline, so let's see, we know that D is dark roast beans and L is the number of kilograms of light roast beans. And then they tell us here, they say she needs 80 kilograms of beans in total. So what we could say is, hey, the number of kilograms of dark roast beans plus the number of kilograms of light roast beans needs to be equal to 80 kilograms. So the number of kilograms of dark roast beans plus the number of kilograms of light roast beans, I'm having trouble saying light roast beans, well, this, what I just underlined here, it says needs to be 80 kilograms in total. So that needs to be 80. So this number of kilograms plus this number of kilograms is going to be equal to your total number of kilograms. All right, so I have one equation with two unknowns."}, {"video_title": "Setting up systems of linear equations example.mp3", "Sentence": "So the number of kilograms of dark roast beans plus the number of kilograms of light roast beans, I'm having trouble saying light roast beans, well, this, what I just underlined here, it says needs to be 80 kilograms in total. So that needs to be 80. So this number of kilograms plus this number of kilograms is going to be equal to your total number of kilograms. All right, so I have one equation with two unknowns. Let's see if we can get another one. So next, they say dark roast beans cost $3 per kilogram. Light roast beans cost $2 per kilogram."}, {"video_title": "Setting up systems of linear equations example.mp3", "Sentence": "All right, so I have one equation with two unknowns. Let's see if we can get another one. So next, they say dark roast beans cost $3 per kilogram. Light roast beans cost $2 per kilogram. And she wants to spend $220 total. So what I just underlined in this aquamarine color, we can set up another equation with. And if you haven't already set up your system of equations, see if you can now do that."}, {"video_title": "Setting up systems of linear equations example.mp3", "Sentence": "Light roast beans cost $2 per kilogram. And she wants to spend $220 total. So what I just underlined in this aquamarine color, we can set up another equation with. And if you haven't already set up your system of equations, see if you can now do that. See if you can set up the second equation. Pause the video. All right, well, the way to think about it is we just have to have an expression for how much does she spend on dark roast beans, how much does she spend on light roast beans, and then we need to add those two together and that needs to be equal to $220 because that's how much she wants to spend in total."}, {"video_title": "Setting up systems of linear equations example.mp3", "Sentence": "And if you haven't already set up your system of equations, see if you can now do that. See if you can set up the second equation. Pause the video. All right, well, the way to think about it is we just have to have an expression for how much does she spend on dark roast beans, how much does she spend on light roast beans, and then we need to add those two together and that needs to be equal to $220 because that's how much she wants to spend in total. So how much does she spend on dark roast beans? Well, it's going to be the number of kilograms of dark roast beans that she buys. And it says that it costs $3 per kilogram."}, {"video_title": "Setting up systems of linear equations example.mp3", "Sentence": "All right, well, the way to think about it is we just have to have an expression for how much does she spend on dark roast beans, how much does she spend on light roast beans, and then we need to add those two together and that needs to be equal to $220 because that's how much she wants to spend in total. So how much does she spend on dark roast beans? Well, it's going to be the number of kilograms of dark roast beans that she buys. And it says that it costs $3 per kilogram. So we're gonna multiply it by three. $3 per kilogram times the number of kilograms of dark roast beans. This is how much she spends on dark roast beans."}, {"video_title": "Setting up systems of linear equations example.mp3", "Sentence": "And it says that it costs $3 per kilogram. So we're gonna multiply it by three. $3 per kilogram times the number of kilograms of dark roast beans. This is how much she spends on dark roast beans. And so how much is she going to spend on light roast beans? Well, she buys L kilograms of light roast beans. They told us that there."}, {"video_title": "Setting up systems of linear equations example.mp3", "Sentence": "This is how much she spends on dark roast beans. And so how much is she going to spend on light roast beans? Well, she buys L kilograms of light roast beans. They told us that there. And they cost $2 per kilogram. So $2 per kilogram times the number of kilograms. This is how much she spends on light roast beans."}, {"video_title": "Setting up systems of linear equations example.mp3", "Sentence": "They told us that there. And they cost $2 per kilogram. So $2 per kilogram times the number of kilograms. This is how much she spends on light roast beans. So you add how much she spends on dark roast to how much she spends on light roast. And so this is going to be $220 in total. And there you have it."}, {"video_title": "Setting up systems of linear equations example.mp3", "Sentence": "This is how much she spends on light roast beans. So you add how much she spends on dark roast to how much she spends on light roast. And so this is going to be $220 in total. And there you have it. We have our two equations with two unknowns. And so now we could go and solve it, but you can do that outside of this video. But the whole point of this video is to understand how to construct these based on the constraints, based on the information that we see in this."}, {"video_title": "Setting up systems of linear equations example.mp3", "Sentence": "And there you have it. We have our two equations with two unknowns. And so now we could go and solve it, but you can do that outside of this video. But the whole point of this video is to understand how to construct these based on the constraints, based on the information that we see in this. So typically, when you're trying to set these up, there's often a sentence or two that will focus on one equation. So this first one is saying, hey, the kilograms, let's add those up for the total number of kilograms. And then there's another sentence or two that'll focus on, in this case, some other equation."}, {"video_title": "Graphs of two-variable inequalities word problem   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "Diana the dog receives five dog biscuits for fetching each frisbee, and three dog biscuits for fetching each ball. That sounds like a pretty good deal. She plans to receive at most D dog biscuits before chasing her tail. That sounds reasonable. The inequality graph below represents the set of all combinations where Diana fetches F frisbees and fetches B balls in order to receive at most, at most D dog biscuits, because at that point she reasonably starts chasing her tail. According to the graph, and we're gonna take a look at the graph in a second, according to the graph, what is the most number of dog biscuits Diana wants to receive before chasing her tail? In other words, what is D?"}, {"video_title": "Graphs of two-variable inequalities word problem   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "That sounds reasonable. The inequality graph below represents the set of all combinations where Diana fetches F frisbees and fetches B balls in order to receive at most, at most D dog biscuits, because at that point she reasonably starts chasing her tail. According to the graph, and we're gonna take a look at the graph in a second, according to the graph, what is the most number of dog biscuits Diana wants to receive before chasing her tail? In other words, what is D? So let's look at, let's interpret this graph properly. So if we look at the horizontal axis right over here, that's F, that's the number of frisbees, number of frisbees she catches, frisbees, and this vertical axis, this is the number of balls, number of, number of balls that she gets, and we know what the total number of biscuits are going to be. The total from catching frisbees, if she catches F frisbees, she gets five biscuits per frisbee."}, {"video_title": "Graphs of two-variable inequalities word problem   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "In other words, what is D? So let's look at, let's interpret this graph properly. So if we look at the horizontal axis right over here, that's F, that's the number of frisbees, number of frisbees she catches, frisbees, and this vertical axis, this is the number of balls, number of, number of balls that she gets, and we know what the total number of biscuits are going to be. The total from catching frisbees, if she catches F frisbees, she gets five biscuits per frisbee. So the total from catching frisbees is 5F, and if she catches, if she catches B balls, or retrieves B balls, so if she gets those B balls, and she gets, what was it, three, three biscuits per ball? Yep, three dog biscuits for fetching each ball, the total number of dog biscuits she gets from fetching B balls is 3B. And so the total number of biscuits she fetches is 5F plus 3B."}, {"video_title": "Graphs of two-variable inequalities word problem   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "The total from catching frisbees, if she catches F frisbees, she gets five biscuits per frisbee. So the total from catching frisbees is 5F, and if she catches, if she catches B balls, or retrieves B balls, so if she gets those B balls, and she gets, what was it, three, three biscuits per ball? Yep, three dog biscuits for fetching each ball, the total number of dog biscuits she gets from fetching B balls is 3B. And so the total number of biscuits she fetches is 5F plus 3B. This is the number of biscuits from frisbees, this is the number of biscuits from balls. And we could see all of the allowable combinations of number of frisbees and number of balls here. And so, for example, if she catches that point right over there, eight, or retrieves eight balls, and catches, well that would be half of a frisbee, so that doesn't seem to make sense."}, {"video_title": "Graphs of two-variable inequalities word problem   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "And so the total number of biscuits she fetches is 5F plus 3B. This is the number of biscuits from frisbees, this is the number of biscuits from balls. And we could see all of the allowable combinations of number of frisbees and number of balls here. And so, for example, if she catches that point right over there, eight, or retrieves eight balls, and catches, well that would be half of a frisbee, so that doesn't seem to make sense. But if she retrieves eight balls and catches one frisbee, well, then that's still okay. She still hasn't met her maximum number of biscuits yet. So how do we think about the maximum number of biscuits?"}, {"video_title": "Graphs of two-variable inequalities word problem   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "And so, for example, if she catches that point right over there, eight, or retrieves eight balls, and catches, well that would be half of a frisbee, so that doesn't seem to make sense. But if she retrieves eight balls and catches one frisbee, well, then that's still okay. She still hasn't met her maximum number of biscuits yet. So how do we think about the maximum number of biscuits? Well, the maximum number of biscuits are any of these points that are sitting on this line. And notice, the solutions, or all of the points that satisfy this inequality are all below this line. So she's hitting a maximum when she's on the line."}, {"video_title": "Graphs of two-variable inequalities word problem   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "So how do we think about the maximum number of biscuits? Well, the maximum number of biscuits are any of these points that are sitting on this line. And notice, the solutions, or all of the points that satisfy this inequality are all below this line. So she's hitting a maximum when she's on the line. And an easy one might be this point right over here, where we see that frisbees, zero frisbees, and 10 balls pretty much maximizes her number of biscuits. So if she catches 10 balls, so let me write this down. So if B is equal to 10, let me write this, F is zero."}, {"video_title": "Graphs of two-variable inequalities word problem   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "So she's hitting a maximum when she's on the line. And an easy one might be this point right over here, where we see that frisbees, zero frisbees, and 10 balls pretty much maximizes her number of biscuits. So if she catches 10 balls, so let me write this down. So if B is equal to 10, let me write this, F is zero. If F is equal to zero and B is equal to 10, well, how many is she going to catch? Or how many biscuits is she going to get? Well, she's going to get, this is going to be zero, and then three times 10 is 30."}, {"video_title": "Graphs of two-variable inequalities word problem   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "So if B is equal to 10, let me write this, F is zero. If F is equal to zero and B is equal to 10, well, how many is she going to catch? Or how many biscuits is she going to get? Well, she's going to get, this is going to be zero, and then three times 10 is 30. That's going to be 30 biscuits. 30 biscuits. So this point right over here, this corresponds with 30 biscuits."}, {"video_title": "Graphs of two-variable inequalities word problem   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "Well, she's going to get, this is going to be zero, and then three times 10 is 30. That's going to be 30 biscuits. 30 biscuits. So this point right over here, this corresponds with 30 biscuits. 30 biscuits. Biscuits. And you could see that any of these points along this blue line actually correspond to 30 biscuits."}, {"video_title": "Graphs of two-variable inequalities word problem   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "So this point right over here, this corresponds with 30 biscuits. 30 biscuits. Biscuits. And you could see that any of these points along this blue line actually correspond to 30 biscuits. If you go over here, where F is six, so let me write it here. If F is equal to six and B is equal to zero, so spends all of her time, she earns her biscuits purely through frisbee catching. So this is a situation where F is equal to six and B is equal to zero."}, {"video_title": "Graphs of two-variable inequalities word problem   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "And you could see that any of these points along this blue line actually correspond to 30 biscuits. If you go over here, where F is six, so let me write it here. If F is equal to six and B is equal to zero, so spends all of her time, she earns her biscuits purely through frisbee catching. So this is a situation where F is equal to six and B is equal to zero. You still have the same scenario. You still have, if F is six, five times six is 30, plus three times zero. Well, that's just going to be, once again, 30 biscuits."}, {"video_title": "Graphs of two-variable inequalities word problem   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "So this is a situation where F is equal to six and B is equal to zero. You still have the same scenario. You still have, if F is six, five times six is 30, plus three times zero. Well, that's just going to be, once again, 30 biscuits. 30 biscuits. So her maximum, or the number of biscuits she needs before she starts chasing her tail, is 30. So D is going to be 30."}, {"video_title": "Graphs of two-variable inequalities word problem   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "Well, that's just going to be, once again, 30 biscuits. 30 biscuits. So her maximum, or the number of biscuits she needs before she starts chasing her tail, is 30. So D is going to be 30. And in fact, you can express this inequality as saying five F plus three B has to be less than or equal to 30. All right. Then they ask us another question."}, {"video_title": "Graphs of two-variable inequalities word problem   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "So D is going to be 30. And in fact, you can express this inequality as saying five F plus three B has to be less than or equal to 30. All right. Then they ask us another question. Can Diana fulfill her plan by fetching, can Diana fulfill her plan by fetching four frisbees and two balls? So let's see. Four frisbees and two balls."}, {"video_title": "Graphs of two-variable inequalities word problem   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "Then they ask us another question. Can Diana fulfill her plan by fetching, can Diana fulfill her plan by fetching four frisbees and two balls? So let's see. Four frisbees and two balls. This is right over here. Four frisbees and two balls. So we're not saying that she has to maximize, that she has to get to 30 biscuits."}, {"video_title": "Graphs of two-variable inequalities word problem   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "Four frisbees and two balls. This is right over here. Four frisbees and two balls. So we're not saying that she has to maximize, that she has to get to 30 biscuits. She just cannot eat any more than 30 biscuits. So it seems, it seems like she can fulfill her plan. Let me see."}, {"video_title": "Example visually evaluating discrete functions.mp3", "Sentence": "What we have here is a visual depiction of a function. And this is a depiction of y is equal to h of x. Now when a lot of people see function notation like this, they see it as somewhat intimidating until you realize what it's saying. All a function is is something that takes an input, in this case, it's taking x as an input, and then the function does something to it, and then it spits out some other value which is going to be equal to y. So for example, what is h of four based on this graph that you see right over here? Pause this video and think through that. Well, all h of four means is when I input four into my function h, what y am I spitting out?"}, {"video_title": "Example visually evaluating discrete functions.mp3", "Sentence": "All a function is is something that takes an input, in this case, it's taking x as an input, and then the function does something to it, and then it spits out some other value which is going to be equal to y. So for example, what is h of four based on this graph that you see right over here? Pause this video and think through that. Well, all h of four means is when I input four into my function h, what y am I spitting out? Or another way to think about it, when x is equal to four, what is y equal to? Well, when x is equal to four, my function spits out that y is equal to three. We know that from this point right over here."}, {"video_title": "Example visually evaluating discrete functions.mp3", "Sentence": "Well, all h of four means is when I input four into my function h, what y am I spitting out? Or another way to think about it, when x is equal to four, what is y equal to? Well, when x is equal to four, my function spits out that y is equal to three. We know that from this point right over here. So y is equal to three, so h of four is equal to three. Let's do another example. What is h of zero?"}, {"video_title": "Example visually evaluating discrete functions.mp3", "Sentence": "We know that from this point right over here. So y is equal to three, so h of four is equal to three. Let's do another example. What is h of zero? Pause the video, try to work that through. Well, all this is saying is if I input x equals zero into the function, what is going to be the corresponding y? Well, when x equals zero, we see that y is equal to four."}, {"video_title": "Example visually evaluating discrete functions.mp3", "Sentence": "What is h of zero? Pause the video, try to work that through. Well, all this is saying is if I input x equals zero into the function, what is going to be the corresponding y? Well, when x equals zero, we see that y is equal to four. So it's as simple as that. Given the input, what is going to be the output? And that's what these points represent."}, {"video_title": "Example visually evaluating discrete functions.mp3", "Sentence": "Well, when x equals zero, we see that y is equal to four. So it's as simple as that. Given the input, what is going to be the output? And that's what these points represent. Each of these points represent a different output for a given input. Now, it's always good to keep in mind one of the things that makes it a function is that for a given x that you input, you only get one y. For example, if we had two dots here, then all of a sudden, we have two dots for x equals six."}, {"video_title": "Example visually evaluating discrete functions.mp3", "Sentence": "And that's what these points represent. Each of these points represent a different output for a given input. Now, it's always good to keep in mind one of the things that makes it a function is that for a given x that you input, you only get one y. For example, if we had two dots here, then all of a sudden, we have two dots for x equals six. Now, all of a sudden, we have a problem at figuring out what h of six would be equal to, because it could be equal to one, or it could be equal to three. So if we had this extra dot here, then this would no longer be a function. In order for it to be a function for any given x, it has to output a unique value."}, {"video_title": "Example visually evaluating discrete functions.mp3", "Sentence": "For example, if we had two dots here, then all of a sudden, we have two dots for x equals six. Now, all of a sudden, we have a problem at figuring out what h of six would be equal to, because it could be equal to one, or it could be equal to three. So if we had this extra dot here, then this would no longer be a function. In order for it to be a function for any given x, it has to output a unique value. It can't output two possible values. Now, the other way is possible. It is possible to have two different x's that output the same value."}, {"video_title": "Example visually evaluating discrete functions.mp3", "Sentence": "In order for it to be a function for any given x, it has to output a unique value. It can't output two possible values. Now, the other way is possible. It is possible to have two different x's that output the same value. For example, if this was circled in, what would h of negative four be? Well, h of negative four, when x is equal to negative four, when you put that into our function, it looks like the function would output two. So h of negative four would be equal to two."}, {"video_title": "Example visually evaluating discrete functions.mp3", "Sentence": "It is possible to have two different x's that output the same value. For example, if this was circled in, what would h of negative four be? Well, h of negative four, when x is equal to negative four, when you put that into our function, it looks like the function would output two. So h of negative four would be equal to two. But h of two is also equal to, we see very clearly there, when we input a two into the function, the corresponding y value is two as well. So it's okay for two different x values to map to the same y value. That works."}, {"video_title": "How to evaluate a piecewise function (example)   Functions   Algebra I   Khan Academy.mp3", "Sentence": "Consider the following piecewise function. And they say f of t is equal to, and they tell us what it's equal to based on what t is. So if t is less than or equal to negative 10, we use this case. If t is between negative 10 and negative two, we use this case. And if t is greater than or equal to negative two, we use this case. And then they ask us, what is the value of f of negative 10? So t is going to be equal to negative 10."}, {"video_title": "How to evaluate a piecewise function (example)   Functions   Algebra I   Khan Academy.mp3", "Sentence": "If t is between negative 10 and negative two, we use this case. And if t is greater than or equal to negative two, we use this case. And then they ask us, what is the value of f of negative 10? So t is going to be equal to negative 10. So which case do we use? So let's see, if t is less than or equal to negative 10, we use this top case right over here. And t is equal to negative 10."}, {"video_title": "How to evaluate a piecewise function (example)   Functions   Algebra I   Khan Academy.mp3", "Sentence": "So t is going to be equal to negative 10. So which case do we use? So let's see, if t is less than or equal to negative 10, we use this top case right over here. And t is equal to negative 10. That's the one that we're trying to evaluate. So we want to use this case right over here. So f of negative 10 is going to be equal to negative 10."}, {"video_title": "How to evaluate a piecewise function (example)   Functions   Algebra I   Khan Academy.mp3", "Sentence": "And t is equal to negative 10. That's the one that we're trying to evaluate. So we want to use this case right over here. So f of negative 10 is going to be equal to negative 10. Everywhere we see a t here, we substitute it with a negative 10. Negative 10 squared minus five times, actually I don't have a denominator there. I don't know why I wrote it so high."}, {"video_title": "How to evaluate a piecewise function (example)   Functions   Algebra I   Khan Academy.mp3", "Sentence": "So f of negative 10 is going to be equal to negative 10. Everywhere we see a t here, we substitute it with a negative 10. Negative 10 squared minus five times, actually I don't have a denominator there. I don't know why I wrote it so high. So it's going to be negative 10 squared minus five times negative 10. So let's see, negative 10 squared, that's positive 100. And then negative, or subtracting five times negative 10, this is going to be subtracting negative 50, or you're going to add 50."}, {"video_title": "How to evaluate a piecewise function (example)   Functions   Algebra I   Khan Academy.mp3", "Sentence": "I don't know why I wrote it so high. So it's going to be negative 10 squared minus five times negative 10. So let's see, negative 10 squared, that's positive 100. And then negative, or subtracting five times negative 10, this is going to be subtracting negative 50, or you're going to add 50. So this is going to be equal to 150. F of negative 10 is 150, because we used this case up here, because t is negative 10. Let's do another one of these examples."}, {"video_title": "How to evaluate a piecewise function (example)   Functions   Algebra I   Khan Academy.mp3", "Sentence": "And then negative, or subtracting five times negative 10, this is going to be subtracting negative 50, or you're going to add 50. So this is going to be equal to 150. F of negative 10 is 150, because we used this case up here, because t is negative 10. Let's do another one of these examples. So here we have, consider the following piecewise function, all right? What is the value of h of negative three? See, when h is negative three, which case do we use?"}, {"video_title": "How to evaluate a piecewise function (example)   Functions   Algebra I   Khan Academy.mp3", "Sentence": "Let's do another one of these examples. So here we have, consider the following piecewise function, all right? What is the value of h of negative three? See, when h is negative three, which case do we use? We use this case if our x is between negative infinity and zero. And negative three is between negative infinity and zero, so we're going to use this case right over here. If it was positive three, we would use this case."}, {"video_title": "How to evaluate a piecewise function (example)   Functions   Algebra I   Khan Academy.mp3", "Sentence": "See, when h is negative three, which case do we use? We use this case if our x is between negative infinity and zero. And negative three is between negative infinity and zero, so we're going to use this case right over here. If it was positive three, we would use this case. If it was positive 30, we would use this case. So we're going to use the first case again. And so we're going, so for h of negative three, we're going to take negative three to the third power."}, {"video_title": "How to evaluate a piecewise function (example)   Functions   Algebra I   Khan Academy.mp3", "Sentence": "If it was positive three, we would use this case. If it was positive 30, we would use this case. So we're going to use the first case again. And so we're going, so for h of negative three, we're going to take negative three to the third power. So let's see, h of negative three is going to be negative three to the third power, which is negative 27. And we're done, that's h of negative three. Because we are using this case, you could almost just ignore these second two cases right over here."}, {"video_title": "How to evaluate a piecewise function (example)   Functions   Algebra I   Khan Academy.mp3", "Sentence": "And so we're going, so for h of negative three, we're going to take negative three to the third power. So let's see, h of negative three is going to be negative three to the third power, which is negative 27. And we're done, that's h of negative three. Because we are using this case, you could almost just ignore these second two cases right over here. Let's do one more example. This one's a little bit different. Below is a graph of the step function g of x."}, {"video_title": "How to evaluate a piecewise function (example)   Functions   Algebra I   Khan Academy.mp3", "Sentence": "Because we are using this case, you could almost just ignore these second two cases right over here. Let's do one more example. This one's a little bit different. Below is a graph of the step function g of x. So we can see g of x right over here. It starts when x equals negative nine, it's at three, and then it jumps up, and then it jumps down. Match each expression with its value."}, {"video_title": "How to evaluate a piecewise function (example)   Functions   Algebra I   Khan Academy.mp3", "Sentence": "Below is a graph of the step function g of x. So we can see g of x right over here. It starts when x equals negative nine, it's at three, and then it jumps up, and then it jumps down. Match each expression with its value. So g of negative 3.0001, so negative 3.0001, so that's right over here, and g of that we see is equal to three. So this is going to be equal to three right over here. G of 3.99999, 3.99999, almost four."}, {"video_title": "How to evaluate a piecewise function (example)   Functions   Algebra I   Khan Academy.mp3", "Sentence": "Match each expression with its value. So g of negative 3.0001, so negative 3.0001, so that's right over here, and g of that we see is equal to three. So this is going to be equal to three right over here. G of 3.99999, 3.99999, almost four. So let's draw the dotted line right here, it's going to be almost four. Well, g of 3.99999 is going to be seven. We see that right over there."}, {"video_title": "How to evaluate a piecewise function (example)   Functions   Algebra I   Khan Academy.mp3", "Sentence": "G of 3.99999, 3.99999, almost four. So let's draw the dotted line right here, it's going to be almost four. Well, g of 3.99999 is going to be seven. We see that right over there. So that is equal to seven. G of 4.00001. So g of four is still seven, but as soon as we go above four, we drop down over here."}, {"video_title": "How to evaluate a piecewise function (example)   Functions   Algebra I   Khan Academy.mp3", "Sentence": "We see that right over there. So that is equal to seven. G of 4.00001. So g of four is still seven, but as soon as we go above four, we drop down over here. So g of 4.00001 is going to be negative three. I want to, actually, let's focus on that a little bit more. How did I know that?"}, {"video_title": "How to evaluate a piecewise function (example)   Functions   Algebra I   Khan Academy.mp3", "Sentence": "So g of four is still seven, but as soon as we go above four, we drop down over here. So g of 4.00001 is going to be negative three. I want to, actually, let's focus on that a little bit more. How did I know that? Well, I know that g of four is seven and not negative three because we have this dot is circled in up here, and it's hollow down here. But as soon as we get any amount larger than four, then the function drops down to this. So 4.0000, or as many, just slightly above four, the value of our function is going to be negative three."}, {"video_title": "How to evaluate a piecewise function (example)   Functions   Algebra I   Khan Academy.mp3", "Sentence": "How did I know that? Well, I know that g of four is seven and not negative three because we have this dot is circled in up here, and it's hollow down here. But as soon as we get any amount larger than four, then the function drops down to this. So 4.0000, or as many, just slightly above four, the value of our function is going to be negative three. Now let's do g of nine. So g of nine, when x is nine, we go down here. You might be tempted to say it's negative three, but you see at this point right over here, we have an open circle."}, {"video_title": "How to evaluate a piecewise function (example)   Functions   Algebra I   Khan Academy.mp3", "Sentence": "So 4.0000, or as many, just slightly above four, the value of our function is going to be negative three. Now let's do g of nine. So g of nine, when x is nine, we go down here. You might be tempted to say it's negative three, but you see at this point right over here, we have an open circle. So that means that, well, it's not, you can't say that the function is negative three right over there, and there's no other place where we have a filled in circle for x equals nine. So the function g actually isn't defined at x equals nine. So I'm gonna put undefined right over there."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "And the definition, we're going to hopefully get a good working knowledge of it in this video, the definition of it is change in y divided by change in x. This may or may not make some sense to you right now, but as we do more and more examples, I think it'll make a good amount of sense. Let's do this first line right here, line A. Let's figure out its slope. And they've actually drawn two points here that we can use as the reference points. So first of all, let's look at the coordinates of those points. So you have this point right here."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "Let's figure out its slope. And they've actually drawn two points here that we can use as the reference points. So first of all, let's look at the coordinates of those points. So you have this point right here. What's its coordinates? Its x-coordinate is 3, and its y-coordinate is 6. And then down here, this point's x-coordinate is negative 1, and its y-coordinate is negative 6."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "So you have this point right here. What's its coordinates? Its x-coordinate is 3, and its y-coordinate is 6. And then down here, this point's x-coordinate is negative 1, and its y-coordinate is negative 6. So there's a couple of ways we can think about slope. One is we could look at it straight up using the formula. We could say change in y."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "And then down here, this point's x-coordinate is negative 1, and its y-coordinate is negative 6. So there's a couple of ways we can think about slope. One is we could look at it straight up using the formula. We could say change in y. So slope is change in y over change in x. And we can figure it out numerically, and I'll in a second draw it graphically. So what's our change in y?"}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "We could say change in y. So slope is change in y over change in x. And we can figure it out numerically, and I'll in a second draw it graphically. So what's our change in y? Our change in y is literally how much did our y values change going from this point to that point? So how much did our y values change? Our y went from here, y is at negative 6, and it went all the way up to positive 6."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "So what's our change in y? Our change in y is literally how much did our y values change going from this point to that point? So how much did our y values change? Our y went from here, y is at negative 6, and it went all the way up to positive 6. So what's this distance right here? Well, it's going to be your endpoint y value. It's going to be 6 minus your starting point y value."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "Our y went from here, y is at negative 6, and it went all the way up to positive 6. So what's this distance right here? Well, it's going to be your endpoint y value. It's going to be 6 minus your starting point y value. Minus negative 6, or 6 plus 6, which is equal to 12. And you could just count this. You say 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "It's going to be 6 minus your starting point y value. Minus negative 6, or 6 plus 6, which is equal to 12. And you could just count this. You say 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. So when we changed our y value by 12, we had to change our x value by what was our change in x over the same change in y? Well, we went from x is equal to negative 1 to x is equal to 3. Right?"}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "You say 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. So when we changed our y value by 12, we had to change our x value by what was our change in x over the same change in y? Well, we went from x is equal to negative 1 to x is equal to 3. Right? X went from negative 1 to 3. So we do the endpoint, which is 3, minus the starting point, which is negative 1, which is equal to 4. So our change in y over change in x is equal to 12 over 4."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "Right? X went from negative 1 to 3. So we do the endpoint, which is 3, minus the starting point, which is negative 1, which is equal to 4. So our change in y over change in x is equal to 12 over 4. Or if we want to write this in simplest form, this is the same thing as 3. Now, the interpretation of this means that for every 1 we move over, we could view this, let me write it this way. Change in y over change in x is equal to, we could say it's 3, or we could say it's 3 over 1."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "So our change in y over change in x is equal to 12 over 4. Or if we want to write this in simplest form, this is the same thing as 3. Now, the interpretation of this means that for every 1 we move over, we could view this, let me write it this way. Change in y over change in x is equal to, we could say it's 3, or we could say it's 3 over 1. Which tells us that for every 1 we move in the positive x direction, we're going to move up 3, because this is a positive 3, in the y direction. You can see that. When we moved 1 in the x, we moved up 3 in the y."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "Change in y over change in x is equal to, we could say it's 3, or we could say it's 3 over 1. Which tells us that for every 1 we move in the positive x direction, we're going to move up 3, because this is a positive 3, in the y direction. You can see that. When we moved 1 in the x, we moved up 3 in the y. If you move 2 in the x direction, you're going to move 6 in the y. 6 over 2 is the same thing as 3. So this 3 tells us how quickly do we go up as we increase x."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "When we moved 1 in the x, we moved up 3 in the y. If you move 2 in the x direction, you're going to move 6 in the y. 6 over 2 is the same thing as 3. So this 3 tells us how quickly do we go up as we increase x. Let's do the same thing for the second line on this graph. Graph B. Same idea."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "So this 3 tells us how quickly do we go up as we increase x. Let's do the same thing for the second line on this graph. Graph B. Same idea. And I'm going to use the points that they gave us. But you really could use any points on that line. We have one point here, which is the point 0,1."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "Same idea. And I'm going to use the points that they gave us. But you really could use any points on that line. We have one point here, which is the point 0,1. And then the starting point, we could call this the finish point. The starting point right here, we could view it as, let's see, x is negative 6, and y is negative 2. So same idea."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "We have one point here, which is the point 0,1. And then the starting point, we could call this the finish point. The starting point right here, we could view it as, let's see, x is negative 6, and y is negative 2. So same idea. What is the change in y given some change in x? Let's do the change in x first. What is our change in x?"}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "So same idea. What is the change in y given some change in x? Let's do the change in x first. What is our change in x? So in this situation, what is our change in x? Delta x, we could even count it. It's 1, 2, 3, 4, 5, 6."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "What is our change in x? So in this situation, what is our change in x? Delta x, we could even count it. It's 1, 2, 3, 4, 5, 6. It's going to be 6. But if you didn't have a graph to count from, you could literally take your finishing x position. So it's 0, and subtract from that your starting x position."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "It's 1, 2, 3, 4, 5, 6. It's going to be 6. But if you didn't have a graph to count from, you could literally take your finishing x position. So it's 0, and subtract from that your starting x position. 0 minus negative 6. So when your change in x is equal to 6, what is our change in y? And remember, we're taking this as our finishing position."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "So it's 0, and subtract from that your starting x position. 0 minus negative 6. So when your change in x is equal to 6, what is our change in y? And remember, we're taking this as our finishing position. That's our finishing position. This is our starting position. So we took 0 minus negative 6, so then on the y, we have to do 1 minus negative 2."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "And remember, we're taking this as our finishing position. That's our finishing position. This is our starting position. So we took 0 minus negative 6, so then on the y, we have to do 1 minus negative 2. So what's 1 minus negative 2? That's the same thing as 1 plus 2. That is equal to 3."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "So we took 0 minus negative 6, so then on the y, we have to do 1 minus negative 2. So what's 1 minus negative 2? That's the same thing as 1 plus 2. That is equal to 3. So it is 3 sixths or 1 half. So notice, when we moved in the x direction by 6, we moved in the y direction by positive 3. So our change in y was 3 when our change in x was 6."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "That is equal to 3. So it is 3 sixths or 1 half. So notice, when we moved in the x direction by 6, we moved in the y direction by positive 3. So our change in y was 3 when our change in x was 6. Now, one of the things that confuses a lot of people is how do I know what order to do the 0 first and the negative 6 second, and then the 1 first and then the negative 2 second? And the answer is you could have done it in either order as long as you keep them straight. So you could have also have done change in y over change in x."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "So our change in y was 3 when our change in x was 6. Now, one of the things that confuses a lot of people is how do I know what order to do the 0 first and the negative 6 second, and then the 1 first and then the negative 2 second? And the answer is you could have done it in either order as long as you keep them straight. So you could have also have done change in y over change in x. We could have said it's equal to negative 2 minus 1. So negative 2 minus 1. So we're using this coordinate first."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "So you could have also have done change in y over change in x. We could have said it's equal to negative 2 minus 1. So negative 2 minus 1. So we're using this coordinate first. Negative 2 minus 1 for the y over negative 6 minus 0. Notice this is the negative of that. That is the negative of that."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "So we're using this coordinate first. Negative 2 minus 1 for the y over negative 6 minus 0. Notice this is the negative of that. That is the negative of that. But since we have a negative over a negative, they're going to cancel out. So this is going to be equal to negative 3 over negative 6. The negatives cancel out."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "That is the negative of that. But since we have a negative over a negative, they're going to cancel out. So this is going to be equal to negative 3 over negative 6. The negatives cancel out. This is also equal to 1 half. So the important thing is if you use this y coordinate first, then you have to use this x coordinate first as well. If you use this y coordinate first, as we did here, then you have to use this x coordinate first as you did there."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "The negatives cancel out. This is also equal to 1 half. So the important thing is if you use this y coordinate first, then you have to use this x coordinate first as well. If you use this y coordinate first, as we did here, then you have to use this x coordinate first as you did there. You just have to make sure that your change in x and change in y are using the same final and starting points. And just to interpret this, this is saying that for every minus 6 we go in x, so if we go minus 6 in x, so that's going backwards, we're going to go minus 3 in y. But they're essentially saying the same thing."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "If you use this y coordinate first, as we did here, then you have to use this x coordinate first as you did there. You just have to make sure that your change in x and change in y are using the same final and starting points. And just to interpret this, this is saying that for every minus 6 we go in x, so if we go minus 6 in x, so that's going backwards, we're going to go minus 3 in y. But they're essentially saying the same thing. The slope of this line is 1 half, which tells us for every 2 we travel in x, we go up 1 in y. Or if we go back 2 in x, we go down 1 in y. That's what 1 half slope tells us."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "But they're essentially saying the same thing. The slope of this line is 1 half, which tells us for every 2 we travel in x, we go up 1 in y. Or if we go back 2 in x, we go down 1 in y. That's what 1 half slope tells us. And notice, the line with the 1 half slope, it is less steep than the line with a slope of 3. Let's do a couple more of these. Let's do this line C right here."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "That's what 1 half slope tells us. And notice, the line with the 1 half slope, it is less steep than the line with a slope of 3. Let's do a couple more of these. Let's do this line C right here. I'll do it in pink. Let's say that the starting point, I'm just picking this arbitrarily, I'm using these points that they've drawn here. The starting point is at the coordinate negative 1, 6, and that my finishing point is at the point 5, negative 6."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "Let's do this line C right here. I'll do it in pink. Let's say that the starting point, I'm just picking this arbitrarily, I'm using these points that they've drawn here. The starting point is at the coordinate negative 1, 6, and that my finishing point is at the point 5, negative 6. Our slope is going to be equal to change in y over change in x. Sometimes it's said rise over run. Run is how much you're moving in the horizontal direction."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "The starting point is at the coordinate negative 1, 6, and that my finishing point is at the point 5, negative 6. Our slope is going to be equal to change in y over change in x. Sometimes it's said rise over run. Run is how much you're moving in the horizontal direction. Rise is how much you're moving in the vertical direction. And then we could say our change in y is our finishing y point minus our starting y point. This is our finishing y point."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "Run is how much you're moving in the horizontal direction. Rise is how much you're moving in the vertical direction. And then we could say our change in y is our finishing y point minus our starting y point. This is our finishing y point. That's our starting y point over our finishing x point minus our starting x point. And if that confuses you, all I'm saying is it's going to be equal to our finishing y point is negative 6 minus our starting y point, which is 6, over our finishing x point, which is 5, minus our starting x point, which is negative 1. So this is equal to negative 6 minus 6 is negative 12."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "This is our finishing y point. That's our starting y point over our finishing x point minus our starting x point. And if that confuses you, all I'm saying is it's going to be equal to our finishing y point is negative 6 minus our starting y point, which is 6, over our finishing x point, which is 5, minus our starting x point, which is negative 1. So this is equal to negative 6 minus 6 is negative 12. 5 minus negative 1, that is 6. So negative 12 over 6 is the same thing as negative 2. And notice, we have a negative slope here."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "So this is equal to negative 6 minus 6 is negative 12. 5 minus negative 1, that is 6. So negative 12 over 6 is the same thing as negative 2. And notice, we have a negative slope here. That's because every time we increase x by 1, we go down in the y direction. So this is a downward sloping line. It's going from the top left to the bottom right."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "And notice, we have a negative slope here. That's because every time we increase x by 1, we go down in the y direction. So this is a downward sloping line. It's going from the top left to the bottom right. As x increases, the y decreases, and that's why we've got a negative slope. This line over here should have a positive slope. Let's verify it."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "It's going from the top left to the bottom right. As x increases, the y decreases, and that's why we've got a negative slope. This line over here should have a positive slope. Let's verify it. So I'll use the same points that they use right over there. So this is line D. Slope is equal to rise over run. Now how much do we rise when we go from that point to that point?"}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "Let's verify it. So I'll use the same points that they use right over there. So this is line D. Slope is equal to rise over run. Now how much do we rise when we go from that point to that point? Well, let's see. We could do it this way. We are rising."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "Now how much do we rise when we go from that point to that point? Well, let's see. We could do it this way. We are rising. I could just count it out. We are rising 1, 2, 3, 4, 5, 6. We are rising 6."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "We are rising. I could just count it out. We are rising 1, 2, 3, 4, 5, 6. We are rising 6. And how much are we running? We are running, I'll do it in a different color. We're running 1, 2, 3, 4, 5, 6."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "We are rising 6. And how much are we running? We are running, I'll do it in a different color. We're running 1, 2, 3, 4, 5, 6. We're running 6. So our slope is 6 over 6, which is 1. Which tells us that every time we move 1 in the x direction, positive 1 in the x direction, we go positive 1 in the y direction."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "We're running 1, 2, 3, 4, 5, 6. We're running 6. So our slope is 6 over 6, which is 1. Which tells us that every time we move 1 in the x direction, positive 1 in the x direction, we go positive 1 in the y direction. So this is a, for every x, if we go negative 2 in the x direction, we're going to go negative 2 in the y direction. So whatever we do in x, we're going to do the same thing in y in the slope. And notice, that was pretty easy."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "Which tells us that every time we move 1 in the x direction, positive 1 in the x direction, we go positive 1 in the y direction. So this is a, for every x, if we go negative 2 in the x direction, we're going to go negative 2 in the y direction. So whatever we do in x, we're going to do the same thing in y in the slope. And notice, that was pretty easy. If we wanted to do it mathematically, we could figure out this coordinate right there. That we could view as our starting position. Our starting position is negative 2, negative 4."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "And notice, that was pretty easy. If we wanted to do it mathematically, we could figure out this coordinate right there. That we could view as our starting position. Our starting position is negative 2, negative 4. And our finishing position is 4, 2. And so our slope, change in y over change in x, I'll take this point, 2 minus negative 4, over 4 minus negative 2. 2 minus negative 4 is 6."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "Our starting position is negative 2, negative 4. And our finishing position is 4, 2. And so our slope, change in y over change in x, I'll take this point, 2 minus negative 4, over 4 minus negative 2. 2 minus negative 4 is 6. Remember, that was just this distance right there. And then 4 minus negative 2, that's also 6. That's that distance right there."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "2 minus negative 4 is 6. Remember, that was just this distance right there. And then 4 minus negative 2, that's also 6. That's that distance right there. And we get a slope of 1. Let's do another one. Let's do another couple."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "That's that distance right there. And we get a slope of 1. Let's do another one. Let's do another couple. These are interesting. Let's do the line e right here. So change in y over change in x."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "Let's do another couple. These are interesting. Let's do the line e right here. So change in y over change in x. So our change in y, when we go from this point to this point, I'll just count it out. It's 1, 2, 3, 4, 5, 6, 7, 8. It's 8."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "So change in y over change in x. So our change in y, when we go from this point to this point, I'll just count it out. It's 1, 2, 3, 4, 5, 6, 7, 8. It's 8. Or you could even take this y coordinate, 2. This y coordinate, 2, minus negative 6, will give you that distance, 8. And then what's the change in y?"}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "It's 8. Or you could even take this y coordinate, 2. This y coordinate, 2, minus negative 6, will give you that distance, 8. And then what's the change in y? Well, the y value here is, sorry, what's the change in x? The x value here is 4. The x value there is 4. x does not change."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "And then what's the change in y? Well, the y value here is, sorry, what's the change in x? The x value here is 4. The x value there is 4. x does not change. So it's 8 over 0. Well, we don't know. 8 over 0 is undefined."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "The x value there is 4. x does not change. So it's 8 over 0. Well, we don't know. 8 over 0 is undefined. So in this situation, the slope is undefined. When you have a vertical line, you say your slope is undefined because you're dividing by 0. But that tells you that you're dealing probably with a vertical line."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "8 over 0 is undefined. So in this situation, the slope is undefined. When you have a vertical line, you say your slope is undefined because you're dividing by 0. But that tells you that you're dealing probably with a vertical line. Now, finally, let's just do this one. This seems like a pretty straight-up vanilla slope problem right there. You have that point right there, which is the point 3, 1."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "But that tells you that you're dealing probably with a vertical line. Now, finally, let's just do this one. This seems like a pretty straight-up vanilla slope problem right there. You have that point right there, which is the point 3, 1. So this is line F. You have the point 3, 1. And then over here, you have the point negative 6, negative 2. So our slope would be equal to change in y. I'll take this as our ending point, just so you can go in different directions."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "You have that point right there, which is the point 3, 1. So this is line F. You have the point 3, 1. And then over here, you have the point negative 6, negative 2. So our slope would be equal to change in y. I'll take this as our ending point, just so you can go in different directions. So our change in y, so now we're going to go down in that direction. So it's negative 2 minus 1. That's what this distance is right here."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "So our slope would be equal to change in y. I'll take this as our ending point, just so you can go in different directions. So our change in y, so now we're going to go down in that direction. So it's negative 2 minus 1. That's what this distance is right here. Negative 2 minus 1, which is equal to negative 3. Notice we went down 3. And then what is going to be our change in x?"}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "That's what this distance is right here. Negative 2 minus 1, which is equal to negative 3. Notice we went down 3. And then what is going to be our change in x? Well, we're going to go back that amount. What is that amount? Well, that is going to be negative 6."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "And then what is going to be our change in x? Well, we're going to go back that amount. What is that amount? Well, that is going to be negative 6. That's our end point, minus 3. Negative 6 minus 3. That gives us that distance, which is negative 9."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "Well, that is going to be negative 6. That's our end point, minus 3. Negative 6 minus 3. That gives us that distance, which is negative 9. So for every time we go back 9, we're going to go down 3. If we go back 9, we're going to go down 3, which is the same thing as if we go forward 9. We're going to go up 3, all equivalent."}, {"video_title": "Slope and rate of change   Graphing lines and slope   Algebra Basics   Khan Academy.mp3", "Sentence": "That gives us that distance, which is negative 9. So for every time we go back 9, we're going to go down 3. If we go back 9, we're going to go down 3, which is the same thing as if we go forward 9. We're going to go up 3, all equivalent. And we see these cancel out, and you get a slope of 1 3rd, positive 1 3rd. It's an upward sloping line. Every time we run 3, we rise 1."}, {"video_title": "Multiplying and dividing with inequalities example   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "We have negative 5c is less than or equal to 15. So negative 5c is less than or equal to 15. I just rewrote it a little bit bigger. So if we want to solve for c, we just want to isolate this c right over here, maybe on the left-hand side. It's right now being multiplied by negative 5. So the best way to just have a c on the left-hand side is we can multiply both sides of this inequality by the inverse of negative 5, or by negative 1 5th. And let me just..."}, {"video_title": "Multiplying and dividing with inequalities example   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So if we want to solve for c, we just want to isolate this c right over here, maybe on the left-hand side. It's right now being multiplied by negative 5. So the best way to just have a c on the left-hand side is we can multiply both sides of this inequality by the inverse of negative 5, or by negative 1 5th. And let me just... So we want to multiply negative 1 5th times negative 5c, and we also want to multiply 15 times negative 1 5th. I'm just multiplying both sides of the inequality by the inverse of negative 5, because this will cancel out with the negative 5 and leave me just with c. Now, I didn't draw the inequality here, because we have to remember, if we multiply or divide both sides of an inequality by a negative number, you have to flip the inequality. And we are doing that."}, {"video_title": "Multiplying and dividing with inequalities example   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "And let me just... So we want to multiply negative 1 5th times negative 5c, and we also want to multiply 15 times negative 1 5th. I'm just multiplying both sides of the inequality by the inverse of negative 5, because this will cancel out with the negative 5 and leave me just with c. Now, I didn't draw the inequality here, because we have to remember, if we multiply or divide both sides of an inequality by a negative number, you have to flip the inequality. And we are doing that. We are multiplying both sides by negative 1 5th, which is the equivalent of dividing both sides by negative 5. So we need to turn this from a less than or equal to a greater than or equal. And now we can proceed solving for c. So negative 1 5th times negative 5 is 1."}, {"video_title": "Multiplying and dividing with inequalities example   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "And we are doing that. We are multiplying both sides by negative 1 5th, which is the equivalent of dividing both sides by negative 5. So we need to turn this from a less than or equal to a greater than or equal. And now we can proceed solving for c. So negative 1 5th times negative 5 is 1. So the left-hand side is just going to be c, is greater than or equal to 15 times negative 1 5th. That's the same thing as 15 divided by negative 5. And so that is negative 3."}, {"video_title": "Multiplying and dividing with inequalities example   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "And now we can proceed solving for c. So negative 1 5th times negative 5 is 1. So the left-hand side is just going to be c, is greater than or equal to 15 times negative 1 5th. That's the same thing as 15 divided by negative 5. And so that is negative 3. So our solution is c is greater than or equal to negative 3. And let's graph it. So that is my number line."}, {"video_title": "Multiplying and dividing with inequalities example   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "And so that is negative 3. So our solution is c is greater than or equal to negative 3. And let's graph it. So that is my number line. Let's say that is 0, negative 1, negative 2, negative 3. And I could go above 1, 2. And so c is greater than or equal to negative 3."}, {"video_title": "Multiplying and dividing with inequalities example   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So that is my number line. Let's say that is 0, negative 1, negative 2, negative 3. And I could go above 1, 2. And so c is greater than or equal to negative 3. So it can be equal to negative 3. So I'll fill that in right over there. Let me do it in a different color."}, {"video_title": "Multiplying and dividing with inequalities example   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "And so c is greater than or equal to negative 3. So it can be equal to negative 3. So I'll fill that in right over there. Let me do it in a different color. So I'll fill it in right over there. And then it's greater than as well. So it's all of these values I am filling in in green."}, {"video_title": "Multiplying and dividing with inequalities example   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Let me do it in a different color. So I'll fill it in right over there. And then it's greater than as well. So it's all of these values I am filling in in green. And you can verify that it works in the original inequality. Pick something that should work. Well, 0 should work."}, {"video_title": "Multiplying and dividing with inequalities example   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So it's all of these values I am filling in in green. And you can verify that it works in the original inequality. Pick something that should work. Well, 0 should work. 0 is one of the numbers that we filled in. Negative 5 times 0 is 0, which is less than or equal to 15. It's less than 15."}, {"video_title": "Multiplying and dividing with inequalities example   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Well, 0 should work. 0 is one of the numbers that we filled in. Negative 5 times 0 is 0, which is less than or equal to 15. It's less than 15. Now let's try a number that's outside of it. And I haven't drawn it here. I could continue with the number line in this direction."}, {"video_title": "Multiplying and dividing with inequalities example   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "It's less than 15. Now let's try a number that's outside of it. And I haven't drawn it here. I could continue with the number line in this direction. You would have a negative 4 here. Negative 4 should not be included. And let's verify that negative 4 doesn't work."}, {"video_title": "Multiplying and dividing with inequalities example   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "I could continue with the number line in this direction. You would have a negative 4 here. Negative 4 should not be included. And let's verify that negative 4 doesn't work. Negative 4 times negative 5 is positive 20. And positive 20 is not less than 15. So it's good that we did not include negative 4."}, {"video_title": "Multiplying and dividing with inequalities example   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "And let's verify that negative 4 doesn't work. Negative 4 times negative 5 is positive 20. And positive 20 is not less than 15. So it's good that we did not include negative 4. This is our solution, and this is that solution graphed. I wanted to do that in that other green color. Here you go."}, {"video_title": "Subtracting a negative = adding a positive   Pre-Algebra   Khan Academy.mp3", "Sentence": "And Steve is in a bit of a bind. Not only does he not have any money, but he also owes Michael, who is not depicted here, he owes Michael three dollars. So he actually has a negative net worth. Steve has a net worth of negative three. Steve's uncle cares about Steve and feels bad that he doesn't even have a zero net worth. He has a negative net worth. And so Steve's uncle wants to at least take away some of this pain."}, {"video_title": "Subtracting a negative = adding a positive   Pre-Algebra   Khan Academy.mp3", "Sentence": "Steve has a net worth of negative three. Steve's uncle cares about Steve and feels bad that he doesn't even have a zero net worth. He has a negative net worth. And so Steve's uncle wants to at least take away some of this pain. He at least wants to get Steve back to a neutral net worth, a zero net worth. And so he decides to take away Steve's negative, negative net worth. So he wants to take away that negative three."}, {"video_title": "Subtracting a negative = adding a positive   Pre-Algebra   Khan Academy.mp3", "Sentence": "And so Steve's uncle wants to at least take away some of this pain. He at least wants to get Steve back to a neutral net worth, a zero net worth. And so he decides to take away Steve's negative, negative net worth. So he wants to take away that negative three. Well, what happens if you take away that negative three? Well, that should get you back to zero. If you take away anything, if you have something and you take it away, it should go back to zero."}, {"video_title": "Subtracting a negative = adding a positive   Pre-Algebra   Khan Academy.mp3", "Sentence": "So he wants to take away that negative three. Well, what happens if you take away that negative three? Well, that should get you back to zero. If you take away anything, if you have something and you take it away, it should go back to zero. So similarly, if you owe people things and that owing people things is taken away, it also gets back to zero. And another way to think about it is how Steve's uncle would actually take that liability, take that debt away from Steve. Well, the easiest way he could do it, if Steve is starting at negative three net worth, is for his uncle to just give Steve three dollars."}, {"video_title": "How to find equivalent expressions by combining like terms and using the distributive property.mp3", "Sentence": "Which expressions are equivalent to x plus 2y plus x plus 2? Select all that apply. All right, let's see if I can, if I can manipulate this thing a little bit. Let me just rewrite it. So I have x plus, x plus 2y plus x plus 2. So the first thing that jumps out at me before I even look at these choices here, I have an x over here, I have an x over there. Well, if I have one x and then I can add it to another x, that would be 2x."}, {"video_title": "How to find equivalent expressions by combining like terms and using the distributive property.mp3", "Sentence": "Let me just rewrite it. So I have x plus, x plus 2y plus x plus 2. So the first thing that jumps out at me before I even look at these choices here, I have an x over here, I have an x over there. Well, if I have one x and then I can add it to another x, that would be 2x. So I could rewrite this, let me do this in a different color. I could rewrite this x and this x, if I add them together, that's going to be 2x. So this is, let me, actually let me, I don't want to skip any steps."}, {"video_title": "How to find equivalent expressions by combining like terms and using the distributive property.mp3", "Sentence": "Well, if I have one x and then I can add it to another x, that would be 2x. So I could rewrite this, let me do this in a different color. I could rewrite this x and this x, if I add them together, that's going to be 2x. So this is, let me, actually let me, I don't want to skip any steps. So that's x plus x plus 2y, now I'm just switching the order, plus 2, and then these two x's right over here, I can just rewrite that as 2x. So I have 2x plus 2y, plus 2y, plus 2. Now let's see, out of all of my choices, so this one, this is 2x plus 4y, plus 4."}, {"video_title": "How to find equivalent expressions by combining like terms and using the distributive property.mp3", "Sentence": "So this is, let me, actually let me, I don't want to skip any steps. So that's x plus x plus 2y, now I'm just switching the order, plus 2, and then these two x's right over here, I can just rewrite that as 2x. So I have 2x plus 2y, plus 2y, plus 2. Now let's see, out of all of my choices, so this one, this is 2x plus 4y, plus 4. So that's not right, I have 2x plus 2y plus 2, so I can rule this one out. Now this one's interesting, it looks like they have factored out a 2. Let's see, if we factor out a 2 here, what happens?"}, {"video_title": "How to find equivalent expressions by combining like terms and using the distributive property.mp3", "Sentence": "Now let's see, out of all of my choices, so this one, this is 2x plus 4y, plus 4. So that's not right, I have 2x plus 2y plus 2, so I can rule this one out. Now this one's interesting, it looks like they have factored out a 2. Let's see, if we factor out a 2 here, what happens? So we do see that 2 is a factor of that term, it's a factor of that term, and it's a factor of that term. So let's see if we can, if we can factor it out. So this is going to be 2 times x, I'll do the x in that same magenta color, 2 times x, plus, you have just a y left when you factor out the 2, and then if you factor out a 2 here, you're just going to have a 1 left."}, {"video_title": "Example 1 Multiplying a binomial by a binomial   Algebra I   Khan Academy.mp3", "Sentence": "Multiply 3x plus 2 times 5x minus 7. So we're multiplying two binomials. I'm actually going to show you two really equivalent ways of doing this, one that you might hear in a classroom, and it's kind of more of a mechanical, memorizing way of doing it, which might be faster, but you really don't know what you're doing. And then there's the one where you're essentially just applying something that you already know in kind of a logical way. So I'll first do the memorizing way that you might be exposed to. And they'll use something called FOIL. Let me write this down here."}, {"video_title": "Example 1 Multiplying a binomial by a binomial   Algebra I   Khan Academy.mp3", "Sentence": "And then there's the one where you're essentially just applying something that you already know in kind of a logical way. So I'll first do the memorizing way that you might be exposed to. And they'll use something called FOIL. Let me write this down here. So you could immediately see that whenever someone gives you a mnemonic to memorize, that you're doing something pretty mechanical. So FOIL literally stands for First Outside. Let me write it this way."}, {"video_title": "Example 1 Multiplying a binomial by a binomial   Algebra I   Khan Academy.mp3", "Sentence": "Let me write this down here. So you could immediately see that whenever someone gives you a mnemonic to memorize, that you're doing something pretty mechanical. So FOIL literally stands for First Outside. Let me write it this way. Let me write FOIL. Where the F in FOIL stands for First. The O in FOIL stands for Outside."}, {"video_title": "Example 1 Multiplying a binomial by a binomial   Algebra I   Khan Academy.mp3", "Sentence": "Let me write it this way. Let me write FOIL. Where the F in FOIL stands for First. The O in FOIL stands for Outside. The I stands for Inside. And then the L stands for Last. And the reason why I don't like these things is when you're 35 years old, you're not going to remember what FOIL stood for, and then you're not going to remember how to multiply this binomial."}, {"video_title": "Example 1 Multiplying a binomial by a binomial   Algebra I   Khan Academy.mp3", "Sentence": "The O in FOIL stands for Outside. The I stands for Inside. And then the L stands for Last. And the reason why I don't like these things is when you're 35 years old, you're not going to remember what FOIL stood for, and then you're not going to remember how to multiply this binomial. But let's just apply FOIL. So FIRST says just multiply the first terms in each of these binomials. So just multiply the 3x times the 5x."}, {"video_title": "Example 1 Multiplying a binomial by a binomial   Algebra I   Khan Academy.mp3", "Sentence": "And the reason why I don't like these things is when you're 35 years old, you're not going to remember what FOIL stood for, and then you're not going to remember how to multiply this binomial. But let's just apply FOIL. So FIRST says just multiply the first terms in each of these binomials. So just multiply the 3x times the 5x. So 3x times the 5x. The Outside part tells us to multiply the outside terms. So in this case, you have 3x on the outside, and you have negative 7 on the outside."}, {"video_title": "Example 1 Multiplying a binomial by a binomial   Algebra I   Khan Academy.mp3", "Sentence": "So just multiply the 3x times the 5x. So 3x times the 5x. The Outside part tells us to multiply the outside terms. So in this case, you have 3x on the outside, and you have negative 7 on the outside. So that is plus 3x times negative 7. The inside. Well, the inside terms here are 2 and 5x."}, {"video_title": "Example 1 Multiplying a binomial by a binomial   Algebra I   Khan Academy.mp3", "Sentence": "So in this case, you have 3x on the outside, and you have negative 7 on the outside. So that is plus 3x times negative 7. The inside. Well, the inside terms here are 2 and 5x. So plus 2 times 5x. And then finally, you have the last terms. You have the 2 and the negative 7."}, {"video_title": "Example 1 Multiplying a binomial by a binomial   Algebra I   Khan Academy.mp3", "Sentence": "Well, the inside terms here are 2 and 5x. So plus 2 times 5x. And then finally, you have the last terms. You have the 2 and the negative 7. So the last term is 2 times negative 7. 2 times negative 7. So what you're essentially doing is just making sure that you're multiplying each term by every other term here."}, {"video_title": "Example 1 Multiplying a binomial by a binomial   Algebra I   Khan Academy.mp3", "Sentence": "You have the 2 and the negative 7. So the last term is 2 times negative 7. 2 times negative 7. So what you're essentially doing is just making sure that you're multiplying each term by every other term here. What we're essentially doing is multiplying, doing the distributive property twice. We're multiplying the 3x times 5x minus 7. So 3x times 5x minus 7 is 3x times 5x plus 3x minus 7."}, {"video_title": "Example 1 Multiplying a binomial by a binomial   Algebra I   Khan Academy.mp3", "Sentence": "So what you're essentially doing is just making sure that you're multiplying each term by every other term here. What we're essentially doing is multiplying, doing the distributive property twice. We're multiplying the 3x times 5x minus 7. So 3x times 5x minus 7 is 3x times 5x plus 3x minus 7. And we're multiplying the 2 times 5x minus 7 to give us these terms. But anyway, let's just multiply this out just to get our answer. 3x times 5x, the same thing as 3 times 5 times x times x, which is the same thing as 15x squared."}, {"video_title": "Example 1 Multiplying a binomial by a binomial   Algebra I   Khan Academy.mp3", "Sentence": "So 3x times 5x minus 7 is 3x times 5x plus 3x minus 7. And we're multiplying the 2 times 5x minus 7 to give us these terms. But anyway, let's just multiply this out just to get our answer. 3x times 5x, the same thing as 3 times 5 times x times x, which is the same thing as 15x squared. You can use x to the first times x to the first. You multiply the x's, you get x squared. 3 times 5 is 15."}, {"video_title": "Example 1 Multiplying a binomial by a binomial   Algebra I   Khan Academy.mp3", "Sentence": "3x times 5x, the same thing as 3 times 5 times x times x, which is the same thing as 15x squared. You can use x to the first times x to the first. You multiply the x's, you get x squared. 3 times 5 is 15. This term right here, 3 times negative 7 is negative 21. And then you have your x right over here. And then you have this term, which is 2 times 5, which is 10 times x, so plus 10x."}, {"video_title": "Example 1 Multiplying a binomial by a binomial   Algebra I   Khan Academy.mp3", "Sentence": "3 times 5 is 15. This term right here, 3 times negative 7 is negative 21. And then you have your x right over here. And then you have this term, which is 2 times 5, which is 10 times x, so plus 10x. And then finally, you have this term here in blue. 2 times negative 7 is negative 14. And we aren't done yet."}, {"video_title": "Example 1 Multiplying a binomial by a binomial   Algebra I   Khan Academy.mp3", "Sentence": "And then you have this term, which is 2 times 5, which is 10 times x, so plus 10x. And then finally, you have this term here in blue. 2 times negative 7 is negative 14. And we aren't done yet. We can simplify this a little bit. We have two like terms here. We have this."}, {"video_title": "Example 1 Multiplying a binomial by a binomial   Algebra I   Khan Academy.mp3", "Sentence": "And we aren't done yet. We can simplify this a little bit. We have two like terms here. We have this. Let me find a new color. We have two terms with an x to the first power, just an x term right over here. So if we have negative 21 of something and you add 10, or another way, if you have 10 of something and you subtract 21 of them, you're going to have negative 11 of that something."}, {"video_title": "Example 1 Multiplying a binomial by a binomial   Algebra I   Khan Academy.mp3", "Sentence": "We have this. Let me find a new color. We have two terms with an x to the first power, just an x term right over here. So if we have negative 21 of something and you add 10, or another way, if you have 10 of something and you subtract 21 of them, you're going to have negative 11 of that something. And we put the other terms here. You have 15x squared, and then you have your minus 14. And we are done."}, {"video_title": "Example 1 Multiplying a binomial by a binomial   Algebra I   Khan Academy.mp3", "Sentence": "So if we have negative 21 of something and you add 10, or another way, if you have 10 of something and you subtract 21 of them, you're going to have negative 11 of that something. And we put the other terms here. You have 15x squared, and then you have your minus 14. And we are done. Now I said I would show you another way to do it. So I want to show you why the distributive property can get us here without having to memorize FOIL. So the distributive property tells us that if we're multiplying something times an expression, you just have to multiply it times every term in the expression."}, {"video_title": "Example 1 Multiplying a binomial by a binomial   Algebra I   Khan Academy.mp3", "Sentence": "And we are done. Now I said I would show you another way to do it. So I want to show you why the distributive property can get us here without having to memorize FOIL. So the distributive property tells us that if we're multiplying something times an expression, you just have to multiply it times every term in the expression. So we can distribute the 5x onto the 3x minus 7, this whole thing, onto the 3x plus 2. Let me just change the order, since we're used to distributing something from the left. So this is the same thing as 5x minus 7 times 3x plus 2."}, {"video_title": "Example 1 Multiplying a binomial by a binomial   Algebra I   Khan Academy.mp3", "Sentence": "So the distributive property tells us that if we're multiplying something times an expression, you just have to multiply it times every term in the expression. So we can distribute the 5x onto the 3x minus 7, this whole thing, onto the 3x plus 2. Let me just change the order, since we're used to distributing something from the left. So this is the same thing as 5x minus 7 times 3x plus 2. I just swapped the two expressions. And we can distribute this whole thing times each of these terms. Now what happens if I take 5x minus 7 times 3x?"}, {"video_title": "Example 1 Multiplying a binomial by a binomial   Algebra I   Khan Academy.mp3", "Sentence": "So this is the same thing as 5x minus 7 times 3x plus 2. I just swapped the two expressions. And we can distribute this whole thing times each of these terms. Now what happens if I take 5x minus 7 times 3x? Well, that's just going to be 3x times 5x minus 7. So I've just distributed the 5x minus 7 times 3x. And to that, I'm going to add 2 times 5x minus 7."}, {"video_title": "Example 1 Multiplying a binomial by a binomial   Algebra I   Khan Academy.mp3", "Sentence": "Now what happens if I take 5x minus 7 times 3x? Well, that's just going to be 3x times 5x minus 7. So I've just distributed the 5x minus 7 times 3x. And to that, I'm going to add 2 times 5x minus 7. I've just distributed the 5x minus 7 onto the 2. Now we can do distributive property again. We can distribute the 3x onto the 5x."}, {"video_title": "Example 1 Multiplying a binomial by a binomial   Algebra I   Khan Academy.mp3", "Sentence": "And to that, I'm going to add 2 times 5x minus 7. I've just distributed the 5x minus 7 onto the 2. Now we can do distributive property again. We can distribute the 3x onto the 5x. And we can distribute the 3x onto the negative 7. We can distribute the 2 onto the 5x over here. And we can distribute the 2 on that negative 7."}, {"video_title": "Example 1 Multiplying a binomial by a binomial   Algebra I   Khan Academy.mp3", "Sentence": "We can distribute the 3x onto the 5x. And we can distribute the 3x onto the negative 7. We can distribute the 2 onto the 5x over here. And we can distribute the 2 on that negative 7. Now if we do it like this, what do we get? 3x times 5x. That's this right over here."}, {"video_title": "Example 1 Multiplying a binomial by a binomial   Algebra I   Khan Academy.mp3", "Sentence": "And we can distribute the 2 on that negative 7. Now if we do it like this, what do we get? 3x times 5x. That's this right over here. If we do 3x times negative 7, that's this term right over here. If you do 2 times 5x, that's this term right over here. If you do 2 times negative 7, that is this term right over here."}, {"video_title": "Example 1 Multiplying a binomial by a binomial   Algebra I   Khan Academy.mp3", "Sentence": "That's this right over here. If we do 3x times negative 7, that's this term right over here. If you do 2 times 5x, that's this term right over here. If you do 2 times negative 7, that is this term right over here. So we got the exact same result that we got with FOIL. Now FOIL can be faster. If you just want to do it, you kind of can skip to this step."}, {"video_title": "System of equations word problem infinite solutions   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "Farmer Jan is a vegetable farmer who divides his field between broccoli crops and spinach crops. Last year, he grew six tons of broccoli per acre, so six tons of broccoli per acre, and nine tons of spinach, nine tons of spinach per acre, for a total of 93 tons of vegetables. This year, he grew two tons of broccoli per acre, so this year he grew two tons of broccoli per acre, and three tons of spinach per acre, and three tons of spinach per acre, for a total of 31 tons of vegetables. How many acres of broccoli crops and how many acres of spinach crops does Farmer Jan have? So let's think about this. So let's say that the number of acres of broccoli crops, let's call that B, and let's say the acres of spinach crops, the acres of spinach crops, let's call that S. So how much broccoli is he going to grow in, I guess you could say, last year? Last year, how much broccoli did he grow?"}, {"video_title": "System of equations word problem infinite solutions   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "How many acres of broccoli crops and how many acres of spinach crops does Farmer Jan have? So let's think about this. So let's say that the number of acres of broccoli crops, let's call that B, and let's say the acres of spinach crops, the acres of spinach crops, let's call that S. So how much broccoli is he going to grow in, I guess you could say, last year? Last year, how much broccoli did he grow? And let me just write this down. This is last year. Last year."}, {"video_title": "System of equations word problem infinite solutions   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "Last year, how much broccoli did he grow? And let me just write this down. This is last year. Last year. And they tell us, last year, he grew six tons of broccoli per acre. So if he grew six tons of broccoli per acre, and he has B acres, well, that means he grew six tons per acre times B acres. So he grew six B tons of broccoli last year, and by the same logic, he grew how much spinach?"}, {"video_title": "System of equations word problem infinite solutions   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "Last year. And they tell us, last year, he grew six tons of broccoli per acre. So if he grew six tons of broccoli per acre, and he has B acres, well, that means he grew six tons per acre times B acres. So he grew six B tons of broccoli last year, and by the same logic, he grew how much spinach? Well, nine tons of spinach per acre times S acres. So nine S tons of spinach, and then the total is 93 tons. So for a total of 93 tons of vegetables."}, {"video_title": "System of equations word problem infinite solutions   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "So he grew six B tons of broccoli last year, and by the same logic, he grew how much spinach? Well, nine tons of spinach per acre times S acres. So nine S tons of spinach, and then the total is 93 tons. So for a total of 93 tons of vegetables. So this is going to be equal to 93. So now let's think about this year. And in general, when you tackle these, just think about, well, set the variables, in general, to what they're asking for, and then how can you use the information that they're giving us?"}, {"video_title": "System of equations word problem infinite solutions   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "So for a total of 93 tons of vegetables. So this is going to be equal to 93. So now let's think about this year. And in general, when you tackle these, just think about, well, set the variables, in general, to what they're asking for, and then how can you use the information that they're giving us? So this year, how much broccoli would he have grown? Well, he grew two tons of broccoli per acre. So he grew two tons per acre."}, {"video_title": "System of equations word problem infinite solutions   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "And in general, when you tackle these, just think about, well, set the variables, in general, to what they're asking for, and then how can you use the information that they're giving us? So this year, how much broccoli would he have grown? Well, he grew two tons of broccoli per acre. So he grew two tons per acre. He has the same number of acres, we can assume that. So two tons per acre times B acres is gonna be two B tons of broccoli. And by that same logic, it's going to be, he grew three tons of spinach per acre."}, {"video_title": "System of equations word problem infinite solutions   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "So he grew two tons per acre. He has the same number of acres, we can assume that. So two tons per acre times B acres is gonna be two B tons of broccoli. And by that same logic, it's going to be, he grew three tons of spinach per acre. Well, he has S acres. Each of those acres, he's growing three tons of spinach per acre. So it's going to be three S tons of spinach."}, {"video_title": "System of equations word problem infinite solutions   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "And by that same logic, it's going to be, he grew three tons of spinach per acre. Well, he has S acres. Each of those acres, he's growing three tons of spinach per acre. So it's going to be three S tons of spinach. And they tell us what that total is. That total is, for a total of 31 tons of vegetables. So this is going to be equal to 31."}, {"video_title": "System of equations word problem infinite solutions   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "So it's going to be three S tons of spinach. And they tell us what that total is. That total is, for a total of 31 tons of vegetables. So this is going to be equal to 31. And so now we have a system of two equations with two unknowns that we can use to solve for B and S. So let's see, what do we want to solve for first? Well, what we could do, let me rewrite the top equation. So we have six B plus nine S is equal to 93."}, {"video_title": "System of equations word problem infinite solutions   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "So this is going to be equal to 31. And so now we have a system of two equations with two unknowns that we can use to solve for B and S. So let's see, what do we want to solve for first? Well, what we could do, let me rewrite the top equation. So we have six B plus nine S is equal to 93. Is equal to 93. And the second equation, well, let's try to eliminate, let's try to eliminate the Bs. So let's multiply the second equation by negative three."}, {"video_title": "System of equations word problem infinite solutions   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "So we have six B plus nine S is equal to 93. Is equal to 93. And the second equation, well, let's try to eliminate, let's try to eliminate the Bs. So let's multiply the second equation by negative three. So negative, I'm gonna multiply the left-hand side by negative three. And I'm gonna multiply the right-hand side by negative three. And so what am I going to get?"}, {"video_title": "System of equations word problem infinite solutions   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "So let's multiply the second equation by negative three. So negative, I'm gonna multiply the left-hand side by negative three. And I'm gonna multiply the right-hand side by negative three. And so what am I going to get? Negative three times two B is negative six B. That was the whole point, by multiplying it by negative three. Negative three times three S is negative nine S. Negative nine S. And then negative three times 31 is going to be negative 93."}, {"video_title": "System of equations word problem infinite solutions   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "And so what am I going to get? Negative three times two B is negative six B. That was the whole point, by multiplying it by negative three. Negative three times three S is negative nine S. Negative nine S. And then negative three times 31 is going to be negative 93. So what do we get if we now add the two sides of these equations? So on the left-hand side, six B minus six B, that's zero. Nine S minus nine S, that's zero."}, {"video_title": "System of equations word problem infinite solutions   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "Negative three times three S is negative nine S. Negative nine S. And then negative three times 31 is going to be negative 93. So what do we get if we now add the two sides of these equations? So on the left-hand side, six B minus six B, that's zero. Nine S minus nine S, that's zero. We're gonna get just zero. On the right-hand side, we get 93 minus 93. Well, that's still just going to be zero."}, {"video_title": "System of equations word problem infinite solutions   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "Nine S minus nine S, that's zero. We're gonna get just zero. On the right-hand side, we get 93 minus 93. Well, that's still just going to be zero. So we have this situation where we get zero equals zero, which is going to be true no matter what X and Y are. And so this is a system with an infinite number of solutions. So this has infinite, infinite, infinite number of solutions."}, {"video_title": "System of equations word problem infinite solutions   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "Well, that's still just going to be zero. So we have this situation where we get zero equals zero, which is going to be true no matter what X and Y are. And so this is a system with an infinite number of solutions. So this has infinite, infinite, infinite number of solutions. So one way we could think about it is, these two constraints, they're not giving us enough information. There's an infinite number of B and S combinations that would satisfy these equations. So they're not giving us enough information to say how exactly what B and S are."}, {"video_title": "How to find the domain and range of a piecewise function   Functions   Algebra I   Khan Academy.mp3", "Sentence": "I have a piecewise defined function here, and my goal is to figure out its domain and its range. So first let's think about the domain. And just as a bit of review, the domain is the set of all inputs for which our function is defined. And over here our input variable is x. So you can think about it as a set of all the values that x can take on and actually have this function be defined, actually be able to figure out what f of x is. And when we look at this, we see, okay, if zero is less than x is less than or equal to two, we're in this clause, as x crosses two and is greater than two, we fall into this clause. And as we approach six, but right when we get to six, we fall into this clause right over here, all the way up to and including 11."}, {"video_title": "How to find the domain and range of a piecewise function   Functions   Algebra I   Khan Academy.mp3", "Sentence": "And over here our input variable is x. So you can think about it as a set of all the values that x can take on and actually have this function be defined, actually be able to figure out what f of x is. And when we look at this, we see, okay, if zero is less than x is less than or equal to two, we're in this clause, as x crosses two and is greater than two, we fall into this clause. And as we approach six, but right when we get to six, we fall into this clause right over here, all the way up to and including 11. But if we get larger than 11, the function is no longer defined. I don't know which of these to use. And if we're at zero or less, the function is no longer defined as well."}, {"video_title": "How to find the domain and range of a piecewise function   Functions   Algebra I   Khan Academy.mp3", "Sentence": "And as we approach six, but right when we get to six, we fall into this clause right over here, all the way up to and including 11. But if we get larger than 11, the function is no longer defined. I don't know which of these to use. And if we're at zero or less, the function is no longer defined as well. So in order for this to be defined, x has to be greater than zero, or we have to say zero is less than x. And you see that part right over there. And x has to be less than or equal to 11."}, {"video_title": "How to find the domain and range of a piecewise function   Functions   Algebra I   Khan Academy.mp3", "Sentence": "And if we're at zero or less, the function is no longer defined as well. So in order for this to be defined, x has to be greater than zero, or we have to say zero is less than x. And you see that part right over there. And x has to be less than or equal to 11. And x has to be less than or equal to 11. It's defined for everything in between. As we, once again, as we get to two, we're here."}, {"video_title": "How to find the domain and range of a piecewise function   Functions   Algebra I   Khan Academy.mp3", "Sentence": "And x has to be less than or equal to 11. And x has to be less than or equal to 11. It's defined for everything in between. As we, once again, as we get to two, we're here. As we cross two, between two and six, we're here. And at six, from six to 11, we're over here. So we're defined for all real numbers in this interval."}, {"video_title": "How to find the domain and range of a piecewise function   Functions   Algebra I   Khan Academy.mp3", "Sentence": "As we, once again, as we get to two, we're here. As we cross two, between two and six, we're here. And at six, from six to 11, we're over here. So we're defined for all real numbers in this interval. So our domain is, actually let me write this, all real values, are all real, let me write all real values, maybe all, let me write it that way, all real values such that, such that zero is less than x is less than or equal to 11. So now let's think about the range. Let's think about the range of this piecewise defined function."}, {"video_title": "How to find the domain and range of a piecewise function   Functions   Algebra I   Khan Academy.mp3", "Sentence": "So we're defined for all real numbers in this interval. So our domain is, actually let me write this, all real values, are all real, let me write all real values, maybe all, let me write it that way, all real values such that, such that zero is less than x is less than or equal to 11. So now let's think about the range. Let's think about the range of this piecewise defined function. And that's the set of all values that this function can actually take on. And this one is maybe deceptively simple, because there's only three values that this function can take on. It can take on, f of x can be equal to one, it can be equal to five, or it could be equal to negative seven."}, {"video_title": "How to find the domain and range of a piecewise function   Functions   Algebra I   Khan Academy.mp3", "Sentence": "Let's think about the range of this piecewise defined function. And that's the set of all values that this function can actually take on. And this one is maybe deceptively simple, because there's only three values that this function can take on. It can take on, f of x can be equal to one, it can be equal to five, or it could be equal to negative seven. So the range here, we could say that f of x needs to be a member of, this is just a fancy mathy symbol, which would say this is a member of the set one, five, negative seven. f of x is going to take, is going to take on one of these three values. Another way to say it is that f of x is going to be equal to one, five, or negative seven."}, {"video_title": "What is a variable   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Let's say that I'm working in a restaurant and I'm making $10 per hour, but on top of my hourly wage, I also get tips each hour. So this entire expression, you can view this, is how much I might make in a given hour. Now, you might also realize that the number of tips or the amount of tips I might make in an hour can change dramatically from hour to hour. It can vary. One hour, it might be lunchtime, get a lot of tips, people might get some big ticket items. The next hour, I might not have any customers, and then my tips might be really low. So the tips part right over here, we consider that, the entire word, we consider that to be a variable."}, {"video_title": "What is a variable   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "It can vary. One hour, it might be lunchtime, get a lot of tips, people might get some big ticket items. The next hour, I might not have any customers, and then my tips might be really low. So the tips part right over here, we consider that, the entire word, we consider that to be a variable. From scenario to scenario, it can change. So for example, in one scenario, maybe it's lunchtime, I'm getting really big tips. So tips is equal to, let's say it's equal to $30."}, {"video_title": "What is a variable   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "So the tips part right over here, we consider that, the entire word, we consider that to be a variable. From scenario to scenario, it can change. So for example, in one scenario, maybe it's lunchtime, I'm getting really big tips. So tips is equal to, let's say it's equal to $30. And so the total amount I might make in that hour is going, we can go back to this expression right over here, it's going to be 10 plus, instead of writing tips here, I'll write 30, because that's what my tips are in that hour. And so that is going to be equal to, it's going to be equal to 40. Let me do that, let me do that in yellow color."}, {"video_title": "What is a variable   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "So tips is equal to, let's say it's equal to $30. And so the total amount I might make in that hour is going, we can go back to this expression right over here, it's going to be 10 plus, instead of writing tips here, I'll write 30, because that's what my tips are in that hour. And so that is going to be equal to, it's going to be equal to 40. Let me do that, let me do that in yellow color. It's going to be equal to $40. But let's say right after that, the restaurant slows down, we're out of the lunch hour for whatever reason, maybe the restaurant next door has a big sale or something. And so the next hour, my tips go down dramatically."}, {"video_title": "What is a variable   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Let me do that, let me do that in yellow color. It's going to be equal to $40. But let's say right after that, the restaurant slows down, we're out of the lunch hour for whatever reason, maybe the restaurant next door has a big sale or something. And so the next hour, my tips go down dramatically. My tips go down to $5 for that hour. Now I go back to this expression. The total I make is my hourly wage plus the $5 in tips, plus the $5 in tips, which is equal to $15."}, {"video_title": "What is a variable   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And so the next hour, my tips go down dramatically. My tips go down to $5 for that hour. Now I go back to this expression. The total I make is my hourly wage plus the $5 in tips, plus the $5 in tips, which is equal to $15. As you see, this entire expression, the 10 plus tips, it changed depending on what the value of the variable tips is. Now, you won't see whole words typically used in algebra as variables. We get lazy, and so instead, we tend to use just easier to write symbols."}, {"video_title": "What is a variable   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "The total I make is my hourly wage plus the $5 in tips, plus the $5 in tips, which is equal to $15. As you see, this entire expression, the 10 plus tips, it changed depending on what the value of the variable tips is. Now, you won't see whole words typically used in algebra as variables. We get lazy, and so instead, we tend to use just easier to write symbols. And so in this context, instead of writing tips, maybe we could have just written 10 plus t, where t represents the tips that we get in an hour. And so then we would say, okay, what happens when t is equal to 30? Well, if t is equal to 30, then we'd have, let me write, so what happens when t is equal to 30?"}, {"video_title": "What is a variable   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "We get lazy, and so instead, we tend to use just easier to write symbols. And so in this context, instead of writing tips, maybe we could have just written 10 plus t, where t represents the tips that we get in an hour. And so then we would say, okay, what happens when t is equal to 30? Well, if t is equal to 30, then we'd have, let me write, so what happens when t is equal to 30? Well, then we have a situation, t is equal to 30. This evaluates to 10 plus 30, which would be 40. What would happen if t is equal to five?"}, {"video_title": "What is a variable   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Well, if t is equal to 30, then we'd have, let me write, so what happens when t is equal to 30? Well, then we have a situation, t is equal to 30. This evaluates to 10 plus 30, which would be 40. What would happen if t is equal to five? Well, then this would evaluate to 10 plus five, which is equal to 15. Now I wanna be clear, we didn't even have to use t. We didn't even really have to use a letter, although in traditional algebra, you almost do use a letter. We could have written it as 10 plus x, where x is your tips per hour."}, {"video_title": "What is a variable   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "What would happen if t is equal to five? Well, then this would evaluate to 10 plus five, which is equal to 15. Now I wanna be clear, we didn't even have to use t. We didn't even really have to use a letter, although in traditional algebra, you almost do use a letter. We could have written it as 10 plus x, where x is your tips per hour. X might not be as natural. It's not the first letter in the word tips. Or you could have even written 10 plus, you could have even written 10 plus star, where you could say star represents the number of tips in an hour, but it just might have not made as much intuitive sense."}, {"video_title": "What is a variable   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "We could have written it as 10 plus x, where x is your tips per hour. X might not be as natural. It's not the first letter in the word tips. Or you could have even written 10 plus, you could have even written 10 plus star, where you could say star represents the number of tips in an hour, but it just might have not made as much intuitive sense. But hopefully this gives you a general idea of just what a variable is. All it is is a symbol. All it is is a symbol that represents different, a symbol that represents varying values, and that's why we call it a variable."}, {"video_title": "Worked example non-equivalent systems of equations   High School Math   Khan Academy.mp3", "Sentence": "So this is the teacher system, this is what Scarlett got after taking some steps, this is what Hansel got. Which of them obtained a system that is equivalent to the teacher's system? And just to remind ourselves, an equivalent system is a system that has, or at least for our purposes, is a system that has the same solution, or the same solution set. So if there's a certain xy that satisfies this system, in order for Scarlett's system to be, for it to be equivalent, it needs to have the same solution. So let's look at this. So Scarlett, let's see, let's see if we can match these up. So her second equation here, so this is interesting, her second equation, 14x minus seven y is equal to two."}, {"video_title": "Worked example non-equivalent systems of equations   High School Math   Khan Academy.mp3", "Sentence": "So if there's a certain xy that satisfies this system, in order for Scarlett's system to be, for it to be equivalent, it needs to have the same solution. So let's look at this. So Scarlett, let's see, let's see if we can match these up. So her second equation here, so this is interesting, her second equation, 14x minus seven y is equal to two. Over here, the teacher has an equation, 14x minus seven y is equal to seven. So this is interesting, because the ratio between x and y is the same, but then your constant term, the constant term is going to be different. And I would make the claim that this alone tells you that Scarlett's system is not equivalent to the teacher."}, {"video_title": "Worked example non-equivalent systems of equations   High School Math   Khan Academy.mp3", "Sentence": "So her second equation here, so this is interesting, her second equation, 14x minus seven y is equal to two. Over here, the teacher has an equation, 14x minus seven y is equal to seven. So this is interesting, because the ratio between x and y is the same, but then your constant term, the constant term is going to be different. And I would make the claim that this alone tells you that Scarlett's system is not equivalent to the teacher. And you're saying, well, how can I say that? Well, these two equations, if you were to write them into slope-intercept form, you would see, because the ratio between x and y, the x and y terms, is the same, you're going to have the same slope, but you're going to have different y-intercepts. In fact, we can actually solve for that."}, {"video_title": "Worked example non-equivalent systems of equations   High School Math   Khan Academy.mp3", "Sentence": "And I would make the claim that this alone tells you that Scarlett's system is not equivalent to the teacher. And you're saying, well, how can I say that? Well, these two equations, if you were to write them into slope-intercept form, you would see, because the ratio between x and y, the x and y terms, is the same, you're going to have the same slope, but you're going to have different y-intercepts. In fact, we can actually solve for that. So this equation right over here, we can write it as, if we, let's see, if we subtract 14x from both sides, you get negative seven y minus 14, whoops, negative seven y is equal to, is equal to negative 14x plus seven. And we could divide both sides by negative seven. You get y is equal to two x minus one."}, {"video_title": "Worked example non-equivalent systems of equations   High School Math   Khan Academy.mp3", "Sentence": "In fact, we can actually solve for that. So this equation right over here, we can write it as, if we, let's see, if we subtract 14x from both sides, you get negative seven y minus 14, whoops, negative seven y is equal to, is equal to negative 14x plus seven. And we could divide both sides by negative seven. You get y is equal to two x minus one. So that's this, all I did is algebraically manipulate this, this is this line, and I could even try to graph it. So let's do that. So I'll draw a quick coordinate, this is just going to be very rough, quick coordinate axis right over there."}, {"video_title": "Worked example non-equivalent systems of equations   High School Math   Khan Academy.mp3", "Sentence": "You get y is equal to two x minus one. So that's this, all I did is algebraically manipulate this, this is this line, and I could even try to graph it. So let's do that. So I'll draw a quick coordinate, this is just going to be very rough, quick coordinate axis right over there. And then this line, this line, would look something like this. So its y-intercept is negative one, and it has a slope of two. So let me draw a line with a slope of, a line with a slope of two, might look something, something like that."}, {"video_title": "Worked example non-equivalent systems of equations   High School Math   Khan Academy.mp3", "Sentence": "So I'll draw a quick coordinate, this is just going to be very rough, quick coordinate axis right over there. And then this line, this line, would look something like this. So its y-intercept is negative one, and it has a slope of two. So let me draw a line with a slope of, a line with a slope of two, might look something, something like that. So that's this line right over here, or this one right over there. And let's see, this one over here is going to be, if we do the same algebra, we're going to have negative seven y is equal to negative 14x plus two, or y is equal to, I'm just dividing everything by negative seven, two x minus 2 7ths. So this is going to look something like this."}, {"video_title": "Worked example non-equivalent systems of equations   High School Math   Khan Academy.mp3", "Sentence": "So let me draw a line with a slope of, a line with a slope of two, might look something, something like that. So that's this line right over here, or this one right over there. And let's see, this one over here is going to be, if we do the same algebra, we're going to have negative seven y is equal to negative 14x plus two, or y is equal to, I'm just dividing everything by negative seven, two x minus 2 7ths. So this is going to look something like this. Its y-intercept is minus 2 7ths, that's like right over there. So this line is going to look something like, I'm going to draw my best attempt at drawing it. It's going to look something, it's going to look, actually that's not quite right."}, {"video_title": "Worked example non-equivalent systems of equations   High School Math   Khan Academy.mp3", "Sentence": "So this is going to look something like this. Its y-intercept is minus 2 7ths, that's like right over there. So this line is going to look something like, I'm going to draw my best attempt at drawing it. It's going to look something, it's going to look, actually that's not quite right. It's going to look something like, actually I'll just start right over here. It's going to look something like, something like this. It's going to have the same slope, and obviously it goes in this direction as well."}, {"video_title": "Worked example non-equivalent systems of equations   High School Math   Khan Academy.mp3", "Sentence": "It's going to look something, it's going to look, actually that's not quite right. It's going to look something like, actually I'll just start right over here. It's going to look something like, something like this. It's going to have the same slope, and obviously it goes in this direction as well. Actually let me just draw that. It's going to have the same slope, but a different y-intercepts, and that doesn't look right, but you get the idea. These two lines are parallel."}, {"video_title": "Worked example non-equivalent systems of equations   High School Math   Khan Academy.mp3", "Sentence": "It's going to have the same slope, and obviously it goes in this direction as well. Actually let me just draw that. It's going to have the same slope, but a different y-intercepts, and that doesn't look right, but you get the idea. These two lines are parallel. So these two lines are parallel. So any coordinate that satisfies this one is not going to satisfy this one. They have no points in common."}, {"video_title": "Worked example non-equivalent systems of equations   High School Math   Khan Academy.mp3", "Sentence": "These two lines are parallel. So these two lines are parallel. So any coordinate that satisfies this one is not going to satisfy this one. They have no points in common. They are parallel. That's the definition of parallel. Since this and this have no points in common, there's no way that some solution set that satisfies this would satisfy this, because any xy that satisfies this can't satisfy this, or vice versa."}, {"video_title": "Worked example non-equivalent systems of equations   High School Math   Khan Academy.mp3", "Sentence": "They have no points in common. They are parallel. That's the definition of parallel. Since this and this have no points in common, there's no way that some solution set that satisfies this would satisfy this, because any xy that satisfies this can't satisfy this, or vice versa. They're parallel. There are no points. These two things will never intersect."}, {"video_title": "Worked example non-equivalent systems of equations   High School Math   Khan Academy.mp3", "Sentence": "Since this and this have no points in common, there's no way that some solution set that satisfies this would satisfy this, because any xy that satisfies this can't satisfy this, or vice versa. They're parallel. There are no points. These two things will never intersect. So Scarlett does not have an equivalent system. Now what about Hansel? Well we see Hansel has the same thing going on here."}, {"video_title": "Worked example non-equivalent systems of equations   High School Math   Khan Academy.mp3", "Sentence": "These two things will never intersect. So Scarlett does not have an equivalent system. Now what about Hansel? Well we see Hansel has the same thing going on here. Five x minus y, five x minus y, but then the constant term is different. Negative six, positive three. So this and this also represent parallel lines."}, {"video_title": "Worked example non-equivalent systems of equations   High School Math   Khan Academy.mp3", "Sentence": "Well we see Hansel has the same thing going on here. Five x minus y, five x minus y, but then the constant term is different. Negative six, positive three. So this and this also represent parallel lines. Any xy pair that satisfies this, there's no way that it's going to satisfy this. These two lines don't intersect. They are parallel."}, {"video_title": "How to find the domain and the range of a function given its graph (example)   Khan Academy.mp3", "Sentence": "What is its domain? So the way it's graphed right over here, we could assume that this is the entire function definition for f of x. So for example, if we say, well, what does f of x equal when x is equal to negative 9? Well, we go up here. We don't see it's graphed here. It's not defined for x equals negative 9, or x equals negative 8 and 1 half, or x equals negative 8. It's not defined for any of these values."}, {"video_title": "How to find the domain and the range of a function given its graph (example)   Khan Academy.mp3", "Sentence": "Well, we go up here. We don't see it's graphed here. It's not defined for x equals negative 9, or x equals negative 8 and 1 half, or x equals negative 8. It's not defined for any of these values. It only starts getting defined at x equals negative 6. At x equals negative 6, f of x is equal to 5. And then it keeps getting defined."}, {"video_title": "How to find the domain and the range of a function given its graph (example)   Khan Academy.mp3", "Sentence": "It's not defined for any of these values. It only starts getting defined at x equals negative 6. At x equals negative 6, f of x is equal to 5. And then it keeps getting defined. f of x is defined for x all the way from x equals negative 6 all the way to x equals 7. When x equals 7, f of x is equal to 5. You can take any x value between negative 6, including negative 6, and positive 7, including positive 7, and you just have to move up above that number wherever you are to find out what the value of the function is at that point."}, {"video_title": "How to find the domain and the range of a function given its graph (example)   Khan Academy.mp3", "Sentence": "And then it keeps getting defined. f of x is defined for x all the way from x equals negative 6 all the way to x equals 7. When x equals 7, f of x is equal to 5. You can take any x value between negative 6, including negative 6, and positive 7, including positive 7, and you just have to move up above that number wherever you are to find out what the value of the function is at that point. So the domain of this function definition, well, f of x is defined for any x that is greater than or equal to negative 6. Or we could say negative 6 is less than or equal to x, which is less than or equal to 7. If x satisfies this condition right over here, the function is defined."}, {"video_title": "How to find the domain and the range of a function given its graph (example)   Khan Academy.mp3", "Sentence": "You can take any x value between negative 6, including negative 6, and positive 7, including positive 7, and you just have to move up above that number wherever you are to find out what the value of the function is at that point. So the domain of this function definition, well, f of x is defined for any x that is greater than or equal to negative 6. Or we could say negative 6 is less than or equal to x, which is less than or equal to 7. If x satisfies this condition right over here, the function is defined. So that's its domain. So let's check our answer. Let's do a few more of these."}, {"video_title": "How to find the domain and the range of a function given its graph (example)   Khan Academy.mp3", "Sentence": "If x satisfies this condition right over here, the function is defined. So that's its domain. So let's check our answer. Let's do a few more of these. The function f of x is graphed. What is its domain? Well, exact similar argument."}, {"video_title": "How to find the domain and the range of a function given its graph (example)   Khan Academy.mp3", "Sentence": "Let's do a few more of these. The function f of x is graphed. What is its domain? Well, exact similar argument. This function is not defined for x's negative 9, negative 8, all the way up, I should say, to negative 1. At negative 1, it starts getting defined. f of negative 1 is negative 5."}, {"video_title": "How to find the domain and the range of a function given its graph (example)   Khan Academy.mp3", "Sentence": "Well, exact similar argument. This function is not defined for x's negative 9, negative 8, all the way up, I should say, to negative 1. At negative 1, it starts getting defined. f of negative 1 is negative 5. So it's defined for negative 1 is less than or equal to x. And it's defined all the way up to x equals 7, including x equals 7. So this right over here, negative 1 is less than or equal to x is less than or equal to 7."}, {"video_title": "How to find the domain and the range of a function given its graph (example)   Khan Academy.mp3", "Sentence": "f of negative 1 is negative 5. So it's defined for negative 1 is less than or equal to x. And it's defined all the way up to x equals 7, including x equals 7. So this right over here, negative 1 is less than or equal to x is less than or equal to 7. The function is defined for any x that satisfies this double inequality right over here. Let's do a few more. The function f of x is graphed."}, {"video_title": "How to find the domain and the range of a function given its graph (example)   Khan Academy.mp3", "Sentence": "So this right over here, negative 1 is less than or equal to x is less than or equal to 7. The function is defined for any x that satisfies this double inequality right over here. Let's do a few more. The function f of x is graphed. What is its range? So now we're not thinking about the x's for which this function is defined. We're thinking about the set of y values."}, {"video_title": "How to find the domain and the range of a function given its graph (example)   Khan Academy.mp3", "Sentence": "The function f of x is graphed. What is its range? So now we're not thinking about the x's for which this function is defined. We're thinking about the set of y values. Where do all of the y values fall into? Well, let's see. The lowest possible y value, or the lowest possible value of f of x that we get here, looks like it's 0."}, {"video_title": "How to find the domain and the range of a function given its graph (example)   Khan Academy.mp3", "Sentence": "We're thinking about the set of y values. Where do all of the y values fall into? Well, let's see. The lowest possible y value, or the lowest possible value of f of x that we get here, looks like it's 0. The function never goes below 0. So f of x, so 0 is less than or equal to f of x. It does equal 0 right over here."}, {"video_title": "How to find the domain and the range of a function given its graph (example)   Khan Academy.mp3", "Sentence": "The lowest possible y value, or the lowest possible value of f of x that we get here, looks like it's 0. The function never goes below 0. So f of x, so 0 is less than or equal to f of x. It does equal 0 right over here. f of negative 4 is 0. And then the highest y value, or the highest value that f of x obtains in this function definition is 8. f of 7 is 8. It never gets above 8, but it does equal 8 right over here when x is equal to 7."}, {"video_title": "How to find the domain and the range of a function given its graph (example)   Khan Academy.mp3", "Sentence": "It does equal 0 right over here. f of negative 4 is 0. And then the highest y value, or the highest value that f of x obtains in this function definition is 8. f of 7 is 8. It never gets above 8, but it does equal 8 right over here when x is equal to 7. So 0 is less than f of x, which is less than or equal to 8. So that's its range. Let's do a few more."}, {"video_title": "How to find the domain and the range of a function given its graph (example)   Khan Academy.mp3", "Sentence": "It never gets above 8, but it does equal 8 right over here when x is equal to 7. So 0 is less than f of x, which is less than or equal to 8. So that's its range. Let's do a few more. This is kind of fun. The function f of x is graphed. What is its domain?"}, {"video_title": "How to find the domain and the range of a function given its graph (example)   Khan Academy.mp3", "Sentence": "Let's do a few more. This is kind of fun. The function f of x is graphed. What is its domain? So once again, this function is defined for negative 2 is less than or equal to x, which is less than or equal to 5. If you give me an x anywhere in between negative 2 and 5, I can look at this graph to see where the function is defined. f of negative 2 is negative 4. f of negative 1 is negative 3, so on and so forth."}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "In this video, I want to do a bunch of examples involving exponent properties. But before I even do that, let's have a little bit of a review of what an exponent even is. So let's say I had 2 to the third power. You might be tempted to say, oh, is that 6? And I would say, no, it is not 6. This means 2 times itself 3 times. So this is going to be equal to 2 times 2 times 2, which is equal to 2 times 2 is 4."}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "You might be tempted to say, oh, is that 6? And I would say, no, it is not 6. This means 2 times itself 3 times. So this is going to be equal to 2 times 2 times 2, which is equal to 2 times 2 is 4. 4 times 2 is equal to 8. If I were to ask you what 3 to the second power is, or 3 squared, this is equal to 3 times itself 2 times. This is equal to 3 times 3, which is equal to 9."}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "So this is going to be equal to 2 times 2 times 2, which is equal to 2 times 2 is 4. 4 times 2 is equal to 8. If I were to ask you what 3 to the second power is, or 3 squared, this is equal to 3 times itself 2 times. This is equal to 3 times 3, which is equal to 9. Let's do one more of these. I think you're getting the general sense if you've never seen these before. Let's say I have 5 to the seventh power."}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "This is equal to 3 times 3, which is equal to 9. Let's do one more of these. I think you're getting the general sense if you've never seen these before. Let's say I have 5 to the seventh power. That's equal to 5 times itself 7 times. 5 times 5 times 5 times 5 times 5 times 5 times 5. That's 7, right?"}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "Let's say I have 5 to the seventh power. That's equal to 5 times itself 7 times. 5 times 5 times 5 times 5 times 5 times 5 times 5. That's 7, right? 1, 2, 3, 4, 5, 6, 7. This is going to be a really, really, really large number. I'm not going to calculate it right now."}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "That's 7, right? 1, 2, 3, 4, 5, 6, 7. This is going to be a really, really, really large number. I'm not going to calculate it right now. If you want to do it by hand, feel free to do so or use a calculator. But this is a really, really, really large number. One thing that you might appreciate very quickly is that exponents increase very rapidly."}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "I'm not going to calculate it right now. If you want to do it by hand, feel free to do so or use a calculator. But this is a really, really, really large number. One thing that you might appreciate very quickly is that exponents increase very rapidly. 5 to the 17th would be even a way, way more massive number. But anyway, that's a review of exponents. Let's get a little bit steeped in algebra using exponents."}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "One thing that you might appreciate very quickly is that exponents increase very rapidly. 5 to the 17th would be even a way, way more massive number. But anyway, that's a review of exponents. Let's get a little bit steeped in algebra using exponents. What would 3x times 3x times 3x be? One thing you need to remember about multiplication is it doesn't matter what order you do the multiplication in. This is going to be the same thing as 3 times 3 times 3 times x times x times x."}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "Let's get a little bit steeped in algebra using exponents. What would 3x times 3x times 3x be? One thing you need to remember about multiplication is it doesn't matter what order you do the multiplication in. This is going to be the same thing as 3 times 3 times 3 times x times x times x. Based on what we reviewed just here, that part right there, 3 times 3, 3 times, that's 3 to the 3rd power. This right here, x times itself 3 times, that's x to the 3rd power. This whole thing can be written as 3 to the 3rd times x to the 3rd."}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "This is going to be the same thing as 3 times 3 times 3 times x times x times x. Based on what we reviewed just here, that part right there, 3 times 3, 3 times, that's 3 to the 3rd power. This right here, x times itself 3 times, that's x to the 3rd power. This whole thing can be written as 3 to the 3rd times x to the 3rd. Or if you know what 3 to the 3rd is, this is 9 times 3, which is 27. This is 27x to the 3rd power. Now you might have said, hey, wasn't 3x times 3x times 3x, wasn't that 3x to the 3rd power?"}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "This whole thing can be written as 3 to the 3rd times x to the 3rd. Or if you know what 3 to the 3rd is, this is 9 times 3, which is 27. This is 27x to the 3rd power. Now you might have said, hey, wasn't 3x times 3x times 3x, wasn't that 3x to the 3rd power? You're multiplying 3x times itself 3 times. And I would say, yes it is. So this right here, you could interpret that as 3x to the 3rd power."}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "Now you might have said, hey, wasn't 3x times 3x times 3x, wasn't that 3x to the 3rd power? You're multiplying 3x times itself 3 times. And I would say, yes it is. So this right here, you could interpret that as 3x to the 3rd power. And just like that, we stumbled on one of our exponent properties. Notice this, when I have something times something and that whole thing is to the 3rd power, that equals each of those things to the 3rd power times each other. So 3x to the 3rd is the same thing as 3 to the 3rd times x to the 3rd, which is 27 to the 3rd power."}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "So this right here, you could interpret that as 3x to the 3rd power. And just like that, we stumbled on one of our exponent properties. Notice this, when I have something times something and that whole thing is to the 3rd power, that equals each of those things to the 3rd power times each other. So 3x to the 3rd is the same thing as 3 to the 3rd times x to the 3rd, which is 27 to the 3rd power. Let's do a couple more examples. What if I were to ask you what 6 to the 3rd times 6 to the 6th power is? This is going to be a really huge number, but I want to write it as a power of 6."}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "So 3x to the 3rd is the same thing as 3 to the 3rd times x to the 3rd, which is 27 to the 3rd power. Let's do a couple more examples. What if I were to ask you what 6 to the 3rd times 6 to the 6th power is? This is going to be a really huge number, but I want to write it as a power of 6. Let me write the 6 to the 6th actually in a different color. 6 to the 3rd times 6 to the 6th power. What is this going to be equal to?"}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "This is going to be a really huge number, but I want to write it as a power of 6. Let me write the 6 to the 6th actually in a different color. 6 to the 3rd times 6 to the 6th power. What is this going to be equal to? 6 to the 3rd, we know that's 6 times itself 3 times. So it's 6 times 6 times 6. And then that's going to be times, the times here is in green, so I'll do it in green."}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "What is this going to be equal to? 6 to the 3rd, we know that's 6 times itself 3 times. So it's 6 times 6 times 6. And then that's going to be times, the times here is in green, so I'll do it in green. Maybe I'll make both of them in orange. That is going to be times 6 to the 6th power. What's 6 to the 6th power?"}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "And then that's going to be times, the times here is in green, so I'll do it in green. Maybe I'll make both of them in orange. That is going to be times 6 to the 6th power. What's 6 to the 6th power? That's 6 times itself 6 times. So it's 6 times 6 times 6 times 6 times 6. Then you get one more times 6."}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "What's 6 to the 6th power? That's 6 times itself 6 times. So it's 6 times 6 times 6 times 6 times 6. Then you get one more times 6. What is this whole number going to be? This whole thing, we're multiplying 6 times itself how many times? 1, 2, 3, 4, 5, 6, 7, 8, 9 times."}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "Then you get one more times 6. What is this whole number going to be? This whole thing, we're multiplying 6 times itself how many times? 1, 2, 3, 4, 5, 6, 7, 8, 9 times. 3 times here and then another 6 times here. We're multiplying 6 times itself 9 times. 3 plus 6."}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "1, 2, 3, 4, 5, 6, 7, 8, 9 times. 3 times here and then another 6 times here. We're multiplying 6 times itself 9 times. 3 plus 6. So this is equal to 6 to the 3 plus 6th power or 6 to the 9th power. And just like that, we've stumbled on another exponent property. When we take exponents, in this case 6 to the 3rd, the number 6 is the base."}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "3 plus 6. So this is equal to 6 to the 3 plus 6th power or 6 to the 9th power. And just like that, we've stumbled on another exponent property. When we take exponents, in this case 6 to the 3rd, the number 6 is the base. We're taking the base to the exponent of 3. When you have the same base and you're multiplying 2 exponents with the same base, you can add the exponents. If I have, let me do several more examples of this."}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "When we take exponents, in this case 6 to the 3rd, the number 6 is the base. We're taking the base to the exponent of 3. When you have the same base and you're multiplying 2 exponents with the same base, you can add the exponents. If I have, let me do several more examples of this. If I have, let's do it in magenta. Let's say I had 2 squared times 2 to the 4th times 2 to the 6th. I have the same base in all of these, so I can add the exponents."}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "If I have, let me do several more examples of this. If I have, let's do it in magenta. Let's say I had 2 squared times 2 to the 4th times 2 to the 6th. I have the same base in all of these, so I can add the exponents. This is going to be equal to 2 to the 2 plus 4 plus 6, which is equal to 2 to the 12th power. Hopefully that makes sense because this is going to be 2 times itself 2 times, 2 times itself 4 times, 2 times itself 6 times. When you multiply them all out, it's going to be 2 times itself 12 times or 2 to the 12th power."}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "I have the same base in all of these, so I can add the exponents. This is going to be equal to 2 to the 2 plus 4 plus 6, which is equal to 2 to the 12th power. Hopefully that makes sense because this is going to be 2 times itself 2 times, 2 times itself 4 times, 2 times itself 6 times. When you multiply them all out, it's going to be 2 times itself 12 times or 2 to the 12th power. Let's do it in a little bit more abstract way using some variables, but it's the same exact idea. What is x to the squared or x squared times x to the 4th? We could use the property we just learned."}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "When you multiply them all out, it's going to be 2 times itself 12 times or 2 to the 12th power. Let's do it in a little bit more abstract way using some variables, but it's the same exact idea. What is x to the squared or x squared times x to the 4th? We could use the property we just learned. We have the exact same base, x. It's going to be x to the 2 plus 4 power. It's going to be x to the 6th power."}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "We could use the property we just learned. We have the exact same base, x. It's going to be x to the 2 plus 4 power. It's going to be x to the 6th power. If you don't believe me, what is x squared? x squared is equal to x times x. If you're going to multiply that times x to the 4th, you're multiplying it by x times itself 4 times, x times x times x times x."}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "It's going to be x to the 6th power. If you don't believe me, what is x squared? x squared is equal to x times x. If you're going to multiply that times x to the 4th, you're multiplying it by x times itself 4 times, x times x times x times x. How many times are you now multiplying x by itself? 1, 2, 3, 4, 5, 6 times x to the 6th power. Let's do another one of these."}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "If you're going to multiply that times x to the 4th, you're multiplying it by x times itself 4 times, x times x times x times x. How many times are you now multiplying x by itself? 1, 2, 3, 4, 5, 6 times x to the 6th power. Let's do another one of these. The more examples you see, I figure the better. Let me do the other property just to mix and match it. Let's say I have a to the 3rd to the 4th power."}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "Let's do another one of these. The more examples you see, I figure the better. Let me do the other property just to mix and match it. Let's say I have a to the 3rd to the 4th power. I'll tell you the property here, and then I'll show you why it makes sense. When you have something to an exponent, and then you raise that to an exponent, you can multiply the exponent. This is going to be a to the 3 times 4 power, or a to the 12th power."}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "Let's say I have a to the 3rd to the 4th power. I'll tell you the property here, and then I'll show you why it makes sense. When you have something to an exponent, and then you raise that to an exponent, you can multiply the exponent. This is going to be a to the 3 times 4 power, or a to the 12th power. Why does that make sense? This right here is a to the 3rd times itself 4 times. This is equal to a to the 3rd times a to the 3rd times a to the 3rd times a to the 3rd."}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "This is going to be a to the 3 times 4 power, or a to the 12th power. Why does that make sense? This right here is a to the 3rd times itself 4 times. This is equal to a to the 3rd times a to the 3rd times a to the 3rd times a to the 3rd. We have the same base, so we can add the exponents. This is going to be a to the 3 times 4. This is equal to a to the 3 plus 3 plus 3 plus 3 power, which is the same thing as a to the 3 times 4 power, or a to the 12th power."}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "This is equal to a to the 3rd times a to the 3rd times a to the 3rd times a to the 3rd. We have the same base, so we can add the exponents. This is going to be a to the 3 times 4. This is equal to a to the 3 plus 3 plus 3 plus 3 power, which is the same thing as a to the 3 times 4 power, or a to the 12th power. Just to review the properties we've learned so far in this video, besides just a review of what an exponent is, if I have something, let's say I have x to the a power times x to the b power, this is going to be equal to x to the a plus b power. We saw that right here. x squared times x to the 4th is equal to x to the 6, 2 plus 4."}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "This is equal to a to the 3 plus 3 plus 3 plus 3 power, which is the same thing as a to the 3 times 4 power, or a to the 12th power. Just to review the properties we've learned so far in this video, besides just a review of what an exponent is, if I have something, let's say I have x to the a power times x to the b power, this is going to be equal to x to the a plus b power. We saw that right here. x squared times x to the 4th is equal to x to the 6, 2 plus 4. We also saw that if I have x times y to the a power, this is the same thing as x to the a power times y to the a power. We saw that early on in this video. We saw that over here."}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "x squared times x to the 4th is equal to x to the 6, 2 plus 4. We also saw that if I have x times y to the a power, this is the same thing as x to the a power times y to the a power. We saw that early on in this video. We saw that over here. 3x to the 3rd is the same thing as 3 to the 3rd times x to the 3rd. That's what this is saying right here. 3x to the 3rd is the same thing as 3 to the 3rd times x to the 3rd."}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "We saw that over here. 3x to the 3rd is the same thing as 3 to the 3rd times x to the 3rd. That's what this is saying right here. 3x to the 3rd is the same thing as 3 to the 3rd times x to the 3rd. The last property, which we just stumbled upon, is if you have x to the a, and then you raise that to the b power, that's equal to x to the a times b. We saw that right there. a to the 3rd, and then raise that to the 4th power, is the same thing as a to the 3 times 4, or a to the 12th power."}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "3x to the 3rd is the same thing as 3 to the 3rd times x to the 3rd. The last property, which we just stumbled upon, is if you have x to the a, and then you raise that to the b power, that's equal to x to the a times b. We saw that right there. a to the 3rd, and then raise that to the 4th power, is the same thing as a to the 3 times 4, or a to the 12th power. Let's use these properties to do a handful of what we could call more complex problems. Let's say we had 2xy squared times negative x squared y squared times 3x squared y squared. We wanted to simplify this."}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "a to the 3rd, and then raise that to the 4th power, is the same thing as a to the 3 times 4, or a to the 12th power. Let's use these properties to do a handful of what we could call more complex problems. Let's say we had 2xy squared times negative x squared y squared times 3x squared y squared. We wanted to simplify this. A good place to start, maybe we could simplify this. This you could view as negative 1 times x squared times y squared. Just this part right here, if we take this whole thing to the squared power, this is like raising each of these to the second power."}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "We wanted to simplify this. A good place to start, maybe we could simplify this. This you could view as negative 1 times x squared times y squared. Just this part right here, if we take this whole thing to the squared power, this is like raising each of these to the second power. This part right here could be simplified as negative 1 squared times x squared squared times y squared. If we were to simplify that, negative 1 squared is just 1. x squared squared, remember, you can just multiply the exponents. That's going to be x to the 4th y squared."}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "Just this part right here, if we take this whole thing to the squared power, this is like raising each of these to the second power. This part right here could be simplified as negative 1 squared times x squared squared times y squared. If we were to simplify that, negative 1 squared is just 1. x squared squared, remember, you can just multiply the exponents. That's going to be x to the 4th y squared. That's what this middle part simplifies to. Let's see if we can merge it with the other parts. The other parts to remember were 2xy squared and then 3x squared y squared."}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "That's going to be x to the 4th y squared. That's what this middle part simplifies to. Let's see if we can merge it with the other parts. The other parts to remember were 2xy squared and then 3x squared y squared. Now we're going ahead and just straight up multiplying everything. We learned in multiplication that it doesn't matter which order you multiply things in. I can just rearrange."}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "The other parts to remember were 2xy squared and then 3x squared y squared. Now we're going ahead and just straight up multiplying everything. We learned in multiplication that it doesn't matter which order you multiply things in. I can just rearrange. We're just going and multiplying 2 times x times y squared times x to the 4th times y squared times 3 times x squared times y squared. I can rearrange this, and I will rearrange it, so that it's in a way that's easy to simplify. I can multiply 2 times 3, and then I can worry about the x terms."}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "I can just rearrange. We're just going and multiplying 2 times x times y squared times x to the 4th times y squared times 3 times x squared times y squared. I can rearrange this, and I will rearrange it, so that it's in a way that's easy to simplify. I can multiply 2 times 3, and then I can worry about the x terms. Let me do it in this color. Then I have times x times x to the 4th times x squared. Then I have to worry about the y terms."}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "I can multiply 2 times 3, and then I can worry about the x terms. Let me do it in this color. Then I have times x times x to the 4th times x squared. Then I have to worry about the y terms. Times y squared times another y squared times another y squared. What are these equal to? 2 times 3, you knew how to do that."}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "Then I have to worry about the y terms. Times y squared times another y squared times another y squared. What are these equal to? 2 times 3, you knew how to do that. That's equal to 6. What is x times x to the 4th times x squared? One thing to remember is x is the same thing as x to the 1st power."}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "2 times 3, you knew how to do that. That's equal to 6. What is x times x to the 4th times x squared? One thing to remember is x is the same thing as x to the 1st power. Anything to the 1st power is just that number. 2 to the 1st power is just 2. 3 to the 1st power is just 3."}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "One thing to remember is x is the same thing as x to the 1st power. Anything to the 1st power is just that number. 2 to the 1st power is just 2. 3 to the 1st power is just 3. What is this going to be equal to? This is going to be equal to, we have the same base, x, we can add the exponents. x to the 1 plus 4 plus 2 power."}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "3 to the 1st power is just 3. What is this going to be equal to? This is going to be equal to, we have the same base, x, we can add the exponents. x to the 1 plus 4 plus 2 power. I will add it in the next step. Then on the y's, this is times y to the 2 plus 2 plus 2 power. What does that give us?"}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "x to the 1 plus 4 plus 2 power. I will add it in the next step. Then on the y's, this is times y to the 2 plus 2 plus 2 power. What does that give us? That gives us 6x to the 7th power, y to the 6th power. I will leave you with something you might already know. It's pretty interesting."}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "What does that give us? That gives us 6x to the 7th power, y to the 6th power. I will leave you with something you might already know. It's pretty interesting. That's the question of what happens when you take something to the 0th power. If I say 7 to the 0th power, what does that equal? I will tell you right now, this might seem very counterintuitive, this is equal to 1."}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "It's pretty interesting. That's the question of what happens when you take something to the 0th power. If I say 7 to the 0th power, what does that equal? I will tell you right now, this might seem very counterintuitive, this is equal to 1. 1 to the 0th power is also equal to 1. Anything to the 0th power, any non-zero number to the 0th power is going to be equal to 1. Just to give you a little bit of intuition on why that is, think about it this way."}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "I will tell you right now, this might seem very counterintuitive, this is equal to 1. 1 to the 0th power is also equal to 1. Anything to the 0th power, any non-zero number to the 0th power is going to be equal to 1. Just to give you a little bit of intuition on why that is, think about it this way. 3 to the 1st power, let me write the powers. 3 to the 1st, 2nd, 3rd, we'll just do it with the number 3. 3 to the 1st power is 3, I think that makes sense."}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "Just to give you a little bit of intuition on why that is, think about it this way. 3 to the 1st power, let me write the powers. 3 to the 1st, 2nd, 3rd, we'll just do it with the number 3. 3 to the 1st power is 3, I think that makes sense. 3 to the 2nd power is 9. 3 to the 3rd power is 27. We're trying to figure out what should 3 to the 0th power be."}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "3 to the 1st power is 3, I think that makes sense. 3 to the 2nd power is 9. 3 to the 3rd power is 27. We're trying to figure out what should 3 to the 0th power be. Think about it, every time you decrement the exponent, every time you take the exponent down by 1, you're dividing by 3. To go from 27 to 9, you divide by 3. To go from 9 to 3, you divide by 3."}, {"video_title": "Exponent properties involving products   Numbers and operations   8th grade   Khan Academy.mp3", "Sentence": "We're trying to figure out what should 3 to the 0th power be. Think about it, every time you decrement the exponent, every time you take the exponent down by 1, you're dividing by 3. To go from 27 to 9, you divide by 3. To go from 9 to 3, you divide by 3. To go from this exponent to that exponent, maybe we should divide by 3 again. That's why anything to the 0th power, in this case 3 to the 0th power, is 1. See you in the next video."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "So he steps in and says, OK, if you and this bird are so smart, how about you tackle the riddle of the fruit prices? And the king says, yes, that is something that we haven't been able to figure out, the fruit prices. Arbegla tell them the riddle of the fruit prices. And so Arbegla says, well, we want to keep track of how much our fruit costs, but we forgot to actually log how much it costs when we went to the market. But we know how much in total we spent. We know how much we got. We know that one week ago, when we went to the fruit market, we bought two pounds of apples and one pound of bananas."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "And so Arbegla says, well, we want to keep track of how much our fruit costs, but we forgot to actually log how much it costs when we went to the market. But we know how much in total we spent. We know how much we got. We know that one week ago, when we went to the fruit market, we bought two pounds of apples and one pound of bananas. And the total cost that time was $3. So there was $3 in total cost. And then when we went the time before that, we bought six pounds of apples and three pounds of bananas."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "We know that one week ago, when we went to the fruit market, we bought two pounds of apples and one pound of bananas. And the total cost that time was $3. So there was $3 in total cost. And then when we went the time before that, we bought six pounds of apples and three pounds of bananas. And the total cost at that point was $15. So what is the cost of apples and bananas? So you look at the bird."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "And then when we went the time before that, we bought six pounds of apples and three pounds of bananas. And the total cost at that point was $15. So what is the cost of apples and bananas? So you look at the bird. The bird looks at you. The bird whispers into the king's ear. And the king says, well, the bird says, well, just start defining some variables here so we can express this thing algebraically."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "So you look at the bird. The bird looks at you. The bird whispers into the king's ear. And the king says, well, the bird says, well, just start defining some variables here so we can express this thing algebraically. So you go about doing that. What we want to figure out is the cost of apples and the cost of bananas per pound. So we set some variables."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "And the king says, well, the bird says, well, just start defining some variables here so we can express this thing algebraically. So you go about doing that. What we want to figure out is the cost of apples and the cost of bananas per pound. So we set some variables. So let's let A equal the cost of apples per pound. And let's let B equal the cost of bananas per pound. So how could we interpret this first information right over here?"}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "So we set some variables. So let's let A equal the cost of apples per pound. And let's let B equal the cost of bananas per pound. So how could we interpret this first information right over here? 2 pounds of apples and a pound of bananas cost $3. Well, how much are the apples going to cost? Well, it's going to cost 2 pounds times the cost per pound times A."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "So how could we interpret this first information right over here? 2 pounds of apples and a pound of bananas cost $3. Well, how much are the apples going to cost? Well, it's going to cost 2 pounds times the cost per pound times A. That's going to be the total cost of the apples in this scenario. And what's the total cost of the bananas? Well, it's 1 pound times the cost per pound."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "Well, it's going to cost 2 pounds times the cost per pound times A. That's going to be the total cost of the apples in this scenario. And what's the total cost of the bananas? Well, it's 1 pound times the cost per pound. So you're just going to have B. That's going to be the total cost of the bananas, because we know we bought 1 pound. So the total cost of the apples and bananas are going to be 2A plus B."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "Well, it's 1 pound times the cost per pound. So you're just going to have B. That's going to be the total cost of the bananas, because we know we bought 1 pound. So the total cost of the apples and bananas are going to be 2A plus B. And we know what that total cost is. It is $3. Now let's do the same thing for the other time that we went to the market."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "So the total cost of the apples and bananas are going to be 2A plus B. And we know what that total cost is. It is $3. Now let's do the same thing for the other time that we went to the market. 6 pounds of apples, the total cost is going to be 6 pounds times A dollars per pound. And the total cost of bananas is going to be, well, we bought 3 pounds of bananas. And the cost per pound is B."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "Now let's do the same thing for the other time that we went to the market. 6 pounds of apples, the total cost is going to be 6 pounds times A dollars per pound. And the total cost of bananas is going to be, well, we bought 3 pounds of bananas. And the cost per pound is B. And so the total cost of the apples and bananas, this scenario, is going to be equal to $15. So let's think about how we might want to solve it. We could use elimination."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "And the cost per pound is B. And so the total cost of the apples and bananas, this scenario, is going to be equal to $15. So let's think about how we might want to solve it. We could use elimination. We could use substitution. Whatever we want. We could even do it graphically."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "We could use elimination. We could use substitution. Whatever we want. We could even do it graphically. Let's try it first with elimination. So the first thing I might want to do is maybe I want to eliminate the A variable right over here. So I have 2A over here."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "We could even do it graphically. Let's try it first with elimination. So the first thing I might want to do is maybe I want to eliminate the A variable right over here. So I have 2A over here. I have 6A over here. So if I multiply this entire white equation by negative 3, then this 2A would become a negative 6A. And then it might be able to cancel out with that."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "So I have 2A over here. I have 6A over here. So if I multiply this entire white equation by negative 3, then this 2A would become a negative 6A. And then it might be able to cancel out with that. So let me do that. Let me multiply this entire equation, the entire equation, times negative 3. So negative 3 times 2A is negative 6A."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "And then it might be able to cancel out with that. So let me do that. Let me multiply this entire equation, the entire equation, times negative 3. So negative 3 times 2A is negative 6A. Negative 3 times B is negative 3B. And then negative 3 times 3 is negative 9. And now we can essentially add the two equations, or add the left side of this equation to the left side of that, and the right side of this equation to the right side of that."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "So negative 3 times 2A is negative 6A. Negative 3 times B is negative 3B. And then negative 3 times 3 is negative 9. And now we can essentially add the two equations, or add the left side of this equation to the left side of that, and the right side of this equation to the right side of that. We're essentially adding the same thing to both sides of this green equation, because we know that this is equal to that. So let's do that. Let's do it."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "And now we can essentially add the two equations, or add the left side of this equation to the left side of that, and the right side of this equation to the right side of that. We're essentially adding the same thing to both sides of this green equation, because we know that this is equal to that. So let's do that. Let's do it. So on the left-hand side, 6A and 6A cancel out. But something else interesting happens. The 3B and the 3B cancels out as well."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "Let's do it. So on the left-hand side, 6A and 6A cancel out. But something else interesting happens. The 3B and the 3B cancels out as well. So we're just left with 0 on the left-hand side. And on the right-hand side, what do we have? 15 minus 9 is equal to 6."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "The 3B and the 3B cancels out as well. So we're just left with 0 on the left-hand side. And on the right-hand side, what do we have? 15 minus 9 is equal to 6. So we get this bizarre statement. All of our variables have gone away, and we're left with this bizarre, nonsensical statement that 0 is equal to 6, which we know is definitely not the case. So what's going on over here?"}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "15 minus 9 is equal to 6. So we get this bizarre statement. All of our variables have gone away, and we're left with this bizarre, nonsensical statement that 0 is equal to 6, which we know is definitely not the case. So what's going on over here? What's going on? And then you say, what's going on? And you look at the bird, because the bird seems to be the most knowledgeable person in the room, or at least the most knowledgeable vertebrate in the room."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "So what's going on over here? What's going on? And then you say, what's going on? And you look at the bird, because the bird seems to be the most knowledgeable person in the room, or at least the most knowledgeable vertebrate in the room. And so the bird whispers into the king's ear. And the king says, well, he says that there's no solution, and you should at least try to graph it to see why. And so you say, well, the bird seems to know what he's talking about."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "And you look at the bird, because the bird seems to be the most knowledgeable person in the room, or at least the most knowledgeable vertebrate in the room. And so the bird whispers into the king's ear. And the king says, well, he says that there's no solution, and you should at least try to graph it to see why. And so you say, well, the bird seems to know what he's talking about. So let me attempt to graph these two equations and see what's going on. And so what you do is you take each of the equation. And when you graph it, you like to put it in kind of the y-intercept form or slope-intercept form."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "And so you say, well, the bird seems to know what he's talking about. So let me attempt to graph these two equations and see what's going on. And so what you do is you take each of the equation. And when you graph it, you like to put it in kind of the y-intercept form or slope-intercept form. And so you do that. So you say, well, let me solve both of these for b. So if you want to solve this first equation for b, you just subtract 2a from both sides."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "And when you graph it, you like to put it in kind of the y-intercept form or slope-intercept form. And so you do that. So you say, well, let me solve both of these for b. So if you want to solve this first equation for b, you just subtract 2a from both sides. If you subtract 2a from both sides of this first equation, you get b is equal to negative 2a plus 3. Now let's solve this second equation for b. So the first thing you might want to do is subtract 6a from both sides."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "So if you want to solve this first equation for b, you just subtract 2a from both sides. If you subtract 2a from both sides of this first equation, you get b is equal to negative 2a plus 3. Now let's solve this second equation for b. So the first thing you might want to do is subtract 6a from both sides. So you would get 3b is equal to negative 6a plus 15. And then you can divide both sides by 3. You get b is equal to negative 2a plus 5."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "So the first thing you might want to do is subtract 6a from both sides. So you would get 3b is equal to negative 6a plus 15. And then you can divide both sides by 3. You get b is equal to negative 2a plus 5. So the second equation, let me revert back to that other shade of green, is b is equal to negative 2a plus 5. And we haven't even graphed it yet, but it looks like something interesting is going on. They both have the exact same slope when you solve for b, but they seem to have different, I guess you could call them b-intercepts."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "You get b is equal to negative 2a plus 5. So the second equation, let me revert back to that other shade of green, is b is equal to negative 2a plus 5. And we haven't even graphed it yet, but it looks like something interesting is going on. They both have the exact same slope when you solve for b, but they seem to have different, I guess you could call them b-intercepts. Let's graph it to actually see what's going on. So let me draw some axes over here. So let's call that my b-axis."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "They both have the exact same slope when you solve for b, but they seem to have different, I guess you could call them b-intercepts. Let's graph it to actually see what's going on. So let me draw some axes over here. So let's call that my b-axis. And then this could be my a-axis. And this first equation has a b-intercept of positive 3. So let's see, 1, 2, 3, 4, 5."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "So let's call that my b-axis. And then this could be my a-axis. And this first equation has a b-intercept of positive 3. So let's see, 1, 2, 3, 4, 5. The first one has a b-intercept of positive 3, and then has a slope of negative 2. So you go down, or you go to the right one, you go down 2. So the line looks something like this."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "So let's see, 1, 2, 3, 4, 5. The first one has a b-intercept of positive 3, and then has a slope of negative 2. So you go down, or you go to the right one, you go down 2. So the line looks something like this. I'm trying my best to draw it straight. So it looks something like that. And now let's draw this green one."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "So the line looks something like this. I'm trying my best to draw it straight. So it looks something like that. And now let's draw this green one. This green one, our b-intercept is 5, so it's right over here. But we have the exact same slope, a slope of negative 2. So it looks something like that right over there."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "And now let's draw this green one. This green one, our b-intercept is 5, so it's right over here. But we have the exact same slope, a slope of negative 2. So it looks something like that right over there. And you immediately see now that the bird was right. There is no solution, because these two constraints can be represented by lines that don't intersect. So the lines don't intersect."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "So it looks something like that right over there. And you immediately see now that the bird was right. There is no solution, because these two constraints can be represented by lines that don't intersect. So the lines don't intersect. And so the bird is right. There's no solution. There's no x and y that can make this statement equal true, or that can make 0 equal 6."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "So the lines don't intersect. And so the bird is right. There's no solution. There's no x and y that can make this statement equal true, or that can make 0 equal 6. There is no possible. There is no overlap between these two things. And so something gets into your brain."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "There's no x and y that can make this statement equal true, or that can make 0 equal 6. There is no possible. There is no overlap between these two things. And so something gets into your brain. You realize that our bagel is trying to stump you. And you say, our bagel, you have given me inconsistent information. This is an inconsistent system of equations."}, {"video_title": "Inconsistent systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "And so something gets into your brain. You realize that our bagel is trying to stump you. And you say, our bagel, you have given me inconsistent information. This is an inconsistent system of equations. Which happens to be the word that is sometimes used to refer to a system that has no solutions, where the lines do not intersect. And therefore, this information is incorrect. We cannot assume that the apple or banana, either you are lying, which is possible, or you accounted for it wrong, or maybe the prices of apples and bananas actually changed between the two visits to the market."}, {"video_title": "Example two-step equation with numerator x   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So really just need to isolate the x variable on one side of this equation, and the best way to do that is first to isolate it, isolate this whole x over 4 term from all of the other terms. So in order to do that, let's get rid of this 2. And the best way to get rid of that 2 is to subtract it. But if we want to subtract it from the right-hand side, we also have to subtract it from the left-hand side, because this is an equation. If this is equal to that, anything we do to that, we also have to do to this. So let's subtract 2 from both sides. So you subtract 2 from the right, subtract 2 from the left, and we get on the left-hand side, negative 16 minus 2 is negative 18, and then that is equal to x over 4, and then we have positive 2 minus 2, which is just going to be 0, so we don't even have to write that."}, {"video_title": "Example two-step equation with numerator x   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "But if we want to subtract it from the right-hand side, we also have to subtract it from the left-hand side, because this is an equation. If this is equal to that, anything we do to that, we also have to do to this. So let's subtract 2 from both sides. So you subtract 2 from the right, subtract 2 from the left, and we get on the left-hand side, negative 16 minus 2 is negative 18, and then that is equal to x over 4, and then we have positive 2 minus 2, which is just going to be 0, so we don't even have to write that. I could write just a plus 0, but I think that's a little unnecessary. And so we have negative 18 is equal to x over 4, and our whole goal here is to isolate the x, to solve for the x. And the best way we could do that, if we have x over 4 here, if we multiply that by 4, we're just going to have an x."}, {"video_title": "Example two-step equation with numerator x   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So you subtract 2 from the right, subtract 2 from the left, and we get on the left-hand side, negative 16 minus 2 is negative 18, and then that is equal to x over 4, and then we have positive 2 minus 2, which is just going to be 0, so we don't even have to write that. I could write just a plus 0, but I think that's a little unnecessary. And so we have negative 18 is equal to x over 4, and our whole goal here is to isolate the x, to solve for the x. And the best way we could do that, if we have x over 4 here, if we multiply that by 4, we're just going to have an x. So we can multiply that by 4, but once again, this is an equation. Anything you do to the right-hand side, you have to do to the left-hand side, and vice versa. So if we multiply the right-hand side by 4, we also have to multiply the left-hand side by 4."}, {"video_title": "Example two-step equation with numerator x   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "And the best way we could do that, if we have x over 4 here, if we multiply that by 4, we're just going to have an x. So we can multiply that by 4, but once again, this is an equation. Anything you do to the right-hand side, you have to do to the left-hand side, and vice versa. So if we multiply the right-hand side by 4, we also have to multiply the left-hand side by 4. So we get 4 times negative 18 is equal to x over 4 times 4. The x over 4 times 4, that cancels out. You divide something by 4 and multiply by 4."}, {"video_title": "Example two-step equation with numerator x   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So if we multiply the right-hand side by 4, we also have to multiply the left-hand side by 4. So we get 4 times negative 18 is equal to x over 4 times 4. The x over 4 times 4, that cancels out. You divide something by 4 and multiply by 4. You're just going to be left with an x. And on the other side, 4 times negative 18, let's see, that's 40. Well, let's just write it out."}, {"video_title": "Example two-step equation with numerator x   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "You divide something by 4 and multiply by 4. You're just going to be left with an x. And on the other side, 4 times negative 18, let's see, that's 40. Well, let's just write it out. So 18 times 4, if we were to multiply 18 times 4, 4 times 8 is 32. 4 times 1 is 4, plus 1 is 72. But this is negative 18 times 4, so it's negative 72."}, {"video_title": "Example two-step equation with numerator x   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Well, let's just write it out. So 18 times 4, if we were to multiply 18 times 4, 4 times 8 is 32. 4 times 1 is 4, plus 1 is 72. But this is negative 18 times 4, so it's negative 72. So x is equal to negative 72. And if we want to check it, we can just substitute it back into that original equation. So let's do that."}, {"video_title": "Example two-step equation with numerator x   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "But this is negative 18 times 4, so it's negative 72. So x is equal to negative 72. And if we want to check it, we can just substitute it back into that original equation. So let's do that. Let's substitute this into the original equation. So the original equation was negative 16 is equal to, instead of writing x, I'm going to write negative 72, is equal to negative 72 over 4 plus 2. Let's see if this is actually true."}, {"video_title": "Example two-step equation with numerator x   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So let's do that. Let's substitute this into the original equation. So the original equation was negative 16 is equal to, instead of writing x, I'm going to write negative 72, is equal to negative 72 over 4 plus 2. Let's see if this is actually true. So this right-hand side simplifies to negative 72 divided by 4. We already know that that is negative 18. So this is equal to negative 18 plus 2."}, {"video_title": "Example two-step equation with numerator x   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Let's see if this is actually true. So this right-hand side simplifies to negative 72 divided by 4. We already know that that is negative 18. So this is equal to negative 18 plus 2. This is what the equation becomes. And then the right-hand side, negative 18 plus 2, that's negative 16. So it all comes out true."}, {"video_title": "Slope-intercept equation from slope and point   Algebra I   Khan Academy.mp3", "Sentence": "What is the equation of this line in slope intercept form? So any line can be represented in slope intercept form as y is equal to mx plus b. Where this m right over here, that is the slope of the line. And this b over here, this is the y intercept of the line. Let me draw a quick line here just so that we can visualize that a little bit. So that is my y axis and then that is my x axis. Let me draw a line."}, {"video_title": "Slope-intercept equation from slope and point   Algebra I   Khan Academy.mp3", "Sentence": "And this b over here, this is the y intercept of the line. Let me draw a quick line here just so that we can visualize that a little bit. So that is my y axis and then that is my x axis. Let me draw a line. Since our line here has a negative slope, I'll draw a downward sloping line. Let's say our line looks something like that. Hopefully we're a little familiar with the slope already."}, {"video_title": "Slope-intercept equation from slope and point   Algebra I   Khan Academy.mp3", "Sentence": "Let me draw a line. Since our line here has a negative slope, I'll draw a downward sloping line. Let's say our line looks something like that. Hopefully we're a little familiar with the slope already. The slope essentially tells us start at some point on the line and go to some other point on the line. Measure how much you have to move in the x direction, that is your run. And then measure how much you have to move in the y direction, that is your rise."}, {"video_title": "Slope-intercept equation from slope and point   Algebra I   Khan Academy.mp3", "Sentence": "Hopefully we're a little familiar with the slope already. The slope essentially tells us start at some point on the line and go to some other point on the line. Measure how much you have to move in the x direction, that is your run. And then measure how much you have to move in the y direction, that is your rise. And our slope is equal to rise over run. What you're going to see over here would be downward sloping. Because if you move in the positive x direction, we have to go down."}, {"video_title": "Slope-intercept equation from slope and point   Algebra I   Khan Academy.mp3", "Sentence": "And then measure how much you have to move in the y direction, that is your rise. And our slope is equal to rise over run. What you're going to see over here would be downward sloping. Because if you move in the positive x direction, we have to go down. If our run is positive, our rise here is negative. So this would be a negative over positive, give you a negative number. That makes sense because we're downward sloping."}, {"video_title": "Slope-intercept equation from slope and point   Algebra I   Khan Academy.mp3", "Sentence": "Because if you move in the positive x direction, we have to go down. If our run is positive, our rise here is negative. So this would be a negative over positive, give you a negative number. That makes sense because we're downward sloping. The more we go down in this situation, for every step we move to the right, the more downward sloping we'll be, the more of a negative slope we'll have. That's slope. The y intercept just tells us where we intercept the y axis."}, {"video_title": "Slope-intercept equation from slope and point   Algebra I   Khan Academy.mp3", "Sentence": "That makes sense because we're downward sloping. The more we go down in this situation, for every step we move to the right, the more downward sloping we'll be, the more of a negative slope we'll have. That's slope. The y intercept just tells us where we intercept the y axis. The y intercept, this point right over here, this is where the line intersects with the y axis. This will be the point 0, b. This actually just falls straight out of this equation."}, {"video_title": "Slope-intercept equation from slope and point   Algebra I   Khan Academy.mp3", "Sentence": "The y intercept just tells us where we intercept the y axis. The y intercept, this point right over here, this is where the line intersects with the y axis. This will be the point 0, b. This actually just falls straight out of this equation. When x is equal to 0, let's evaluate this equation when x is equal to 0. y will be equal to m times 0 plus b. Anything times 0 is 0. y is equal to 0 plus b, or y will be equal to b when x is equal to 0. When x is equal to 0."}, {"video_title": "Slope-intercept equation from slope and point   Algebra I   Khan Academy.mp3", "Sentence": "This actually just falls straight out of this equation. When x is equal to 0, let's evaluate this equation when x is equal to 0. y will be equal to m times 0 plus b. Anything times 0 is 0. y is equal to 0 plus b, or y will be equal to b when x is equal to 0. When x is equal to 0. This is the point 0, b. They tell us what the slope of this line is. They tell us a line has a slope of negative 3 4ths."}, {"video_title": "Slope-intercept equation from slope and point   Algebra I   Khan Academy.mp3", "Sentence": "When x is equal to 0. This is the point 0, b. They tell us what the slope of this line is. They tell us a line has a slope of negative 3 4ths. We know that our slope is negative 3 4ths. They tell us that the line goes through the point 0, 8. They tell us we go through the point 0, 8."}, {"video_title": "Slope-intercept equation from slope and point   Algebra I   Khan Academy.mp3", "Sentence": "They tell us a line has a slope of negative 3 4ths. We know that our slope is negative 3 4ths. They tell us that the line goes through the point 0, 8. They tell us we go through the point 0, 8. Notice x is 0. We're on the y axis. When x is 0, we're on the y axis."}, {"video_title": "Slope-intercept equation from slope and point   Algebra I   Khan Academy.mp3", "Sentence": "They tell us we go through the point 0, 8. Notice x is 0. We're on the y axis. When x is 0, we're on the y axis. This is our y intercept. Our y intercept is the point 0, 8. We could say that b is equal to 8."}, {"video_title": "Slope-intercept equation from slope and point   Algebra I   Khan Academy.mp3", "Sentence": "When x is 0, we're on the y axis. This is our y intercept. Our y intercept is the point 0, 8. We could say that b is equal to 8. We know m is equal to negative 3 4ths. b is equal to 8. We can write the equation of this line in slope intercept form."}, {"video_title": "Solving equations with the distributive property   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "We have the equation negative 9 minus this whole expression, 9x minus 6, this whole thing is being subtracted from negative 9, is equal to 3 times this whole expression, 4x plus 6. Now a good place to start is to just get rid of these parentheses, and the best way to get rid of these parentheses is to kind of multiply them out. So this has a negative 1, you just see a minus here, but it's really the same thing as having a negative 1 times this quantity, and here you have a 3 times this quantity. So let's multiply it out using the distributive property. So the left hand side of our equation we have our negative 9, and then we want to multiply the negative 1 times each of these terms. So negative 1 times 9x is negative 9x, and then negative 1 times negative 6 is plus 6, or positive 6. And then that is going to be equal to, let's distribute the 3, 3 times 4x is 12x, and then 3 times 6 is 18."}, {"video_title": "Solving equations with the distributive property   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So let's multiply it out using the distributive property. So the left hand side of our equation we have our negative 9, and then we want to multiply the negative 1 times each of these terms. So negative 1 times 9x is negative 9x, and then negative 1 times negative 6 is plus 6, or positive 6. And then that is going to be equal to, let's distribute the 3, 3 times 4x is 12x, and then 3 times 6 is 18. Now what we want to do, let's combine our constant terms if we can, we have a negative 9 and a 6 here, on this side we've combined all of our like terms, we can't combine a 12x and an 18. So let's combine this, let's combine the negative 9 and the 6, our two constant terms on the left hand side of the equation. So we're going to have this negative 9x, so we're going to have negative 9x plus, let's see we have a negative 9 and then plus 6, so negative 9 plus 6 is negative 3."}, {"video_title": "Solving equations with the distributive property   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "And then that is going to be equal to, let's distribute the 3, 3 times 4x is 12x, and then 3 times 6 is 18. Now what we want to do, let's combine our constant terms if we can, we have a negative 9 and a 6 here, on this side we've combined all of our like terms, we can't combine a 12x and an 18. So let's combine this, let's combine the negative 9 and the 6, our two constant terms on the left hand side of the equation. So we're going to have this negative 9x, so we're going to have negative 9x plus, let's see we have a negative 9 and then plus 6, so negative 9 plus 6 is negative 3. So we're going to have a negative 9x and then we have a negative 3, so minus 3 right here, that's the negative 9 plus the 6, and that is equal to 12x plus 18. Now we want to group all of the x terms on one side of the equation and all of the constant terms, the negative 3 and the positive 18 on the other side. I like to always have my x terms on the left hand side if I can, you don't have to have them on the left, so let's do that."}, {"video_title": "Solving equations with the distributive property   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So we're going to have this negative 9x, so we're going to have negative 9x plus, let's see we have a negative 9 and then plus 6, so negative 9 plus 6 is negative 3. So we're going to have a negative 9x and then we have a negative 3, so minus 3 right here, that's the negative 9 plus the 6, and that is equal to 12x plus 18. Now we want to group all of the x terms on one side of the equation and all of the constant terms, the negative 3 and the positive 18 on the other side. I like to always have my x terms on the left hand side if I can, you don't have to have them on the left, so let's do that. So if I want all my x terms on the left, I have to get rid of this 12x from the right, and the best way to do that is to subtract 12x from both sides of the equation. So let me subtract 12x from the right, and subtract 12x from the left. Now on the left hand side I have negative 9x minus 12x, so negative 9 minus 12, that's negative 21x minus 3 is equal to 12x minus 12x, well that's just nothing, that's 0, so I could write a 0 here but I don't have to write anything, that was the whole point of subtracting the 12x from the left hand side."}, {"video_title": "Solving equations with the distributive property   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "I like to always have my x terms on the left hand side if I can, you don't have to have them on the left, so let's do that. So if I want all my x terms on the left, I have to get rid of this 12x from the right, and the best way to do that is to subtract 12x from both sides of the equation. So let me subtract 12x from the right, and subtract 12x from the left. Now on the left hand side I have negative 9x minus 12x, so negative 9 minus 12, that's negative 21x minus 3 is equal to 12x minus 12x, well that's just nothing, that's 0, so I could write a 0 here but I don't have to write anything, that was the whole point of subtracting the 12x from the left hand side. And that is going to be equal to, so on the right hand side we just are left with an 18, we are just left with that 18 here, these guys cancelled out. Now let's get rid of this negative 3 from the left hand side, so on the left hand side we only have x terms, on the right hand side we only have constant terms. So the best way to cancel out a negative 3 is to add 3, so it cancels out to 0, so we're going to add 3 to the left, let's add 3 to the right."}, {"video_title": "Solving equations with the distributive property   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Now on the left hand side I have negative 9x minus 12x, so negative 9 minus 12, that's negative 21x minus 3 is equal to 12x minus 12x, well that's just nothing, that's 0, so I could write a 0 here but I don't have to write anything, that was the whole point of subtracting the 12x from the left hand side. And that is going to be equal to, so on the right hand side we just are left with an 18, we are just left with that 18 here, these guys cancelled out. Now let's get rid of this negative 3 from the left hand side, so on the left hand side we only have x terms, on the right hand side we only have constant terms. So the best way to cancel out a negative 3 is to add 3, so it cancels out to 0, so we're going to add 3 to the left, let's add 3 to the right. And we get the left hand side of the equation, we have the negative 21x, no other x term to add or subtract here, so we have negative 21x, the negative 3 and the plus 3 or the positive 3 cancel out, that was the whole point, equals, what's 18 plus 3? 18 plus 3 is 21, so now we have negative 21x is equal to 21 and we want to solve for x. So if you have something times x and you just want it to be an x, let's divide by that something, and in this case that something is negative 21, so let's divide both sides of this equation by negative 21."}, {"video_title": "Solving equations with the distributive property   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So the best way to cancel out a negative 3 is to add 3, so it cancels out to 0, so we're going to add 3 to the left, let's add 3 to the right. And we get the left hand side of the equation, we have the negative 21x, no other x term to add or subtract here, so we have negative 21x, the negative 3 and the plus 3 or the positive 3 cancel out, that was the whole point, equals, what's 18 plus 3? 18 plus 3 is 21, so now we have negative 21x is equal to 21 and we want to solve for x. So if you have something times x and you just want it to be an x, let's divide by that something, and in this case that something is negative 21, so let's divide both sides of this equation by negative 21. Divide both sides by negative 21, the left hand side, negative 21 divided by negative 21, you're just left with an x, that was the whole point behind dividing by negative 21. And we get x is equal to, what's 21 divided by negative 21? Well that's just negative 1, right?"}, {"video_title": "Solving equations with the distributive property   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So if you have something times x and you just want it to be an x, let's divide by that something, and in this case that something is negative 21, so let's divide both sides of this equation by negative 21. Divide both sides by negative 21, the left hand side, negative 21 divided by negative 21, you're just left with an x, that was the whole point behind dividing by negative 21. And we get x is equal to, what's 21 divided by negative 21? Well that's just negative 1, right? You have the positive version divided by the negative version of itself, so it's just negative 1. So that is our answer. Now let's verify that this actually works for that original equation, so let's substitute negative 1 into that original equation."}, {"video_title": "Solving equations with the distributive property   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Well that's just negative 1, right? You have the positive version divided by the negative version of itself, so it's just negative 1. So that is our answer. Now let's verify that this actually works for that original equation, so let's substitute negative 1 into that original equation. So we have negative 9, I'll do it over here, I'll do it in a different color than we've been using, we have negative 9 minus, that 1 wasn't there originally, it's there implicitly, minus 9 times negative 1. 9 times, I'll put negative 1 in parentheses, minus 6 is equal to, well actually let me just solve for the left hand side when I substitute a negative 1 there. So the left hand side becomes negative 9 minus, 9 times negative 1 is negative 9 minus 6, and so this is negative 9 minus, in parentheses, negative 9 minus 6 is negative 15."}, {"video_title": "Solving equations with the distributive property   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Now let's verify that this actually works for that original equation, so let's substitute negative 1 into that original equation. So we have negative 9, I'll do it over here, I'll do it in a different color than we've been using, we have negative 9 minus, that 1 wasn't there originally, it's there implicitly, minus 9 times negative 1. 9 times, I'll put negative 1 in parentheses, minus 6 is equal to, well actually let me just solve for the left hand side when I substitute a negative 1 there. So the left hand side becomes negative 9 minus, 9 times negative 1 is negative 9 minus 6, and so this is negative 9 minus, in parentheses, negative 9 minus 6 is negative 15. So this is equal to negative 15, and so we get negative 9, let me make sure I did that, negative 9 minus 6, yep, negative 15. So I have negative 9 minus negative 15, that's the same thing as negative 9 plus 15, which is 6. So that's what we get on the left hand side of the equation when we substitute x is equal to negative 1, we get that it equals 6."}, {"video_title": "Solving equations with the distributive property   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So the left hand side becomes negative 9 minus, 9 times negative 1 is negative 9 minus 6, and so this is negative 9 minus, in parentheses, negative 9 minus 6 is negative 15. So this is equal to negative 15, and so we get negative 9, let me make sure I did that, negative 9 minus 6, yep, negative 15. So I have negative 9 minus negative 15, that's the same thing as negative 9 plus 15, which is 6. So that's what we get on the left hand side of the equation when we substitute x is equal to negative 1, we get that it equals 6. So let's see what happens when we substitute negative 1 to the right hand side of the equation. I'll do it in green. We get 3 times 4 times negative 1, plus 6, so that is 3 times negative 4 plus 6, negative 4 plus 6 is 2, so it's 3 times 2, which is also 6."}, {"video_title": "Simplifying square roots comment response   Algebra I   Khan Academy.mp3", "Sentence": "What I want to do in this video is re-simplify this expression, 3 times the principal root of 500 times x to the third. And take into consideration some of the comments that we got out on YouTube that actually give some interesting perspective on how you can simplify this. So just as a quick review of what we did in the last video, we said that this is the same thing as 3 times the principal root of 500. And I'm going to do it a little bit different than I did in the last video, just to make it interesting. This is 3 times the principal root of 500 times the principal root of x to the third. And 500, we can rewrite it, because 500 is not a perfect square. We can rewrite 500 as 100 times 5, or even better, we can rewrite that as 10 squared times 5."}, {"video_title": "Simplifying square roots comment response   Algebra I   Khan Academy.mp3", "Sentence": "And I'm going to do it a little bit different than I did in the last video, just to make it interesting. This is 3 times the principal root of 500 times the principal root of x to the third. And 500, we can rewrite it, because 500 is not a perfect square. We can rewrite 500 as 100 times 5, or even better, we can rewrite that as 10 squared times 5. 10 squared is the same thing as 100. So we can rewrite this first part over here as 3 times the principal root of 10 squared times 5 times the principal root of x squared times x. That's the same thing as x to the third."}, {"video_title": "Simplifying square roots comment response   Algebra I   Khan Academy.mp3", "Sentence": "We can rewrite 500 as 100 times 5, or even better, we can rewrite that as 10 squared times 5. 10 squared is the same thing as 100. So we can rewrite this first part over here as 3 times the principal root of 10 squared times 5 times the principal root of x squared times x. That's the same thing as x to the third. Now, the one thing I'm going to do here, actually I won't talk about it just yet, of how we're going to do it differently than we did in the last video. This radical right here can be rewritten as, so this is going to be 3 times the square root, or the principal root I should say, of 10 squared times the square root of 5. If we take the square root of the product of two things, it's the same thing as taking the square root of each of them and taking the product."}, {"video_title": "Simplifying square roots comment response   Algebra I   Khan Academy.mp3", "Sentence": "That's the same thing as x to the third. Now, the one thing I'm going to do here, actually I won't talk about it just yet, of how we're going to do it differently than we did in the last video. This radical right here can be rewritten as, so this is going to be 3 times the square root, or the principal root I should say, of 10 squared times the square root of 5. If we take the square root of the product of two things, it's the same thing as taking the square root of each of them and taking the product. And so this over here is going to be times the square root of, or the principal root of x squared times the principal root of x. And the principal root of 10 squared is 10. And then what I said in the last video is that the principal root of x squared is going to be the absolute value of x, just in case x itself is a negative number."}, {"video_title": "Simplifying square roots comment response   Algebra I   Khan Academy.mp3", "Sentence": "If we take the square root of the product of two things, it's the same thing as taking the square root of each of them and taking the product. And so this over here is going to be times the square root of, or the principal root of x squared times the principal root of x. And the principal root of 10 squared is 10. And then what I said in the last video is that the principal root of x squared is going to be the absolute value of x, just in case x itself is a negative number. And so then if you simplify all of this, you get 3 times 10, which is 30, times, and I'm just going to switch the order here, times the absolute value of x. And then you have the square root of 5, or the principal root of 5, times the principal root of x. And this is just going to be equal to the principal root of 5x."}, {"video_title": "Simplifying square roots comment response   Algebra I   Khan Academy.mp3", "Sentence": "And then what I said in the last video is that the principal root of x squared is going to be the absolute value of x, just in case x itself is a negative number. And so then if you simplify all of this, you get 3 times 10, which is 30, times, and I'm just going to switch the order here, times the absolute value of x. And then you have the square root of 5, or the principal root of 5, times the principal root of x. And this is just going to be equal to the principal root of 5x. Taking the square root of something and multiplying that times the square root of something else is the same thing as just taking the square root of 5x. So all of this simplified down to 30 times the absolute value of x times the principal root of 5x. And this is what we got in the last video."}, {"video_title": "Simplifying square roots comment response   Algebra I   Khan Academy.mp3", "Sentence": "And this is just going to be equal to the principal root of 5x. Taking the square root of something and multiplying that times the square root of something else is the same thing as just taking the square root of 5x. So all of this simplified down to 30 times the absolute value of x times the principal root of 5x. And this is what we got in the last video. And the interesting thing here is if we assume we're only dealing with real numbers, the domain of x right over here, the x's that will make this expression defined in the real numbers, then x has to be greater than or equal to 0. So let me, so maybe I could write it this way. The domain here is that x is, x, any real number greater than or equal to 0."}, {"video_title": "Simplifying square roots comment response   Algebra I   Khan Academy.mp3", "Sentence": "And this is what we got in the last video. And the interesting thing here is if we assume we're only dealing with real numbers, the domain of x right over here, the x's that will make this expression defined in the real numbers, then x has to be greater than or equal to 0. So let me, so maybe I could write it this way. The domain here is that x is, x, any real number greater than or equal to 0. And the reason why I say that is if you put a negative number in here, if you put a negative number in here and you cube it, you're going to get another negative number. And then it doesn't make, at least in the real numbers, you won't get an actual value. You'll get a square root of a negative number here."}, {"video_title": "Simplifying square roots comment response   Algebra I   Khan Academy.mp3", "Sentence": "The domain here is that x is, x, any real number greater than or equal to 0. And the reason why I say that is if you put a negative number in here, if you put a negative number in here and you cube it, you're going to get another negative number. And then it doesn't make, at least in the real numbers, you won't get an actual value. You'll get a square root of a negative number here. So if you make this, if you assume this right here, we're dealing with the real numbers. We're not dealing with any complex numbers. When you open up the complex numbers, then you can expand the domain more broadly."}, {"video_title": "Simplifying square roots comment response   Algebra I   Khan Academy.mp3", "Sentence": "You'll get a square root of a negative number here. So if you make this, if you assume this right here, we're dealing with the real numbers. We're not dealing with any complex numbers. When you open up the complex numbers, then you can expand the domain more broadly. But if you're dealing with real numbers, you can say that x is going to be greater than or equal to 0. And then the absolute value of x is just going to be x because it's not going to be a negative number. And so if we're assuming that the domain of x is, or if this expression is going to be a valuable, or it's going to have a positive number, then this can be written as 30x times the square root of 5x."}, {"video_title": "Simplifying square roots comment response   Algebra I   Khan Academy.mp3", "Sentence": "When you open up the complex numbers, then you can expand the domain more broadly. But if you're dealing with real numbers, you can say that x is going to be greater than or equal to 0. And then the absolute value of x is just going to be x because it's not going to be a negative number. And so if we're assuming that the domain of x is, or if this expression is going to be a valuable, or it's going to have a positive number, then this can be written as 30x times the square root of 5x. If you had the situation where we were dealing with complex numbers, then you would. So numbers that were, and if you don't know what a complex number is or an imaginary number, don't worry too much about it. But if you were dealing with those, then you would have to keep the absolute value of x there because then this would be defined for numbers that are less than 0."}, {"video_title": "Example 2 Factoring a difference of squares with leading coefficient other than 1   Khan Academy.mp3", "Sentence": "So whenever I see something like this, I have a second degree term here, I have a subtraction sign, my temptation is to look at this as a difference of squares. We've already seen this multiple times. We've already seen that if we have something of the form a squared minus b squared, that this can be factored as a plus b times a minus b. So let's look over here. Well, over here it's not obvious that this right over here is a perfect square. Neither is it obvious that this right over here is a perfect square. So it's not clear to me that this is a difference of squares."}, {"video_title": "Example 2 Factoring a difference of squares with leading coefficient other than 1   Khan Academy.mp3", "Sentence": "So let's look over here. Well, over here it's not obvious that this right over here is a perfect square. Neither is it obvious that this right over here is a perfect square. So it's not clear to me that this is a difference of squares. But what is interesting is that both 45 and 125 have some factors in common. And the one that jumps out at me is 5. So let's see if we can factor out a 5."}, {"video_title": "Example 2 Factoring a difference of squares with leading coefficient other than 1   Khan Academy.mp3", "Sentence": "So it's not clear to me that this is a difference of squares. But what is interesting is that both 45 and 125 have some factors in common. And the one that jumps out at me is 5. So let's see if we can factor out a 5. And by doing that, whether we can get something that's a little bit closer to this pattern right over here. So if we factor out a 5, this becomes 5 times 45x squared divided by 5 is going to be 9x squared. And then 125 divided by 5 is 25."}, {"video_title": "Example 2 Factoring a difference of squares with leading coefficient other than 1   Khan Academy.mp3", "Sentence": "So let's see if we can factor out a 5. And by doing that, whether we can get something that's a little bit closer to this pattern right over here. So if we factor out a 5, this becomes 5 times 45x squared divided by 5 is going to be 9x squared. And then 125 divided by 5 is 25. Now this is interesting. 9x squared, that's a perfect square. If we call this a squared, then that tells us that a would be equal to 3x."}, {"video_title": "Example 2 Factoring a difference of squares with leading coefficient other than 1   Khan Academy.mp3", "Sentence": "And then 125 divided by 5 is 25. Now this is interesting. 9x squared, that's a perfect square. If we call this a squared, then that tells us that a would be equal to 3x. 3x, the whole thing squared is 9x squared. Similarly, I can never say similarly correctly, 25 is clearly just 5 squared. So in this case, if we're looking at this template, b would be equal to 5."}, {"video_title": "Example 2 Factoring a difference of squares with leading coefficient other than 1   Khan Academy.mp3", "Sentence": "If we call this a squared, then that tells us that a would be equal to 3x. 3x, the whole thing squared is 9x squared. Similarly, I can never say similarly correctly, 25 is clearly just 5 squared. So in this case, if we're looking at this template, b would be equal to 5. So now this is a difference of squares. And we can factor it completely. So we can't forget our 5 out front that we factored out."}, {"video_title": "Example 2 Factoring a difference of squares with leading coefficient other than 1   Khan Academy.mp3", "Sentence": "So in this case, if we're looking at this template, b would be equal to 5. So now this is a difference of squares. And we can factor it completely. So we can't forget our 5 out front that we factored out. So it's going to be 5 times a plus b. So let me write this. So it's going to be 5 times a plus b times a minus b."}, {"video_title": "Example 2 Factoring a difference of squares with leading coefficient other than 1   Khan Academy.mp3", "Sentence": "So we can't forget our 5 out front that we factored out. So it's going to be 5 times a plus b. So let me write this. So it's going to be 5 times a plus b times a minus b. So it's 5 times a plus b times a minus b. So let me write the b's down. Plus b and minus b."}, {"video_title": "Proportion word problem (example 1)   7th grade   Khan Academy.mp3", "Sentence": "A recipe for oatmeal cookies calls for two cups of flour for every three cups of oatmeal. How much flour is needed for a big batch of cookie that uses nine cups of oatmeal? So let's think about what they're saying. They're saying two cups of flour. So two cups of flour for every three cups of oatmeal. For every three cups of oatmeal. And so they're saying how much flour is needed for a big batch of cookies that uses nine cups of oatmeal."}, {"video_title": "Proportion word problem (example 1)   7th grade   Khan Academy.mp3", "Sentence": "They're saying two cups of flour. So two cups of flour for every three cups of oatmeal. For every three cups of oatmeal. And so they're saying how much flour is needed for a big batch of cookies that uses nine cups of oatmeal. So now we're going to a situation where we are using nine cups of oatmeal. Let me write it this way. Nine cups of oatmeal."}, {"video_title": "Proportion word problem (example 1)   7th grade   Khan Academy.mp3", "Sentence": "And so they're saying how much flour is needed for a big batch of cookies that uses nine cups of oatmeal. So now we're going to a situation where we are using nine cups of oatmeal. Let me write it this way. Nine cups of oatmeal. And I'll show you a couple of different ways to think about it. And whatever works for you, that works. So first of all, one way to think about it, so we're wondering, we're going to say, look, we know if we have three cups of oatmeal, we should use two cups of flour."}, {"video_title": "Proportion word problem (example 1)   7th grade   Khan Academy.mp3", "Sentence": "Nine cups of oatmeal. And I'll show you a couple of different ways to think about it. And whatever works for you, that works. So first of all, one way to think about it, so we're wondering, we're going to say, look, we know if we have three cups of oatmeal, we should use two cups of flour. But what we don't know is if we have nine cups of oatmeal, how many cups of flour do we use? That's what they're asking us. But if we're going from three cups of oatmeal to nine cups of oatmeal, how much more oatmeal are we using?"}, {"video_title": "Proportion word problem (example 1)   7th grade   Khan Academy.mp3", "Sentence": "So first of all, one way to think about it, so we're wondering, we're going to say, look, we know if we have three cups of oatmeal, we should use two cups of flour. But what we don't know is if we have nine cups of oatmeal, how many cups of flour do we use? That's what they're asking us. But if we're going from three cups of oatmeal to nine cups of oatmeal, how much more oatmeal are we using? Well, we're using three times more oatmeal. We're multiplying by three. Three cups of oatmeal to nine cups of oatmeal, we're using three times the oatmeal."}, {"video_title": "Proportion word problem (example 1)   7th grade   Khan Academy.mp3", "Sentence": "But if we're going from three cups of oatmeal to nine cups of oatmeal, how much more oatmeal are we using? Well, we're using three times more oatmeal. We're multiplying by three. Three cups of oatmeal to nine cups of oatmeal, we're using three times the oatmeal. Well, if we want to use flour in the same proportion, we have to use three times the flour. So then we're also going to have to multiply the flour times three. So we're going to have to use six cups of flour."}, {"video_title": "Proportion word problem (example 1)   7th grade   Khan Academy.mp3", "Sentence": "Three cups of oatmeal to nine cups of oatmeal, we're using three times the oatmeal. Well, if we want to use flour in the same proportion, we have to use three times the flour. So then we're also going to have to multiply the flour times three. So we're going to have to use six cups of flour. We're going to have to use 6 cups of, ignore that question mark, 6 cups of flour. Another way you could think about it, and that answers their question. That's how much flour we need for a big batch of cookies that uses 9 cups of oatmeal."}, {"video_title": "Proportion word problem (example 1)   7th grade   Khan Academy.mp3", "Sentence": "So we're going to have to use six cups of flour. We're going to have to use 6 cups of, ignore that question mark, 6 cups of flour. Another way you could think about it, and that answers their question. That's how much flour we need for a big batch of cookies that uses 9 cups of oatmeal. The other thing is you could set up a proportion. You could say 2 cups of flour over 3 cups of oatmeal is equal to question mark. And I'll say, instead of writing question mark, I'll put a variable in there."}, {"video_title": "Proportion word problem (example 1)   7th grade   Khan Academy.mp3", "Sentence": "That's how much flour we need for a big batch of cookies that uses 9 cups of oatmeal. The other thing is you could set up a proportion. You could say 2 cups of flour over 3 cups of oatmeal is equal to question mark. And I'll say, instead of writing question mark, I'll put a variable in there. I'll say is equal to a question, actually let me put a question mark there just so you really understand, is equal to a question mark in a box number cups of flour over 9 cups of oatmeal. And so I like this first way we did it because it's really just common sense. If we're tripling the oatmeal, then we're going to have to triple the flour to make the recipe in the same proportion."}, {"video_title": "Proportion word problem (example 1)   7th grade   Khan Academy.mp3", "Sentence": "And I'll say, instead of writing question mark, I'll put a variable in there. I'll say is equal to a question, actually let me put a question mark there just so you really understand, is equal to a question mark in a box number cups of flour over 9 cups of oatmeal. And so I like this first way we did it because it's really just common sense. If we're tripling the oatmeal, then we're going to have to triple the flour to make the recipe in the same proportion. Another way, once you set up an equation like this, is actually to do a little bit of algebra. Some people might call it cross multiplying, but that cross multiplying is still using a little bit of algebra. And I'll show you why they're really the same thing."}, {"video_title": "Proportion word problem (example 1)   7th grade   Khan Academy.mp3", "Sentence": "If we're tripling the oatmeal, then we're going to have to triple the flour to make the recipe in the same proportion. Another way, once you set up an equation like this, is actually to do a little bit of algebra. Some people might call it cross multiplying, but that cross multiplying is still using a little bit of algebra. And I'll show you why they're really the same thing. In cross multiplication, whenever you have a proportion set up like this, people will multiply the diagonals. So when you use cross multiplication, you'll say that 2 times 9 must be equal to question mark times 3. Whatever's in this question mark, the number of cups of flour times 3."}, {"video_title": "Proportion word problem (example 1)   7th grade   Khan Academy.mp3", "Sentence": "And I'll show you why they're really the same thing. In cross multiplication, whenever you have a proportion set up like this, people will multiply the diagonals. So when you use cross multiplication, you'll say that 2 times 9 must be equal to question mark times 3. Whatever's in this question mark, the number of cups of flour times 3. Or we get 18 is equal to whatever our question mark was times 3. So the number of cups of flour we need to use times 3 needs to be equal to 18. What times 3 is equal to 18?"}, {"video_title": "Proportion word problem (example 1)   7th grade   Khan Academy.mp3", "Sentence": "Whatever's in this question mark, the number of cups of flour times 3. Or we get 18 is equal to whatever our question mark was times 3. So the number of cups of flour we need to use times 3 needs to be equal to 18. What times 3 is equal to 18? You might be able to do that in your head. That is 6. Or you could divide both sides by 3 and you will get 6."}, {"video_title": "Proportion word problem (example 1)   7th grade   Khan Academy.mp3", "Sentence": "What times 3 is equal to 18? You might be able to do that in your head. That is 6. Or you could divide both sides by 3 and you will get 6. So we get question mark in a box needs to be equal to 6 cups of flour. Same answer we got through kind of common sense. Now, you might be wondering, hey, this cross multiplying doesn't make any intuitive sense."}, {"video_title": "Proportion word problem (example 1)   7th grade   Khan Academy.mp3", "Sentence": "Or you could divide both sides by 3 and you will get 6. So we get question mark in a box needs to be equal to 6 cups of flour. Same answer we got through kind of common sense. Now, you might be wondering, hey, this cross multiplying doesn't make any intuitive sense. Why does that work? If I have something set up like this, a proportion set up, why does it work that if I take the denominator here and multiply it by the numerator there, that that needs to be equal to the numerator here times the denominator there? And that comes from straight up algebra."}, {"video_title": "Proportion word problem (example 1)   7th grade   Khan Academy.mp3", "Sentence": "Now, you might be wondering, hey, this cross multiplying doesn't make any intuitive sense. Why does that work? If I have something set up like this, a proportion set up, why does it work that if I take the denominator here and multiply it by the numerator there, that that needs to be equal to the numerator here times the denominator there? And that comes from straight up algebra. And to do that, I'm just going to rewrite this part as x, just to simplify the writing a little bit. So we have 2 over 3 is equal to, instead of that question mark, I'll write x over 9. And in algebra, all you're saying is that this quantity over here is equal to this quantity over here."}, {"video_title": "Proportion word problem (example 1)   7th grade   Khan Academy.mp3", "Sentence": "And that comes from straight up algebra. And to do that, I'm just going to rewrite this part as x, just to simplify the writing a little bit. So we have 2 over 3 is equal to, instead of that question mark, I'll write x over 9. And in algebra, all you're saying is that this quantity over here is equal to this quantity over here. So if you do anything to what's on the left, if you want it to still be equal, if the thing on the right still needs to be equal, you'd have to do the same thing to it. And what we want to do is we want to simplify this so all we have on the right-hand side is an x. So what can we multiply this by so that we're just left with an x, so that we've solved for x?"}, {"video_title": "Proportion word problem (example 1)   7th grade   Khan Academy.mp3", "Sentence": "And in algebra, all you're saying is that this quantity over here is equal to this quantity over here. So if you do anything to what's on the left, if you want it to still be equal, if the thing on the right still needs to be equal, you'd have to do the same thing to it. And what we want to do is we want to simplify this so all we have on the right-hand side is an x. So what can we multiply this by so that we're just left with an x, so that we've solved for x? Well, if we multiply this times 9, the 9's are going to cancel out. So let's multiply the right by 9. But of course, if we multiply the right by 9, we have to still multiply the left by 9."}, {"video_title": "Proportion word problem (example 1)   7th grade   Khan Academy.mp3", "Sentence": "So what can we multiply this by so that we're just left with an x, so that we've solved for x? Well, if we multiply this times 9, the 9's are going to cancel out. So let's multiply the right by 9. But of course, if we multiply the right by 9, we have to still multiply the left by 9. Otherwise, they still wouldn't be equal. If they were equal before being multiplied by 9, for them to still be equal, you have to multiply 9 times both sides. On the right-hand side, the 9's cancel out, so you're just left with an x."}, {"video_title": "Proportion word problem (example 1)   7th grade   Khan Academy.mp3", "Sentence": "But of course, if we multiply the right by 9, we have to still multiply the left by 9. Otherwise, they still wouldn't be equal. If they were equal before being multiplied by 9, for them to still be equal, you have to multiply 9 times both sides. On the right-hand side, the 9's cancel out, so you're just left with an x. On the left-hand side, you have 9 times 2 thirds, or 9 over 1 times 2 thirds. Or this is equal to 18 over 3. And we know that 18 over 3 is the same thing as 6."}, {"video_title": "Proportion word problem (example 1)   7th grade   Khan Academy.mp3", "Sentence": "On the right-hand side, the 9's cancel out, so you're just left with an x. On the left-hand side, you have 9 times 2 thirds, or 9 over 1 times 2 thirds. Or this is equal to 18 over 3. And we know that 18 over 3 is the same thing as 6. So these are all legitimate ways to do it. I want you to understand that what I'm doing right here is algebra. That's actually the reasoning why cross multiplication works."}, {"video_title": "How to use the distributive property with variables   6th grade   Khan Academy.mp3", "Sentence": "We're asked to apply the distributive property. We have 1 half times the expression 2a minus 6b plus 8. So to figure this out, I've actually already copy and pasted this problem onto my scratch pad. I have it right over here. 1 half times 2a minus 6b plus 8. Actually, let me just rewrite it. So I'm gonna take, and let me color code it too, just for fun."}, {"video_title": "How to use the distributive property with variables   6th grade   Khan Academy.mp3", "Sentence": "I have it right over here. 1 half times 2a minus 6b plus 8. Actually, let me just rewrite it. So I'm gonna take, and let me color code it too, just for fun. So this is going to be 1 half times, give myself some space, 1 half times 2a minus 6b. So 2a minus 6b minus 6, let me write it this way, minus 6b, and then we have plus 8. Plus, and I will do 8 in this color."}, {"video_title": "How to use the distributive property with variables   6th grade   Khan Academy.mp3", "Sentence": "So I'm gonna take, and let me color code it too, just for fun. So this is going to be 1 half times, give myself some space, 1 half times 2a minus 6b. So 2a minus 6b minus 6, let me write it this way, minus 6b, and then we have plus 8. Plus, and I will do 8 in this color. Plus 8. And so I just need to distribute the 1 half. If I'm multiplying 1 half times this entire expression, that means I multiply 1 half times each of these terms."}, {"video_title": "How to use the distributive property with variables   6th grade   Khan Academy.mp3", "Sentence": "Plus, and I will do 8 in this color. Plus 8. And so I just need to distribute the 1 half. If I'm multiplying 1 half times this entire expression, that means I multiply 1 half times each of these terms. So I'm gonna multiply 1 half times this, 1 half times this, and 1 half times that. So 1 half times 2a, so it's gonna be 1 half times 2a times, let me do that same color so you see where the 2a came from. 1 half times 2a minus 1 half times 6b minus 1 half times 6b times 6b plus 1 half times 8."}, {"video_title": "How to use the distributive property with variables   6th grade   Khan Academy.mp3", "Sentence": "If I'm multiplying 1 half times this entire expression, that means I multiply 1 half times each of these terms. So I'm gonna multiply 1 half times this, 1 half times this, and 1 half times that. So 1 half times 2a, so it's gonna be 1 half times 2a times, let me do that same color so you see where the 2a came from. 1 half times 2a minus 1 half times 6b minus 1 half times 6b times 6b plus 1 half times 8. 1 half times 8. And so what's this going to be? Well, let's see, I have 1 half times 2a."}, {"video_title": "How to use the distributive property with variables   6th grade   Khan Academy.mp3", "Sentence": "1 half times 2a minus 1 half times 6b minus 1 half times 6b times 6b plus 1 half times 8. 1 half times 8. And so what's this going to be? Well, let's see, I have 1 half times 2a. 1 half times 2 is just 1, so you're just gonna be left with a. And then you have minus 1 half times 6b. Well, we can just think about what 1 half times 6 is going to be."}, {"video_title": "How to use the distributive property with variables   6th grade   Khan Academy.mp3", "Sentence": "Well, let's see, I have 1 half times 2a. 1 half times 2 is just 1, so you're just gonna be left with a. And then you have minus 1 half times 6b. Well, we can just think about what 1 half times 6 is going to be. 1 half times 6 is going to be 3, and then you still are multiplying by this b. So it's gonna be 3b, and then we have plus 1 half times 8. 1 half of 8 is 4, or as you can see, 8 halves is equal to 4 wholes."}, {"video_title": "How to use the distributive property with variables   6th grade   Khan Academy.mp3", "Sentence": "Well, we can just think about what 1 half times 6 is going to be. 1 half times 6 is going to be 3, and then you still are multiplying by this b. So it's gonna be 3b, and then we have plus 1 half times 8. 1 half of 8 is 4, or as you can see, 8 halves is equal to 4 wholes. All right, so this is going to be 4. So it's a minus 3b plus 4. a minus 3b plus 4. So let's type that in."}, {"video_title": "How to use the distributive property with variables   6th grade   Khan Academy.mp3", "Sentence": "1 half of 8 is 4, or as you can see, 8 halves is equal to 4 wholes. All right, so this is going to be 4. So it's a minus 3b plus 4. a minus 3b plus 4. So let's type that in. It's gonna be a minus 3b plus 4. And notice, it's literally half of each of these terms. Half of 2a is a, half of 6b is 3b, so we have minus 6b, so it's gonna be minus 3b, and then plus 8, instead of that, half of that, plus 4."}, {"video_title": "How to use the distributive property with variables   6th grade   Khan Academy.mp3", "Sentence": "So let's type that in. It's gonna be a minus 3b plus 4. And notice, it's literally half of each of these terms. Half of 2a is a, half of 6b is 3b, so we have minus 6b, so it's gonna be minus 3b, and then plus 8, instead of that, half of that, plus 4. So let's check our answer. And we got it right. Let's do another one of these."}, {"video_title": "How to use the distributive property with variables   6th grade   Khan Academy.mp3", "Sentence": "Half of 2a is a, half of 6b is 3b, so we have minus 6b, so it's gonna be minus 3b, and then plus 8, instead of that, half of that, plus 4. So let's check our answer. And we got it right. Let's do another one of these. So let's say, so they say apply the distributive property to factor out the greatest common factor. And here we have 60m minus 40. So let me get my scratch pad out again."}, {"video_title": "How to use the distributive property with variables   6th grade   Khan Academy.mp3", "Sentence": "Let's do another one of these. So let's say, so they say apply the distributive property to factor out the greatest common factor. And here we have 60m minus 40. So let me get my scratch pad out again. So, I'm running out of space that way. So we have, all right, like this. We have 60m minus 40."}, {"video_title": "How to use the distributive property with variables   6th grade   Khan Academy.mp3", "Sentence": "So let me get my scratch pad out again. So, I'm running out of space that way. So we have, all right, like this. We have 60m minus 40. Minus 40. So what is the greatest common factor of 60m and 40? Well, 10 might jump out at us."}, {"video_title": "How to use the distributive property with variables   6th grade   Khan Academy.mp3", "Sentence": "We have 60m minus 40. Minus 40. So what is the greatest common factor of 60m and 40? Well, 10 might jump out at us. We might say, okay, look, you know, 60 is 10, so we could say this is the same thing as 10 times 6, and actually, and then of course you have the m there, so you could view this 10 times 6m, and then you could view, you could view this as 10 times 4. But 10 still isn't the greatest common factor, and you say, well, Sal, how do you know that? Well, because 4 and 6 still share a factor in common."}, {"video_title": "How to use the distributive property with variables   6th grade   Khan Academy.mp3", "Sentence": "Well, 10 might jump out at us. We might say, okay, look, you know, 60 is 10, so we could say this is the same thing as 10 times 6, and actually, and then of course you have the m there, so you could view this 10 times 6m, and then you could view, you could view this as 10 times 4. But 10 still isn't the greatest common factor, and you say, well, Sal, how do you know that? Well, because 4 and 6 still share a factor in common. They still share 2. So if you're actually factoring out the greatest common factor, what's left should not share a factor with each other. So let me think even harder about what a greatest common factor of 60 and 40 is."}, {"video_title": "How to use the distributive property with variables   6th grade   Khan Academy.mp3", "Sentence": "Well, because 4 and 6 still share a factor in common. They still share 2. So if you're actually factoring out the greatest common factor, what's left should not share a factor with each other. So let me think even harder about what a greatest common factor of 60 and 40 is. Well, 2 times 10 is 20. So you could actually factor out a 20. So you have 20 and 30m, sorry, 20 and 3m, and 40 can be factored out into 20 and 20 and 2."}, {"video_title": "How to use the distributive property with variables   6th grade   Khan Academy.mp3", "Sentence": "So let me think even harder about what a greatest common factor of 60 and 40 is. Well, 2 times 10 is 20. So you could actually factor out a 20. So you have 20 and 30m, sorry, 20 and 3m, and 40 can be factored out into 20 and 20 and 2. And now 3m and 2, 3m and 2 share no common factor, so you know that you have fully factored these two things out. Now if you think that this is something, kind of a strange art that I just did, one way to think about greatest common factors, you say, okay, 60, you can literally do a prime factorization. You say 60 is 2 times 30, which is 2 times 15, which is 3 times 5."}, {"video_title": "How to use the distributive property with variables   6th grade   Khan Academy.mp3", "Sentence": "So you have 20 and 30m, sorry, 20 and 3m, and 40 can be factored out into 20 and 20 and 2. And now 3m and 2, 3m and 2 share no common factor, so you know that you have fully factored these two things out. Now if you think that this is something, kind of a strange art that I just did, one way to think about greatest common factors, you say, okay, 60, you can literally do a prime factorization. You say 60 is 2 times 30, which is 2 times 15, which is 3 times 5. So that's 60's prime factorization, 2 times 2 times 3 times 5. And then 40's prime factorization is 2 times 20. 20 is 2 times 10."}, {"video_title": "How to use the distributive property with variables   6th grade   Khan Academy.mp3", "Sentence": "You say 60 is 2 times 30, which is 2 times 15, which is 3 times 5. So that's 60's prime factorization, 2 times 2 times 3 times 5. And then 40's prime factorization is 2 times 20. 20 is 2 times 10. 10 is 2 times 5. So that right over here, this is 40's prime factorization. And to get out the greatest common factor, you want to get out as many common prime factors."}, {"video_title": "How to use the distributive property with variables   6th grade   Khan Academy.mp3", "Sentence": "20 is 2 times 10. 10 is 2 times 5. So that right over here, this is 40's prime factorization. And to get out the greatest common factor, you want to get out as many common prime factors. So you have here, you have two 2's and a 5. Here you have two 2's and a 5. You can't go to three 2's and a 5 because there aren't three 2's and a 5 over here."}, {"video_title": "How to use the distributive property with variables   6th grade   Khan Academy.mp3", "Sentence": "And to get out the greatest common factor, you want to get out as many common prime factors. So you have here, you have two 2's and a 5. Here you have two 2's and a 5. You can't go to three 2's and a 5 because there aren't three 2's and a 5 over here. So you have two 2's and a 5 here, two 2's and a 5 here. So 2 times 2 times 5 is going to be the greatest common factor. So 2 times 2 times 5, that's 4 times 5."}, {"video_title": "How to use the distributive property with variables   6th grade   Khan Academy.mp3", "Sentence": "You can't go to three 2's and a 5 because there aren't three 2's and a 5 over here. So you have two 2's and a 5 here, two 2's and a 5 here. So 2 times 2 times 5 is going to be the greatest common factor. So 2 times 2 times 5, that's 4 times 5. 4 times 5, that is 20. That's one way of kind of very systematically figuring out a greatest common factor. But anyway, now that we know that 20 is the greatest common factor, let's factor it out."}, {"video_title": "How to use the distributive property with variables   6th grade   Khan Academy.mp3", "Sentence": "So 2 times 2 times 5, that's 4 times 5. 4 times 5, that is 20. That's one way of kind of very systematically figuring out a greatest common factor. But anyway, now that we know that 20 is the greatest common factor, let's factor it out. So this is going to be equal to 20 times, so 60m divided by 20, you're just going to be left with 3m. Just going to be left with 3m. And then minus, minus, 40 divided by 20, you're just left with a 2."}, {"video_title": "How to use the distributive property with variables   6th grade   Khan Academy.mp3", "Sentence": "But anyway, now that we know that 20 is the greatest common factor, let's factor it out. So this is going to be equal to 20 times, so 60m divided by 20, you're just going to be left with 3m. Just going to be left with 3m. And then minus, minus, 40 divided by 20, you're just left with a 2. Minus 2. Minus 2. So let's type that in."}, {"video_title": "How to use the distributive property with variables   6th grade   Khan Academy.mp3", "Sentence": "And then minus, minus, 40 divided by 20, you're just left with a 2. Minus 2. Minus 2. So let's type that in. So this is going to be 20 times, 20 times 3m, 3m minus 2. And once again, we feel good that we, literally, we did take out the greatest common factor because 3m and 2, especially 3 and 2, are now relatively prime. Relatively prime just means they don't share any factors in common other than 1."}, {"video_title": "Pythagorean theorem   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "So a right triangle is a triangle that has a 90 degree angle in it. So the way I drew it right here, this is our 90 degree angle. If you've never seen a 90 degree angle before, the way to think about it is if this side goes straight left to right, this side goes straight up and down. These sides are perpendicular, or the angle between them is 90 degrees, or it is a right angle. And the Pythagorean Theorem tells us that if we're dealing with a right triangle, let me write that down, if we're dealing with a right triangle, not a wrong triangle, if we're dealing with a right triangle, which is a triangle that has a right angle or a 90 degree angle in it, then the relationship between their sides is this. So if this side is A, this side is B, and this side is C, and remember, the C that we're dealing with right here is the side opposite the 90 degree angle. It's important to keep track of which side is which."}, {"video_title": "Pythagorean theorem   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "These sides are perpendicular, or the angle between them is 90 degrees, or it is a right angle. And the Pythagorean Theorem tells us that if we're dealing with a right triangle, let me write that down, if we're dealing with a right triangle, not a wrong triangle, if we're dealing with a right triangle, which is a triangle that has a right angle or a 90 degree angle in it, then the relationship between their sides is this. So if this side is A, this side is B, and this side is C, and remember, the C that we're dealing with right here is the side opposite the 90 degree angle. It's important to keep track of which side is which. The Pythagorean Theorem tells us that if and only if this is a right triangle, then A squared plus B squared is going to be equal to C squared. And we can use this information. If we know two of these, we can then use this theorem, this formula, to solve for the third."}, {"video_title": "Pythagorean theorem   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "It's important to keep track of which side is which. The Pythagorean Theorem tells us that if and only if this is a right triangle, then A squared plus B squared is going to be equal to C squared. And we can use this information. If we know two of these, we can then use this theorem, this formula, to solve for the third. I'll give you one more piece of terminology here. This long side, the side that is the longest side of our right triangle, the side that is opposite of our right angle, this right here, it's C in this example, this is called a hypotenuse. Very fancy word for a very simple idea."}, {"video_title": "Pythagorean theorem   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "If we know two of these, we can then use this theorem, this formula, to solve for the third. I'll give you one more piece of terminology here. This long side, the side that is the longest side of our right triangle, the side that is opposite of our right angle, this right here, it's C in this example, this is called a hypotenuse. Very fancy word for a very simple idea. The longest side of a right triangle, the side that is opposite the 90 degree angle, is called the hypotenuse. Now that we know the Pythagorean Theorem, let's actually use it. It's one thing to know something, but it's a lot more fun to use it."}, {"video_title": "Pythagorean theorem   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "Very fancy word for a very simple idea. The longest side of a right triangle, the side that is opposite the 90 degree angle, is called the hypotenuse. Now that we know the Pythagorean Theorem, let's actually use it. It's one thing to know something, but it's a lot more fun to use it. So let's say I have the following right triangle. Let me draw a little bit neater than that. The following right triangle."}, {"video_title": "Pythagorean theorem   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "It's one thing to know something, but it's a lot more fun to use it. So let's say I have the following right triangle. Let me draw a little bit neater than that. The following right triangle. This side over here has length 9. This side over here has length 7. And my question is, what is this side over here?"}, {"video_title": "Pythagorean theorem   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "The following right triangle. This side over here has length 9. This side over here has length 7. And my question is, what is this side over here? Maybe we can call that C. Well, C in this case, once again, it is a hypotenuse. It is the longest side. So we know that the sum of the squares of the other side is going to be equal to C squared."}, {"video_title": "Pythagorean theorem   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "And my question is, what is this side over here? Maybe we can call that C. Well, C in this case, once again, it is a hypotenuse. It is the longest side. So we know that the sum of the squares of the other side is going to be equal to C squared. So by the Pythagorean Theorem, 9 squared plus 7 squared is going to be equal to C squared. 9 squared is 81 plus 7 squared is 49. 80 plus 40 is 120."}, {"video_title": "Pythagorean theorem   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "So we know that the sum of the squares of the other side is going to be equal to C squared. So by the Pythagorean Theorem, 9 squared plus 7 squared is going to be equal to C squared. 9 squared is 81 plus 7 squared is 49. 80 plus 40 is 120. Then we're going to have the 1 plus the 9, that's another 10. So this is going to be equal to 130. So this is going to be equal to 130."}, {"video_title": "Pythagorean theorem   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "80 plus 40 is 120. Then we're going to have the 1 plus the 9, that's another 10. So this is going to be equal to 130. So this is going to be equal to 130. And that is equal to C squared. So what's C going to be equal to? Let me rewrite it over here."}, {"video_title": "Pythagorean theorem   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "So this is going to be equal to 130. And that is equal to C squared. So what's C going to be equal to? Let me rewrite it over here. C squared is equal to 130. Or we could say that C is equal to the square root of 130. And notice, I'm only taking the principal root here."}, {"video_title": "Pythagorean theorem   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "Let me rewrite it over here. C squared is equal to 130. Or we could say that C is equal to the square root of 130. And notice, I'm only taking the principal root here. Because C has to be positive. We're dealing with a distance. So we can't take the negative square root."}, {"video_title": "Pythagorean theorem   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "And notice, I'm only taking the principal root here. Because C has to be positive. We're dealing with a distance. So we can't take the negative square root. So we'll only take the principal square root right here. And if we want to simplify this a little bit, we know how to simplify our radicals. 130 is 2 times 65, which is 5 times 13."}, {"video_title": "Pythagorean theorem   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "So we can't take the negative square root. So we'll only take the principal square root right here. And if we want to simplify this a little bit, we know how to simplify our radicals. 130 is 2 times 65, which is 5 times 13. Well, these are all prime numbers, so that's about as simple as I can get. C is equal to the square root of 130. Let's do another one of these."}, {"video_title": "Pythagorean theorem   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "130 is 2 times 65, which is 5 times 13. Well, these are all prime numbers, so that's about as simple as I can get. C is equal to the square root of 130. Let's do another one of these. Maybe I want to keep this Pythagorean theorem right there, just so we always remember what we're referring to. So let's say I have a triangle that looks like this. Let's say it looks like that."}, {"video_title": "Pythagorean theorem   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "Let's do another one of these. Maybe I want to keep this Pythagorean theorem right there, just so we always remember what we're referring to. So let's say I have a triangle that looks like this. Let's say it looks like that. And this is the right angle up here. That's my right angle. Let's say that this side, I'm going to call it A."}, {"video_title": "Pythagorean theorem   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "Let's say it looks like that. And this is the right angle up here. That's my right angle. Let's say that this side, I'm going to call it A. This side, I'm going to call 20, or it's going to have length 21. And this side right here is going to be of length 35. So your instinct to solve for A might say, 21 squared plus 35 squared is going to be equal to A squared."}, {"video_title": "Pythagorean theorem   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "Let's say that this side, I'm going to call it A. This side, I'm going to call 20, or it's going to have length 21. And this side right here is going to be of length 35. So your instinct to solve for A might say, 21 squared plus 35 squared is going to be equal to A squared. But notice, in this situation, 35 is the hypotenuse. 35 is our C. It's the longest side of our right triangle. So what the Pythagorean theorem tells us is that A squared plus the other non-longest side, the other non-hypotenuse squared, so A squared plus 21 squared is going to be equal to 35 squared."}, {"video_title": "Pythagorean theorem   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "So your instinct to solve for A might say, 21 squared plus 35 squared is going to be equal to A squared. But notice, in this situation, 35 is the hypotenuse. 35 is our C. It's the longest side of our right triangle. So what the Pythagorean theorem tells us is that A squared plus the other non-longest side, the other non-hypotenuse squared, so A squared plus 21 squared is going to be equal to 35 squared. You always have to remember, the C squared right here, the C that we're talking about is always going to be the longest side of your right triangle, the side that is opposite of our right angle. This is the side that's opposite of the right angle. So A squared plus 21 squared is equal to 35 squared."}, {"video_title": "Pythagorean theorem   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "So what the Pythagorean theorem tells us is that A squared plus the other non-longest side, the other non-hypotenuse squared, so A squared plus 21 squared is going to be equal to 35 squared. You always have to remember, the C squared right here, the C that we're talking about is always going to be the longest side of your right triangle, the side that is opposite of our right angle. This is the side that's opposite of the right angle. So A squared plus 21 squared is equal to 35 squared. And what do we have here? So 21 squared, I'm tempted to use a calculator, but I won't. So 21 times 21, 1 times 21 is 21, 2 times 21 is 42, it is 441."}, {"video_title": "Pythagorean theorem   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "So A squared plus 21 squared is equal to 35 squared. And what do we have here? So 21 squared, I'm tempted to use a calculator, but I won't. So 21 times 21, 1 times 21 is 21, 2 times 21 is 42, it is 441. 35 squared, once again I'm tempted to use a calculator, but I won't. 35 times 35, 5 times 5 is 25, carry the 2. 5 times 3 is 15, plus 2 is 17, put a 0 here, get rid of that thing."}, {"video_title": "Pythagorean theorem   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "So 21 times 21, 1 times 21 is 21, 2 times 21 is 42, it is 441. 35 squared, once again I'm tempted to use a calculator, but I won't. 35 times 35, 5 times 5 is 25, carry the 2. 5 times 3 is 15, plus 2 is 17, put a 0 here, get rid of that thing. 3 times 5 is 15, 3 times 3 is 9, plus 1 is 10. So it is 11, let me do it in order. 5 plus 5, 0 is 5, 7 plus 5 is 12, 1 plus 1 is 2, bring down the 1, 1225."}, {"video_title": "Pythagorean theorem   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "5 times 3 is 15, plus 2 is 17, put a 0 here, get rid of that thing. 3 times 5 is 15, 3 times 3 is 9, plus 1 is 10. So it is 11, let me do it in order. 5 plus 5, 0 is 5, 7 plus 5 is 12, 1 plus 1 is 2, bring down the 1, 1225. This tells us that A squared plus 440, 441 is going to be equal to 35 squared, which is 1225. Now we could subtract 441 from both sides of this equation. 441, the left hand side just becomes A squared."}, {"video_title": "Pythagorean theorem   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "5 plus 5, 0 is 5, 7 plus 5 is 12, 1 plus 1 is 2, bring down the 1, 1225. This tells us that A squared plus 440, 441 is going to be equal to 35 squared, which is 1225. Now we could subtract 441 from both sides of this equation. 441, the left hand side just becomes A squared. The right hand side, what do we get? 5 minus 1 is 4, we want to, let me write this a little bit neater here. Let me write this a little bit neater."}, {"video_title": "Pythagorean theorem   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "441, the left hand side just becomes A squared. The right hand side, what do we get? 5 minus 1 is 4, we want to, let me write this a little bit neater here. Let me write this a little bit neater. Minus 441, so the left hand side once again they cancel out, A squared is equal to. And then on the right hand side, what do we have to do? That's larger than that, but 2 is not larger than 4, so we're going to have to borrow."}, {"video_title": "Pythagorean theorem   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "Let me write this a little bit neater. Minus 441, so the left hand side once again they cancel out, A squared is equal to. And then on the right hand side, what do we have to do? That's larger than that, but 2 is not larger than 4, so we're going to have to borrow. So that becomes a 12, or regroup depending on how you want to view it. That becomes a 1. 1 is not greater than 4, so we're going to have to borrow again."}, {"video_title": "Pythagorean theorem   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "That's larger than that, but 2 is not larger than 4, so we're going to have to borrow. So that becomes a 12, or regroup depending on how you want to view it. That becomes a 1. 1 is not greater than 4, so we're going to have to borrow again. Get rid of that, and then this becomes an 11. 5 minus 1 is 4, 12 minus 4 is 8, 11 minus 4 is 7. So A squared is equal to 784, and we could write then that A is equal to the square root of 784."}, {"video_title": "Pythagorean theorem   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "1 is not greater than 4, so we're going to have to borrow again. Get rid of that, and then this becomes an 11. 5 minus 1 is 4, 12 minus 4 is 8, 11 minus 4 is 7. So A squared is equal to 784, and we could write then that A is equal to the square root of 784. And once again, I'm very tempted to use a calculator, but let's not. Let's not use it. So this is 2 times 392, and then 390 times 2 is 78."}, {"video_title": "Pythagorean theorem   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "So A squared is equal to 784, and we could write then that A is equal to the square root of 784. And once again, I'm very tempted to use a calculator, but let's not. Let's not use it. So this is 2 times 392, and then 390 times 2 is 78. And then this is 2 times 196. That's right, 190 times 2 is 2 times 196. 196 is 2 times 98."}, {"video_title": "Pythagorean theorem   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "So this is 2 times 392, and then 390 times 2 is 78. And then this is 2 times 196. That's right, 190 times 2 is 2 times 196. 196 is 2 times 98. Let's keep going down here. 98 is 2 times 49. And of course we know what that is."}, {"video_title": "Pythagorean theorem   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "196 is 2 times 98. Let's keep going down here. 98 is 2 times 49. And of course we know what that is. Notice we have 2 times 2 times 2 times 2, so this is 2 to the 4th power, so 16 times 49. So A is equal to the square root of 16 times 49. I picked those numbers because they're both perfect squares."}, {"video_title": "Pythagorean theorem   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "And of course we know what that is. Notice we have 2 times 2 times 2 times 2, so this is 2 to the 4th power, so 16 times 49. So A is equal to the square root of 16 times 49. I picked those numbers because they're both perfect squares. So this is equal to the square root of 16 is 4 times the square root of 49 is 7. It's equal to 28. So this side right here is going to be equal to 28 by the Pythagorean theorem."}, {"video_title": "Pythagorean theorem   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "I picked those numbers because they're both perfect squares. So this is equal to the square root of 16 is 4 times the square root of 49 is 7. It's equal to 28. So this side right here is going to be equal to 28 by the Pythagorean theorem. Let's do one more of these. It can never get enough practice. So let's say I have another triangle."}, {"video_title": "Pythagorean theorem   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "So this side right here is going to be equal to 28 by the Pythagorean theorem. Let's do one more of these. It can never get enough practice. So let's say I have another triangle. I'll draw this one big. There you go. That's my triangle."}, {"video_title": "Pythagorean theorem   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "So let's say I have another triangle. I'll draw this one big. There you go. That's my triangle. That is the right angle. This side is 24. This side is 12."}, {"video_title": "Pythagorean theorem   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "That's my triangle. That is the right angle. This side is 24. This side is 12. We'll call this side right here B. Now, once again, always identify the hypotenuse. That's the longest side, the side opposite the 90-degree angle."}, {"video_title": "Pythagorean theorem   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "This side is 12. We'll call this side right here B. Now, once again, always identify the hypotenuse. That's the longest side, the side opposite the 90-degree angle. You might say, hey, I don't know that's the longest side. I don't know what B is yet. How do I know this is longest?"}, {"video_title": "Pythagorean theorem   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "That's the longest side, the side opposite the 90-degree angle. You might say, hey, I don't know that's the longest side. I don't know what B is yet. How do I know this is longest? And there, in that situation, you just say, well, it's the side opposite the 90-degree angle. So if that's the hypotenuse, that right there is the hypotenuse, then this squared plus that squared is going to be equal to 24 squared. Pythagorean theorem, B squared plus 12 squared is equal to 24 squared."}, {"video_title": "Pythagorean theorem   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "How do I know this is longest? And there, in that situation, you just say, well, it's the side opposite the 90-degree angle. So if that's the hypotenuse, that right there is the hypotenuse, then this squared plus that squared is going to be equal to 24 squared. Pythagorean theorem, B squared plus 12 squared is equal to 24 squared. Or we could subtract 12 squared from both sides. We say B squared is equal to 24 squared minus 12 squared, which we know is 144. And that B is equal to the square root of 24 squared minus 12 squared."}, {"video_title": "Pythagorean theorem   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "Pythagorean theorem, B squared plus 12 squared is equal to 24 squared. Or we could subtract 12 squared from both sides. We say B squared is equal to 24 squared minus 12 squared, which we know is 144. And that B is equal to the square root of 24 squared minus 12 squared. Now, I'm tempted to use the calculator, and I'll give into the temptation. So let's do it. The last one was so painful, I'm still recovering."}, {"video_title": "Pythagorean theorem   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "And that B is equal to the square root of 24 squared minus 12 squared. Now, I'm tempted to use the calculator, and I'll give into the temptation. So let's do it. The last one was so painful, I'm still recovering. So 24 squared minus 12 squared is equal to 24.78. So this actually turns into 24 squared minus 12 squared is equal to 432. So B is equal to the square root of 432."}, {"video_title": "Pythagorean theorem   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "The last one was so painful, I'm still recovering. So 24 squared minus 12 squared is equal to 24.78. So this actually turns into 24 squared minus 12 squared is equal to 432. So B is equal to the square root of 432. And let's factor this again. We saw what the answer is, but maybe we can write it in kind of a simplified radical form. So this is 2 times 216."}, {"video_title": "Pythagorean theorem   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "So B is equal to the square root of 432. And let's factor this again. We saw what the answer is, but maybe we can write it in kind of a simplified radical form. So this is 2 times 216. 216, I believe, is a perfect square. So let me take the square root of 216. Nope, not a perfect square."}, {"video_title": "Pythagorean theorem   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "So this is 2 times 216. 216, I believe, is a perfect square. So let me take the square root of 216. Nope, not a perfect square. So 216, let's just keep going. 216 is 2 times 108. 108 is, we could say, 4 times what?"}, {"video_title": "Pythagorean theorem   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "Nope, not a perfect square. So 216, let's just keep going. 216 is 2 times 108. 108 is, we could say, 4 times what? 25 plus another 2, 4 times 27, which is 9 times 3. So what do we have here? We have 2 times 2 times 4."}, {"video_title": "Pythagorean theorem   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "108 is, we could say, 4 times what? 25 plus another 2, 4 times 27, which is 9 times 3. So what do we have here? We have 2 times 2 times 4. So what we have right here is a 16. 16 times 9 times 3. Is that right?"}, {"video_title": "Pythagorean theorem   Right triangles and trigonometry   Geometry   Khan Academy.mp3", "Sentence": "We have 2 times 2 times 4. So what we have right here is a 16. 16 times 9 times 3. Is that right? I'm using a different calculator. 16 times 9 times 3 is equal to 432. So this is going to be equal to, B is equal to, the square root of 16 times 9 times 3, which is equal to the square root of 16, which is 4 times the square root of 9, times the square root of 3, which is equal to 12 roots of 3."}, {"video_title": "Proof that square root of prime number is irrational   Algebra I   Khan Academy.mp3", "Sentence": "In a previous video, we used a proof by contradiction to show that the square root of 2 is irrational. What I want to do in this video is essentially use the same argument, but do it in a more general way to show that the square root of any prime number is irrational. So let's assume that p is prime. And we're going to set this up to be a proof by contradiction. So we're going to assume that the square root of p is rational. And see if this leads us to any contradiction. So if something is rational, that means that we can represent it as the ratio of two integers."}, {"video_title": "Proof that square root of prime number is irrational   Algebra I   Khan Academy.mp3", "Sentence": "And we're going to set this up to be a proof by contradiction. So we're going to assume that the square root of p is rational. And see if this leads us to any contradiction. So if something is rational, that means that we can represent it as the ratio of two integers. And if we can represent something as the ratio of two integers, that means that we can also represent it as the ratio of two co-prime integers, or two integers that have no factors in common. Or that we can represent it as a fraction that is irreducible. So I'm assuming that a, this fraction that I'm writing right over here, a over b, that this right over here is an irreducible fraction."}, {"video_title": "Proof that square root of prime number is irrational   Algebra I   Khan Academy.mp3", "Sentence": "So if something is rational, that means that we can represent it as the ratio of two integers. And if we can represent something as the ratio of two integers, that means that we can also represent it as the ratio of two co-prime integers, or two integers that have no factors in common. Or that we can represent it as a fraction that is irreducible. So I'm assuming that a, this fraction that I'm writing right over here, a over b, that this right over here is an irreducible fraction. You say, well, how can I do that? Well, this being rational says I can represent square root of p as some fraction, as some ratio of two integers. And if I can represent anything as a ratio of two integers, I can keep dividing both the numerator and the denominator by the common factors until I eventually get to an irreducible fraction."}, {"video_title": "Proof that square root of prime number is irrational   Algebra I   Khan Academy.mp3", "Sentence": "So I'm assuming that a, this fraction that I'm writing right over here, a over b, that this right over here is an irreducible fraction. You say, well, how can I do that? Well, this being rational says I can represent square root of p as some fraction, as some ratio of two integers. And if I can represent anything as a ratio of two integers, I can keep dividing both the numerator and the denominator by the common factors until I eventually get to an irreducible fraction. So I'm assuming that's where we are right here. So this cannot be reduced. And this is important for our proof."}, {"video_title": "Proof that square root of prime number is irrational   Algebra I   Khan Academy.mp3", "Sentence": "And if I can represent anything as a ratio of two integers, I can keep dividing both the numerator and the denominator by the common factors until I eventually get to an irreducible fraction. So I'm assuming that's where we are right here. So this cannot be reduced. And this is important for our proof. Cannot be reduced, which is another way of saying that a and b are co-prime, which is another way of saying that a and b share no common factors other than 1. So let's see if we can manipulate this a little bit. Let's take the square of both sides."}, {"video_title": "Proof that square root of prime number is irrational   Algebra I   Khan Academy.mp3", "Sentence": "And this is important for our proof. Cannot be reduced, which is another way of saying that a and b are co-prime, which is another way of saying that a and b share no common factors other than 1. So let's see if we can manipulate this a little bit. Let's take the square of both sides. We get p is equal to, well, a over b, the whole thing squared, that's the same thing as a squared over b squared. a squared over b squared. We can multiply both sides by b squared, and we get b squared times p is equal to a squared."}, {"video_title": "Proof that square root of prime number is irrational   Algebra I   Khan Academy.mp3", "Sentence": "Let's take the square of both sides. We get p is equal to, well, a over b, the whole thing squared, that's the same thing as a squared over b squared. a squared over b squared. We can multiply both sides by b squared, and we get b squared times p is equal to a squared. Well, what does this tell us about a squared? Well, b is an integer, so b squared must be an integer. So an integer times p is equal to a squared."}, {"video_title": "Proof that square root of prime number is irrational   Algebra I   Khan Academy.mp3", "Sentence": "We can multiply both sides by b squared, and we get b squared times p is equal to a squared. Well, what does this tell us about a squared? Well, b is an integer, so b squared must be an integer. So an integer times p is equal to a squared. Well, that means that p must be a factor of a squared. So let me write this down. So a squared is a multiple of p. Now, what does that tell us about a?"}, {"video_title": "Proof that square root of prime number is irrational   Algebra I   Khan Academy.mp3", "Sentence": "So an integer times p is equal to a squared. Well, that means that p must be a factor of a squared. So let me write this down. So a squared is a multiple of p. Now, what does that tell us about a? Does that tell us that a must also be a multiple of p? Well, to think about that, let's think about the prime factorization of a. Let's say that a can be rewritten, and any number can be rewritten as a product of primes, or any integer, I should say."}, {"video_title": "Proof that square root of prime number is irrational   Algebra I   Khan Academy.mp3", "Sentence": "So a squared is a multiple of p. Now, what does that tell us about a? Does that tell us that a must also be a multiple of p? Well, to think about that, let's think about the prime factorization of a. Let's say that a can be rewritten, and any number can be rewritten as a product of primes, or any integer, I should say. So let's write this out as a product of primes right over here. So let's say that I have my first prime factor times my second prime factor all the way to my nth prime factor. And I don't know how many prime factors a actually has."}, {"video_title": "Proof that square root of prime number is irrational   Algebra I   Khan Academy.mp3", "Sentence": "Let's say that a can be rewritten, and any number can be rewritten as a product of primes, or any integer, I should say. So let's write this out as a product of primes right over here. So let's say that I have my first prime factor times my second prime factor all the way to my nth prime factor. And I don't know how many prime factors a actually has. I'm just saying that a is some integer right over here. So if that's the prime factorization of a, what is the prime factorization of a squared going to be? Well, a squared is just a times a."}, {"video_title": "Proof that square root of prime number is irrational   Algebra I   Khan Academy.mp3", "Sentence": "And I don't know how many prime factors a actually has. I'm just saying that a is some integer right over here. So if that's the prime factorization of a, what is the prime factorization of a squared going to be? Well, a squared is just a times a. Its prime factorization is going to be f1 times f2 all the way to fn. And then that times f1 times f2 times all the way to fn. Or I could rearrange them if I want."}, {"video_title": "Proof that square root of prime number is irrational   Algebra I   Khan Academy.mp3", "Sentence": "Well, a squared is just a times a. Its prime factorization is going to be f1 times f2 all the way to fn. And then that times f1 times f2 times all the way to fn. Or I could rearrange them if I want. f1 times f1 times f2 times f2 all the way to fn times fn. Now, we know that a squared is a multiple of p. p is a prime number. So p must be one of these numbers in the prime factorization."}, {"video_title": "Proof that square root of prime number is irrational   Algebra I   Khan Academy.mp3", "Sentence": "Or I could rearrange them if I want. f1 times f1 times f2 times f2 all the way to fn times fn. Now, we know that a squared is a multiple of p. p is a prime number. So p must be one of these numbers in the prime factorization. p could be f2 or p could be f1. But p needs to be one of these numbers in the prime factorization. So p needs to be one of these factors."}, {"video_title": "Proof that square root of prime number is irrational   Algebra I   Khan Academy.mp3", "Sentence": "So p must be one of these numbers in the prime factorization. p could be f2 or p could be f1. But p needs to be one of these numbers in the prime factorization. So p needs to be one of these factors. Well, if it's, let's say, and I'll just pick one of these arbitrarily. Let's say that p is f2. If p is f2, then that means that p is also a factor of a."}, {"video_title": "Proof that square root of prime number is irrational   Algebra I   Khan Academy.mp3", "Sentence": "So p needs to be one of these factors. Well, if it's, let's say, and I'll just pick one of these arbitrarily. Let's say that p is f2. If p is f2, then that means that p is also a factor of a. So this allows us to deduce that a is a multiple of p. Or another way of saying that is that we can represent a as being some integer times p. Now, why is that interesting? And actually, I'm going to let me box this off because we're going to reuse this part later. But how can we use this?"}, {"video_title": "Proof that square root of prime number is irrational   Algebra I   Khan Academy.mp3", "Sentence": "If p is f2, then that means that p is also a factor of a. So this allows us to deduce that a is a multiple of p. Or another way of saying that is that we can represent a as being some integer times p. Now, why is that interesting? And actually, I'm going to let me box this off because we're going to reuse this part later. But how can we use this? Well, just like we did in the proof of the square root of 2 being irrational, let's now substitute this back into this equation right over here. So we get b squared times p is equal to a squared. Well, a, we're now saying we can represent that as some integer k times p. So we can rewrite that as some integer k times p. And so let's see."}, {"video_title": "Proof that square root of prime number is irrational   Algebra I   Khan Academy.mp3", "Sentence": "But how can we use this? Well, just like we did in the proof of the square root of 2 being irrational, let's now substitute this back into this equation right over here. So we get b squared times p is equal to a squared. Well, a, we're now saying we can represent that as some integer k times p. So we can rewrite that as some integer k times p. And so let's see. If we were to multiply this out, so we get b squared times p, and you probably see where this is going, is equal to k squared times p squared. We could divide both sides by p. And we get b squared is equal to p times k squared. Well, the same argument that we used, if a squared is equal to b squared times p, that let us know that a squared is a multiple of p. So now we have it the other way around."}, {"video_title": "Proof that square root of prime number is irrational   Algebra I   Khan Academy.mp3", "Sentence": "Well, a, we're now saying we can represent that as some integer k times p. So we can rewrite that as some integer k times p. And so let's see. If we were to multiply this out, so we get b squared times p, and you probably see where this is going, is equal to k squared times p squared. We could divide both sides by p. And we get b squared is equal to p times k squared. Well, the same argument that we used, if a squared is equal to b squared times p, that let us know that a squared is a multiple of p. So now we have it the other way around. b squared is equal to some integer squared, which is still going to be an integer times p. So b squared must be a multiple of p. So this lets us know that b squared is a multiple of p. And by the logic that we applied right over here, that lets us know that b is a multiple of p. And that's our contradiction, or this establishes our contradiction, that we assumed at the beginning, we assumed that a and b are co-prime, that they share no factors in common other than 1. We assumed that this cannot be reduced. But we've just established, just from this, we have deduced that a is a multiple of p and b is a multiple of p, which means that this fraction can be reduced."}, {"video_title": "Proof that square root of prime number is irrational   Algebra I   Khan Academy.mp3", "Sentence": "Well, the same argument that we used, if a squared is equal to b squared times p, that let us know that a squared is a multiple of p. So now we have it the other way around. b squared is equal to some integer squared, which is still going to be an integer times p. So b squared must be a multiple of p. So this lets us know that b squared is a multiple of p. And by the logic that we applied right over here, that lets us know that b is a multiple of p. And that's our contradiction, or this establishes our contradiction, that we assumed at the beginning, we assumed that a and b are co-prime, that they share no factors in common other than 1. We assumed that this cannot be reduced. But we've just established, just from this, we have deduced that a is a multiple of p and b is a multiple of p, which means that this fraction can be reduced. We can divide the numerator and the denominator by p. So that is our contradiction. We started assuming it cannot be reduced, but then we show that no, it must be able to be reduced. The numerator and the denominator have a common factor of p. So our contradiction is established."}, {"video_title": "Proof that square root of prime number is irrational   Algebra I   Khan Academy.mp3", "Sentence": "But we've just established, just from this, we have deduced that a is a multiple of p and b is a multiple of p, which means that this fraction can be reduced. We can divide the numerator and the denominator by p. So that is our contradiction. We started assuming it cannot be reduced, but then we show that no, it must be able to be reduced. The numerator and the denominator have a common factor of p. So our contradiction is established. Square root of p cannot be rational. Square root of p is irrational. Let me just write it down."}, {"video_title": "Worked examples slope-intercept intro   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "What is the slope of y is equal to negative four x minus three? So you might already recognize this is in slope-intercept form. Just as a reminder, slope-intercept form is y is equal to mx plus b, where the coefficient on this x term right over here, that is our slope, that is our slope, and then this constant right over here, that is going to give you your y-intercept. So if they're saying what is the slope here, well I just need to figure out what is the coefficient on this x term. And you can see that the coefficient here is a negative four. So that is going to be our m, that is going to be our slope. Now just as a reminder, you have to make sure that it's solved in this way, that it is solved for y."}, {"video_title": "Worked examples slope-intercept intro   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "So if they're saying what is the slope here, well I just need to figure out what is the coefficient on this x term. And you can see that the coefficient here is a negative four. So that is going to be our m, that is going to be our slope. Now just as a reminder, you have to make sure that it's solved in this way, that it is solved for y. It is equal to something times x minus three. So that's our slope. Let's do another one of these."}, {"video_title": "Worked examples slope-intercept intro   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "Now just as a reminder, you have to make sure that it's solved in this way, that it is solved for y. It is equal to something times x minus three. So that's our slope. Let's do another one of these. So we're asked, what is the y-intercept of y is equal to negative three x minus two? So once again, we already have it in slope-intercept form. It's already been solved for y."}, {"video_title": "Worked examples slope-intercept intro   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "Let's do another one of these. So we're asked, what is the y-intercept of y is equal to negative three x minus two? So once again, we already have it in slope-intercept form. It's already been solved for y. It's of the form y is equal to mx plus b, where m, our slope, is given right over here, negative three, but they're not asking for our slope. They're asking for the y-intercept. Well the y-intercept is given by b here, so b is negative two."}, {"video_title": "Worked examples slope-intercept intro   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "It's already been solved for y. It's of the form y is equal to mx plus b, where m, our slope, is given right over here, negative three, but they're not asking for our slope. They're asking for the y-intercept. Well the y-intercept is given by b here, so b is negative two. Pay close attention to the sign here. So b is equal to negative two. But when I look at these choices, I don't see a b is equal to negative two, so what are they talking about?"}, {"video_title": "Worked examples slope-intercept intro   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "Well the y-intercept is given by b here, so b is negative two. Pay close attention to the sign here. So b is equal to negative two. But when I look at these choices, I don't see a b is equal to negative two, so what are they talking about? Well a y-intercept is, what is the y value when x is equal to zero? And you can see that here. If x was equal to zero, then that term goes away and y is equal to b."}, {"video_title": "Worked examples slope-intercept intro   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "But when I look at these choices, I don't see a b is equal to negative two, so what are they talking about? Well a y-intercept is, what is the y value when x is equal to zero? And you can see that here. If x was equal to zero, then that term goes away and y is equal to b. So if you want to know the point where the graph described by this equation intercepts the y-axis, well it's going to be, what is y when x is equal to zero? Well when x is equal to zero, y is equal to negative two. And you can see that in our original equation again."}, {"video_title": "Worked examples slope-intercept intro   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "If x was equal to zero, then that term goes away and y is equal to b. So if you want to know the point where the graph described by this equation intercepts the y-axis, well it's going to be, what is y when x is equal to zero? Well when x is equal to zero, y is equal to negative two. And you can see that in our original equation again. If x was zero, this term would go away and y would be equal to negative two. So zero comma negative two. So it would be that choice right over there."}, {"video_title": "Worked examples slope-intercept intro   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "And you can see that in our original equation again. If x was zero, this term would go away and y would be equal to negative two. So zero comma negative two. So it would be that choice right over there. On Khan Academy, obviously you just have to click on that. You don't have to shade it in. Let's do one more."}, {"video_title": "Worked examples slope-intercept intro   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "So it would be that choice right over there. On Khan Academy, obviously you just have to click on that. You don't have to shade it in. Let's do one more. Complete the equation of the line whose slope is five and y-intercept is zero comma four. So once again, the general form is y is equal to our slope times x, if I want to put it in slope-intercept form, plus our y-intercept. Well they're telling us our slope is five, whose slope is five."}, {"video_title": "Worked examples slope-intercept intro   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "Let's do one more. Complete the equation of the line whose slope is five and y-intercept is zero comma four. So once again, the general form is y is equal to our slope times x, if I want to put it in slope-intercept form, plus our y-intercept. Well they're telling us our slope is five, whose slope is five. So we know that m is going to be five and they tell us that the y-intercept is zero four. So the y-intercept b, that is the value of y when x is equal to zero. So the value of y when x equals zero is this four right over here."}, {"video_title": "Worked examples slope-intercept intro   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "Well they're telling us our slope is five, whose slope is five. So we know that m is going to be five and they tell us that the y-intercept is zero four. So the y-intercept b, that is the value of y when x is equal to zero. So the value of y when x equals zero is this four right over here. So that is going to be four. So I could say y is equal to five times x plus four. And when you're actually entering it on Khan Academy, you would just type it in or if you're using the app, you would use it with your finger."}, {"video_title": "Worked examples slope-intercept intro   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "So the value of y when x equals zero is this four right over here. So that is going to be four. So I could say y is equal to five times x plus four. And when you're actually entering it on Khan Academy, you would just type it in or if you're using the app, you would use it with your finger. And I always make the mistake of writing y equals and I type in y equals five x plus four. They already gave you the y equals right over there. But that's all you have to do."}, {"video_title": "Proof that sum of rational and irrational is irrational   Algebra I   Khan Academy.mp3", "Sentence": "Is the resulting number going to be rational or irrational? Well, to think about this, let's just assume it's going to be rational and then see if this leads to any form of contradiction. So let's assume that this is going to give us a rational number. So let's say that this first rational number we can represent as the ratio of two integers, a and b. Let's call this a rational number. Let's just call this x. And let's call, and their sum gives us another rational number."}, {"video_title": "Proof that sum of rational and irrational is irrational   Algebra I   Khan Academy.mp3", "Sentence": "So let's say that this first rational number we can represent as the ratio of two integers, a and b. Let's call this a rational number. Let's just call this x. And let's call, and their sum gives us another rational number. Let's express that as the ratio of two other integers, m and n. So we're saying that a over b plus x is equal to m over n. Or another way of thinking about it, we could subtract a over b from both sides and we would get our irrational number, x, is equal to m over n minus a over b. Minus a over b, which is the same thing as n times b in the denominator. And then, let's see, m over n is the same thing as mb over nb."}, {"video_title": "Proof that sum of rational and irrational is irrational   Algebra I   Khan Academy.mp3", "Sentence": "And let's call, and their sum gives us another rational number. Let's express that as the ratio of two other integers, m and n. So we're saying that a over b plus x is equal to m over n. Or another way of thinking about it, we could subtract a over b from both sides and we would get our irrational number, x, is equal to m over n minus a over b. Minus a over b, which is the same thing as n times b in the denominator. And then, let's see, m over n is the same thing as mb over nb. So this would be mb. I'm just adding these two fractions. mb minus a over b is the same thing as n times a over n times b."}, {"video_title": "Proof that sum of rational and irrational is irrational   Algebra I   Khan Academy.mp3", "Sentence": "And then, let's see, m over n is the same thing as mb over nb. So this would be mb. I'm just adding these two fractions. mb minus a over b is the same thing as n times a over n times b. So minus n times a. Minus n times a. All I did is I added these two fractions."}, {"video_title": "Proof that sum of rational and irrational is irrational   Algebra I   Khan Academy.mp3", "Sentence": "mb minus a over b is the same thing as n times a over n times b. So minus n times a. Minus n times a. All I did is I added these two fractions. I found a common denominator. So let me make it clear. I multiplied this one, b, and b."}, {"video_title": "Proof that sum of rational and irrational is irrational   Algebra I   Khan Academy.mp3", "Sentence": "All I did is I added these two fractions. I found a common denominator. So let me make it clear. I multiplied this one, b, and b. And then I multiplied this one, n and n. And I just added these two things. And I got this expression right over here. So this denominator is clearly an integer."}, {"video_title": "Proof that sum of rational and irrational is irrational   Algebra I   Khan Academy.mp3", "Sentence": "I multiplied this one, b, and b. And then I multiplied this one, n and n. And I just added these two things. And I got this expression right over here. So this denominator is clearly an integer. I have the product of two integers. That's going to be an integer. That's going to be an integer."}, {"video_title": "Proof that sum of rational and irrational is irrational   Algebra I   Khan Academy.mp3", "Sentence": "So this denominator is clearly an integer. I have the product of two integers. That's going to be an integer. That's going to be an integer. And then this numerator, mb is an integer, na is an integer. The difference of two integers. This whole thing is going to be an integer."}, {"video_title": "Proof that sum of rational and irrational is irrational   Algebra I   Khan Academy.mp3", "Sentence": "That's going to be an integer. And then this numerator, mb is an integer, na is an integer. The difference of two integers. This whole thing is going to be an integer. So it looks like, assuming that the sum is rational, that all of a sudden we have this contradiction. We assumed that x is irrational. We're assuming x is irrational, but all of a sudden because we made that assumption, we're able to assume that we can represent it as the ratio of two integers."}, {"video_title": "Proof that sum of rational and irrational is irrational   Algebra I   Khan Academy.mp3", "Sentence": "This whole thing is going to be an integer. So it looks like, assuming that the sum is rational, that all of a sudden we have this contradiction. We assumed that x is irrational. We're assuming x is irrational, but all of a sudden because we made that assumption, we're able to assume that we can represent it as the ratio of two integers. So this tells us that x must be rational. And that is the contradiction. That is a very large contradiction right over there."}, {"video_title": "Proof that sum of rational and irrational is irrational   Algebra I   Khan Academy.mp3", "Sentence": "We're assuming x is irrational, but all of a sudden because we made that assumption, we're able to assume that we can represent it as the ratio of two integers. So this tells us that x must be rational. And that is the contradiction. That is a very large contradiction right over there. The assumption was that x is irrational. Now we got that x must be rational. So therefore, this cannot be the case."}, {"video_title": "Proof that sum of rational and irrational is irrational   Algebra I   Khan Academy.mp3", "Sentence": "That is a very large contradiction right over there. The assumption was that x is irrational. Now we got that x must be rational. So therefore, this cannot be the case. A rational plus an irrational must, so this is not right. A rational plus an irrational must be irrational. Let me write that down."}, {"video_title": "Slope from equation   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "And what I want to do is figure out what is the slope of the line that this equation describes. And there's a couple of ways that you can approach it. What my brain wants to do is, well, I know a few forms where it's easy to pick out the slope. For example, if I can manipulate that equation to get me in the form y is equal to mx plus b, well then I know that this m here, the coefficient on the x term, well that's going to be my slope. And b is going to be my y intercept. We cover that in many other videos. Another option is to get into point-slope form."}, {"video_title": "Slope from equation   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "For example, if I can manipulate that equation to get me in the form y is equal to mx plus b, well then I know that this m here, the coefficient on the x term, well that's going to be my slope. And b is going to be my y intercept. We cover that in many other videos. Another option is to get into point-slope form. So the general framework or the general template for point-slope form is, if I have an equation of the form y minus y one is equal to m times x minus x one, well then I immediately know that the line that this equation describes is going to have a slope of m, once again. And here the y intercept doesn't jump out at you. Let me make sure you can read this over here."}, {"video_title": "Slope from equation   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "Another option is to get into point-slope form. So the general framework or the general template for point-slope form is, if I have an equation of the form y minus y one is equal to m times x minus x one, well then I immediately know that the line that this equation describes is going to have a slope of m, once again. And here the y intercept doesn't jump out at you. Let me make sure you can read this over here. The y intercept doesn't jump out at you, but you know a point that is on this line. In particular, you know that the point x one, y one is going to be on this line, x one comma y one. So let's look at our original example."}, {"video_title": "Slope from equation   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "Let me make sure you can read this over here. The y intercept doesn't jump out at you, but you know a point that is on this line. In particular, you know that the point x one, y one is going to be on this line, x one comma y one. So let's look at our original example. So it might immediately jump out at you that this is actually in point-slope form. You might say, well, okay, I see I have a minus x one, so x one would be three. I have my slope here, and that answers our question."}, {"video_title": "Slope from equation   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "So let's look at our original example. So it might immediately jump out at you that this is actually in point-slope form. You might say, well, okay, I see I have a minus x one, so x one would be three. I have my slope here, and that answers our question. Our slope would be negative two. But here it says plus two. I have to subtract a y one."}, {"video_title": "Slope from equation   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "I have my slope here, and that answers our question. Our slope would be negative two. But here it says plus two. I have to subtract a y one. Well, you could just rewrite this. So it says, so you have y minus negative two is equal to negative two times x minus three. And then you see it's exactly this point-slope form right over here."}, {"video_title": "Slope from equation   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "I have to subtract a y one. Well, you could just rewrite this. So it says, so you have y minus negative two is equal to negative two times x minus three. And then you see it's exactly this point-slope form right over here. So our slope right over there is negative two. And then if I were to ask you, well, give me a point that sits on this line, you could say, all right, an x one would be three, and a y one would be negative two. This point sits on the line."}, {"video_title": "Slope from equation   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "And then you see it's exactly this point-slope form right over here. So our slope right over there is negative two. And then if I were to ask you, well, give me a point that sits on this line, you could say, all right, an x one would be three, and a y one would be negative two. This point sits on the line. It's not the y intercept, but it's a point on the line, and we know the slope is negative two. Now, another way to approach this is to just manipulate it so that we get into slope-intercept form. So let's do that."}, {"video_title": "Slope from equation   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "This point sits on the line. It's not the y intercept, but it's a point on the line, and we know the slope is negative two. Now, another way to approach this is to just manipulate it so that we get into slope-intercept form. So let's do that. Let's manipulate it so we get into slope-intercept form. So the first thing my brain wants to do is distribute this negative two. And if I do that, I get y plus two is equal to negative two x, negative two times negative three plus six."}, {"video_title": "Slope from equation   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "So let's do that. Let's manipulate it so we get into slope-intercept form. So the first thing my brain wants to do is distribute this negative two. And if I do that, I get y plus two is equal to negative two x, negative two times negative three plus six. And then I can subtract two from both sides. And then I get y is equal to negative two x plus four. And so here I am in slope-y-intercept form."}, {"video_title": "Slope from equation   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "And if I do that, I get y plus two is equal to negative two x, negative two times negative three plus six. And then I can subtract two from both sides. And then I get y is equal to negative two x plus four. And so here I am in slope-y-intercept form. And once again, I could say, all right, my m here, the coefficient on the x term, is my slope. So my slope is negative two. Let's do another example."}, {"video_title": "Slope from equation   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "And so here I am in slope-y-intercept form. And once again, I could say, all right, my m here, the coefficient on the x term, is my slope. So my slope is negative two. Let's do another example. So here, this equation doesn't immediately go into either one of these forms. So let's manipulate it. And if it's in either one of them, I like to get into slope-y-intercept form."}, {"video_title": "Slope from equation   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "Let's do another example. So here, this equation doesn't immediately go into either one of these forms. So let's manipulate it. And if it's in either one of them, I like to get into slope-y-intercept form. It's a little bit easier for my brain to understand. So let's do that. So let us collect, well, let's get the x's, let's just isolate the y on the right-hand side since the two y is already there."}, {"video_title": "Slope from equation   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "And if it's in either one of them, I like to get into slope-y-intercept form. It's a little bit easier for my brain to understand. So let's do that. So let us collect, well, let's get the x's, let's just isolate the y on the right-hand side since the two y is already there. So let's add three to both sides. I'm just trying to get rid of this negative three. So if we add three to both sides, on the left-hand side, we have negative four x plus 10 is equal to two y."}, {"video_title": "Slope from equation   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "So let us collect, well, let's get the x's, let's just isolate the y on the right-hand side since the two y is already there. So let's add three to both sides. I'm just trying to get rid of this negative three. So if we add three to both sides, on the left-hand side, we have negative four x plus 10 is equal to two y. These cancel out, that was the whole point. And now to solve for y, we just have to divide both sides by two. So if we divide everything by two, we get negative two x plus five is equal to y."}, {"video_title": "Slope from equation   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "So if we add three to both sides, on the left-hand side, we have negative four x plus 10 is equal to two y. These cancel out, that was the whole point. And now to solve for y, we just have to divide both sides by two. So if we divide everything by two, we get negative two x plus five is equal to y. So this is in slope-intercept form. I just have the y on the right-hand side instead of the left-hand side. We have y is equal to mx plus b."}, {"video_title": "Slope from equation   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "So if we divide everything by two, we get negative two x plus five is equal to y. So this is in slope-intercept form. I just have the y on the right-hand side instead of the left-hand side. We have y is equal to mx plus b. And so our m is the coefficient on the x term right over here. So our slope is once again is negative two. And here our y-intercept is five in case we wanted to know it."}, {"video_title": "Slope from equation   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "We have y is equal to mx plus b. And so our m is the coefficient on the x term right over here. So our slope is once again is negative two. And here our y-intercept is five in case we wanted to know it. Let's do one more example. One more example. All right, so once again, this is in neither slope-intercept or point-slope form to begin with."}, {"video_title": "Slope from equation   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "And here our y-intercept is five in case we wanted to know it. Let's do one more example. One more example. All right, so once again, this is in neither slope-intercept or point-slope form to begin with. So let's just try to get it to slope-intercept form. And like always, pause the video and see if you can figure it out yourself. All right, so let's get all the y's on the left-hand side isolated and the x's on the right-hand side."}, {"video_title": "Slope from equation   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "All right, so once again, this is in neither slope-intercept or point-slope form to begin with. So let's just try to get it to slope-intercept form. And like always, pause the video and see if you can figure it out yourself. All right, so let's get all the y's on the left-hand side isolated and the x's on the right-hand side. So let me get rid of this negative three x. So I'm gonna add three x to both sides. And let's get rid of this three y right over here."}, {"video_title": "Slope from equation   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "All right, so let's get all the y's on the left-hand side isolated and the x's on the right-hand side. So let me get rid of this negative three x. So I'm gonna add three x to both sides. And let's get rid of this three y right over here. So let's subtract three y from both sides. You could do this undoing two steps at once, but once again, I'm trying to get rid of this three x. So I'm trying to get rid of this negative three x."}, {"video_title": "Slope from equation   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "And let's get rid of this three y right over here. So let's subtract three y from both sides. You could do this undoing two steps at once, but once again, I'm trying to get rid of this three x. So I'm trying to get rid of this negative three x. So I add three x to the left, but I have to do it to the right if I wanna maintain the equality. And if I wanna get rid of this three y, well, I subtract three y from here, but I have to do it on the left-hand side if I wanna maintain the equality. So what do I get?"}, {"video_title": "Slope from equation   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "So I'm trying to get rid of this negative three x. So I add three x to the left, but I have to do it to the right if I wanna maintain the equality. And if I wanna get rid of this three y, well, I subtract three y from here, but I have to do it on the left-hand side if I wanna maintain the equality. So what do I get? That cancels out. Five y minus three y is two y is equal to two x plus three x is five x. And then these two characters cancel out."}, {"video_title": "Slope from equation   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "So what do I get? That cancels out. Five y minus three y is two y is equal to two x plus three x is five x. And then these two characters cancel out. And so if I wanna solve for y, I just divide both sides by two, and I get y is equal to 5 1\u20442 x, and I'm done. And you might say, wait, this doesn't look exactly like slope-intercept form. Where's my b?"}, {"video_title": "Slope from equation   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "And then these two characters cancel out. And so if I wanna solve for y, I just divide both sides by two, and I get y is equal to 5 1\u20442 x, and I'm done. And you might say, wait, this doesn't look exactly like slope-intercept form. Where's my b? Well, your b, if you wanted to see it, you could just write plus zero. B is implicitly zero right over here. So your slope, your slope is going to be the coefficient on the x term."}, {"video_title": "Slope from equation   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "Where's my b? Well, your b, if you wanted to see it, you could just write plus zero. B is implicitly zero right over here. So your slope, your slope is going to be the coefficient on the x term. It's going to be 5 1\u20442. And if you wanna know your y-intercept, well, it's zero. When x is zero, y is zero."}, {"video_title": "Example 1 Completing the square   Quadratic equations   Algebra I   Khan Academy.mp3", "Sentence": "Then write the expression as the square of a binomial. So we have x squared minus 44x plus c. So how do we make this into a perfect square? Well if you just look at the traditional pattern for a perfect square, let's just think of it in terms of x plus a squared. That's the same thing as x plus a times x plus a, and we've seen this before. And if you were to multiply this out, that's x times x, which is x squared, plus x times a, which is ax, plus a times x, which is ax, plus a times a, which is a squared. So it's x squared plus 2ax, these two, you have an ax plus an ax, gives you 2ax, plus a squared. So if we can get this into this pattern, where I have whatever value is here, if I take half of it, this is going to be 2a here, if I take half of it and square it over here, then this will be a perfect square."}, {"video_title": "Example 1 Completing the square   Quadratic equations   Algebra I   Khan Academy.mp3", "Sentence": "That's the same thing as x plus a times x plus a, and we've seen this before. And if you were to multiply this out, that's x times x, which is x squared, plus x times a, which is ax, plus a times x, which is ax, plus a times a, which is a squared. So it's x squared plus 2ax, these two, you have an ax plus an ax, gives you 2ax, plus a squared. So if we can get this into this pattern, where I have whatever value is here, if I take half of it, this is going to be 2a here, if I take half of it and square it over here, then this will be a perfect square. So if we look over here, this thing right here is 2a, if we want to pattern match, if we want to make this look like a perfect square, that has to be 2a. So negative 44 is equal to 2a. And this right here, this c, if we pattern match, that has to be equal to, c has to be equal to a squared."}, {"video_title": "Example 1 Completing the square   Quadratic equations   Algebra I   Khan Academy.mp3", "Sentence": "So if we can get this into this pattern, where I have whatever value is here, if I take half of it, this is going to be 2a here, if I take half of it and square it over here, then this will be a perfect square. So if we look over here, this thing right here is 2a, if we want to pattern match, if we want to make this look like a perfect square, that has to be 2a. So negative 44 is equal to 2a. And this right here, this c, if we pattern match, that has to be equal to, c has to be equal to a squared. So what's a? Well if we know negative 44 is 2a, we can divide both sides of that by 2, and we know that a, or we know that negative 22 has got to be equal to a. a has got to be equal to negative 22. a is half of the coefficient right here, it's half of negative 44. And whenever you complete the square, it's always going to be half of the coefficient right here."}, {"video_title": "Example 1 Completing the square   Quadratic equations   Algebra I   Khan Academy.mp3", "Sentence": "And this right here, this c, if we pattern match, that has to be equal to, c has to be equal to a squared. So what's a? Well if we know negative 44 is 2a, we can divide both sides of that by 2, and we know that a, or we know that negative 22 has got to be equal to a. a has got to be equal to negative 22. a is half of the coefficient right here, it's half of negative 44. And whenever you complete the square, it's always going to be half of the coefficient right here. Now, if that's a, what does c need to be? Well c needs to be a squared in order for this to be a perfect square. So c needs to equal negative 22 squared."}, {"video_title": "Example 1 Completing the square   Quadratic equations   Algebra I   Khan Academy.mp3", "Sentence": "And whenever you complete the square, it's always going to be half of the coefficient right here. Now, if that's a, what does c need to be? Well c needs to be a squared in order for this to be a perfect square. So c needs to equal negative 22 squared. And we can figure out what that is. 22 times 22, we can put the negative later, actually it's going to be the same thing because the negative times the negative is a positive. 2 times 22 is 44, put a 0, 2 times 22 is 44, get a 4, get an 8, get a 4."}, {"video_title": "Example 1 Completing the square   Quadratic equations   Algebra I   Khan Academy.mp3", "Sentence": "So c needs to equal negative 22 squared. And we can figure out what that is. 22 times 22, we can put the negative later, actually it's going to be the same thing because the negative times the negative is a positive. 2 times 22 is 44, put a 0, 2 times 22 is 44, get a 4, get an 8, get a 4. So it's 484, so if we were to rewrite this as, if we were to rewrite this as x squared minus 44x plus 484, then this is a perfect square trinomial. Or we could write it like this, this is x squared minus 2 times, minus 2 times, or maybe I should write it this way, plus 2 times negative 22x plus negative 22 squared. And when you view it that way, it's pretty clear that this is a perfect square and if you were to factor it, it's the same thing as x minus 22 times x minus 22 or x minus 22 squared."}, {"video_title": "Systems of equations word problems example 2   Algebra I   Khan Academy.mp3", "Sentence": "You notice that the person in front of you gets five apples and four oranges for $10. You get five apples and five oranges for $11. Can we solve for the price of an apple and an orange using this information in a system of linear equations in two variables? If yes, what is the solution? If no, what is the reason we cannot? So we're trying to figure out the price of an apple and the price of an orange. So I would use a for apple, but I don't like using o for orange because o looks too much like a 0."}, {"video_title": "Systems of equations word problems example 2   Algebra I   Khan Academy.mp3", "Sentence": "If yes, what is the solution? If no, what is the reason we cannot? So we're trying to figure out the price of an apple and the price of an orange. So I would use a for apple, but I don't like using o for orange because o looks too much like a 0. So I'll just say x for apples. Let's let x equal the price of apples. And let's let y equal the price of oranges."}, {"video_title": "Systems of equations word problems example 2   Algebra I   Khan Academy.mp3", "Sentence": "So I would use a for apple, but I don't like using o for orange because o looks too much like a 0. So I'll just say x for apples. Let's let x equal the price of apples. And let's let y equal the price of oranges. So let's describe what happened to the person in line in front of us. They bought five apples. So how much did they spend on apples?"}, {"video_title": "Systems of equations word problems example 2   Algebra I   Khan Academy.mp3", "Sentence": "And let's let y equal the price of oranges. So let's describe what happened to the person in line in front of us. They bought five apples. So how much did they spend on apples? Well, they bought five apples times x dollars per apple. So they spent 5x dollars on their five apples. And they bought four oranges."}, {"video_title": "Systems of equations word problems example 2   Algebra I   Khan Academy.mp3", "Sentence": "So how much did they spend on apples? Well, they bought five apples times x dollars per apple. So they spent 5x dollars on their five apples. And they bought four oranges. So they bought four oranges times y dollars per orange. So they spent 4y on oranges. So the total amount that they spent is 5x plus 4y."}, {"video_title": "Systems of equations word problems example 2   Algebra I   Khan Academy.mp3", "Sentence": "And they bought four oranges. So they bought four oranges times y dollars per orange. So they spent 4y on oranges. So the total amount that they spent is 5x plus 4y. And they tell us that this is $10. This is equal to $10. Now you get in line and you buy five apples."}, {"video_title": "Systems of equations word problems example 2   Algebra I   Khan Academy.mp3", "Sentence": "So the total amount that they spent is 5x plus 4y. And they tell us that this is $10. This is equal to $10. Now you get in line and you buy five apples. So you buy five apples, just like the guy in front of you. And you paid x dollars per apple. So you're going to pay five apples times the price per apple."}, {"video_title": "Systems of equations word problems example 2   Algebra I   Khan Academy.mp3", "Sentence": "Now you get in line and you buy five apples. So you buy five apples, just like the guy in front of you. And you paid x dollars per apple. So you're going to pay five apples times the price per apple. This is the amount that you spend on apples. And then you buy five oranges. So you're going to pay five oranges times the price per orange, which is y."}, {"video_title": "Systems of equations word problems example 2   Algebra I   Khan Academy.mp3", "Sentence": "So you're going to pay five apples times the price per apple. This is the amount that you spend on apples. And then you buy five oranges. So you're going to pay five oranges times the price per orange, which is y. So this is how much you spend on oranges. This is how much you spend on apples and oranges, this sum. And they tell us that this is going to be $11."}, {"video_title": "Systems of equations word problems example 2   Algebra I   Khan Academy.mp3", "Sentence": "So you're going to pay five oranges times the price per orange, which is y. So this is how much you spend on oranges. This is how much you spend on apples and oranges, this sum. And they tell us that this is going to be $11. So can we solve for an x and a y? And it looks like we can. And a big giveaway right over here is the ratio between the x's and the y's in these two equations are different."}, {"video_title": "Systems of equations word problems example 2   Algebra I   Khan Academy.mp3", "Sentence": "And they tell us that this is going to be $11. So can we solve for an x and a y? And it looks like we can. And a big giveaway right over here is the ratio between the x's and the y's in these two equations are different. So we're getting some information here. If the ratios were exactly the same, if this was 5x plus 4y right over here, we got a different number, then we would be in trouble. Because we bought the same combination, but we got a different price."}, {"video_title": "Systems of equations word problems example 2   Algebra I   Khan Academy.mp3", "Sentence": "And a big giveaway right over here is the ratio between the x's and the y's in these two equations are different. So we're getting some information here. If the ratios were exactly the same, if this was 5x plus 4y right over here, we got a different number, then we would be in trouble. Because we bought the same combination, but we got a different price. But the good thing is that we have a different combination here. So let's see if we can work it out. Now, the most obvious thing that jumps out at me is that I have a 5x here and I have a 5x right over here."}, {"video_title": "Systems of equations word problems example 2   Algebra I   Khan Academy.mp3", "Sentence": "Because we bought the same combination, but we got a different price. But the good thing is that we have a different combination here. So let's see if we can work it out. Now, the most obvious thing that jumps out at me is that I have a 5x here and I have a 5x right over here. So if I could subtract this 5x from that 5x, then I would cancel out all of the x terms. So what I'm going to do is I'm going to multiply this bottom equation by negative 1. So it becomes negative 5x minus 5, or plus negative 5y is equal to negative 11."}, {"video_title": "Systems of equations word problems example 2   Algebra I   Khan Academy.mp3", "Sentence": "Now, the most obvious thing that jumps out at me is that I have a 5x here and I have a 5x right over here. So if I could subtract this 5x from that 5x, then I would cancel out all of the x terms. So what I'm going to do is I'm going to multiply this bottom equation by negative 1. So it becomes negative 5x minus 5, or plus negative 5y is equal to negative 11. And then I'm going to essentially add both of these equations. And I can do that because I'm doing the same thing to both sides. I already know that this thing is equal to this thing."}, {"video_title": "Systems of equations word problems example 2   Algebra I   Khan Academy.mp3", "Sentence": "So it becomes negative 5x minus 5, or plus negative 5y is equal to negative 11. And then I'm going to essentially add both of these equations. And I can do that because I'm doing the same thing to both sides. I already know that this thing is equal to this thing. So I'm just adding those things to either side. So on the left-hand side, I have 5x minus 5x. Well, those cancel out."}, {"video_title": "Systems of equations word problems example 2   Algebra I   Khan Academy.mp3", "Sentence": "I already know that this thing is equal to this thing. So I'm just adding those things to either side. So on the left-hand side, I have 5x minus 5x. Well, those cancel out. And then I have 4y minus 5y. Well, that's negative y. And that's going to be equal to 10 minus 11, which is negative 1."}, {"video_title": "Systems of equations word problems example 2   Algebra I   Khan Academy.mp3", "Sentence": "Well, those cancel out. And then I have 4y minus 5y. Well, that's negative y. And that's going to be equal to 10 minus 11, which is negative 1. And then if we multiply both sides of this times negative 1 or divide both sides by negative 1, we're going to get y is equal to 1. So just like that, we were able to figure out the price of oranges. It's $1 per orange."}, {"video_title": "Systems of equations word problems example 2   Algebra I   Khan Academy.mp3", "Sentence": "And that's going to be equal to 10 minus 11, which is negative 1. And then if we multiply both sides of this times negative 1 or divide both sides by negative 1, we're going to get y is equal to 1. So just like that, we were able to figure out the price of oranges. It's $1 per orange. Now let's figure out, so this is equal to 1. Now let's figure out the price of apples. So we can go back into either one of these equations."}, {"video_title": "Systems of equations word problems example 2   Algebra I   Khan Academy.mp3", "Sentence": "It's $1 per orange. Now let's figure out, so this is equal to 1. Now let's figure out the price of apples. So we can go back into either one of these equations. I'll go back into this first one. So 5 times, so let's go to the person in line in front of us. They bought 5 apples at $x per apple plus 4 oranges at $1 per orange."}, {"video_title": "Systems of equations word problems example 2   Algebra I   Khan Academy.mp3", "Sentence": "So we can go back into either one of these equations. I'll go back into this first one. So 5 times, so let's go to the person in line in front of us. They bought 5 apples at $x per apple plus 4 oranges at $1 per orange. And they spent a total of $10. They spent a total of $10. So this, of course, is just 4."}, {"video_title": "Systems of equations word problems example 2   Algebra I   Khan Academy.mp3", "Sentence": "They bought 5 apples at $x per apple plus 4 oranges at $1 per orange. And they spent a total of $10. They spent a total of $10. So this, of course, is just 4. Let's subtract 4 from both sides. And we get, well, 4 times 1 minus 4, that just cancels out. We're just going to be left with 5x on the left-hand side."}, {"video_title": "Systems of equations word problems example 2   Algebra I   Khan Academy.mp3", "Sentence": "So this, of course, is just 4. Let's subtract 4 from both sides. And we get, well, 4 times 1 minus 4, that just cancels out. We're just going to be left with 5x on the left-hand side. We're just going to have 5x on the left-hand side. And on the right-hand side, we have 10 minus 4, which is equal to 6. And we can just divide both sides by 6 now in order to solve for x."}, {"video_title": "Systems of equations word problems example 2   Algebra I   Khan Academy.mp3", "Sentence": "We're just going to be left with 5x on the left-hand side. We're just going to have 5x on the left-hand side. And on the right-hand side, we have 10 minus 4, which is equal to 6. And we can just divide both sides by 6 now in order to solve for x. And so we get, oh, sorry. We can divide both sides by 5 in order to solve for x. My brain isn't, it's late in the day."}, {"video_title": "Systems of equations word problems example 2   Algebra I   Khan Academy.mp3", "Sentence": "And we can just divide both sides by 6 now in order to solve for x. And so we get, oh, sorry. We can divide both sides by 5 in order to solve for x. My brain isn't, it's late in the day. Brain isn't working. So dividing by 6 wouldn't have done anything. We would have gotten 5, 6x."}, {"video_title": "Systems of equations word problems example 2   Algebra I   Khan Academy.mp3", "Sentence": "My brain isn't, it's late in the day. Brain isn't working. So dividing by 6 wouldn't have done anything. We would have gotten 5, 6x. We just want to get an x here. So dividing both sides by 5, we get x is equal to $6 fifths. Or you could say that x is equal to 6 fifths, which is the same thing as 1 and 1 fifth, which is the same thing as $1.20."}, {"video_title": "Constructing linear equation from context.mp3", "Sentence": "Tara was hiking up a mountain. She started her hike at an elevation of 1,200 meters and ascended at a constant rate. After four hours, she reached an elevation of 1,700 meters. Let y represent Tara's elevation in meters after x hours. And they ask us, and this is from an exercise on Khan Academy, it says complete the equation for the relationship between the elevation and the number of hours. And if you're on Khan Academy, you would type it in, but we could do it by hand. So pause this video and work it out on some paper and let's see if we get to the same place."}, {"video_title": "Constructing linear equation from context.mp3", "Sentence": "Let y represent Tara's elevation in meters after x hours. And they ask us, and this is from an exercise on Khan Academy, it says complete the equation for the relationship between the elevation and the number of hours. And if you're on Khan Academy, you would type it in, but we could do it by hand. So pause this video and work it out on some paper and let's see if we get to the same place. All right, now let's do this together. So first of all, they tell us that she's ascending at a constant rate. So that's a pretty good indication that we could describe her elevation based on the number of hours she travels with a linear equation."}, {"video_title": "Constructing linear equation from context.mp3", "Sentence": "So pause this video and work it out on some paper and let's see if we get to the same place. All right, now let's do this together. So first of all, they tell us that she's ascending at a constant rate. So that's a pretty good indication that we could describe her elevation based on the number of hours she travels with a linear equation. And we could even figure out that constant rate. It says that she goes from 1,200 meters to 1,700 meters in four hours. So we could say her rate is going to be her change in elevation over change in time."}, {"video_title": "Constructing linear equation from context.mp3", "Sentence": "So that's a pretty good indication that we could describe her elevation based on the number of hours she travels with a linear equation. And we could even figure out that constant rate. It says that she goes from 1,200 meters to 1,700 meters in four hours. So we could say her rate is going to be her change in elevation over change in time. So her change in elevation is 1,700 meters minus 1,200 meters. And she does this over four hours. Over, her change in time is four hours."}, {"video_title": "Constructing linear equation from context.mp3", "Sentence": "So we could say her rate is going to be her change in elevation over change in time. So her change in elevation is 1,700 meters minus 1,200 meters. And she does this over four hours. Over, her change in time is four hours. So her constant rate in the numerator here, 1,700 minus 1,200 is 500 meters. She's able to go up 500 meters in four hours. If we divide 500 by four, this is 125 meters per hour."}, {"video_title": "Constructing linear equation from context.mp3", "Sentence": "Over, her change in time is four hours. So her constant rate in the numerator here, 1,700 minus 1,200 is 500 meters. She's able to go up 500 meters in four hours. If we divide 500 by four, this is 125 meters per hour. And so we could use this now to think about what our equation would be. Our elevation, y, would be equal to, well, where is she starting? Well, it's starting at 1,200 meters."}, {"video_title": "Constructing linear equation from context.mp3", "Sentence": "If we divide 500 by four, this is 125 meters per hour. And so we could use this now to think about what our equation would be. Our elevation, y, would be equal to, well, where is she starting? Well, it's starting at 1,200 meters. So she's starting at 1,200 meters. And then to that, we're going to add how much she climbs based on how many hours she's traveled. So it's going to be this rate, 125 meters per hour, times the number of hours she has been hiking."}, {"video_title": "Constructing linear equation from context.mp3", "Sentence": "Well, it's starting at 1,200 meters. So she's starting at 1,200 meters. And then to that, we're going to add how much she climbs based on how many hours she's traveled. So it's going to be this rate, 125 meters per hour, times the number of hours she has been hiking. So the number of hours is x times x. So this right over here is an equation for the relationship between the elevation and the number of hours. Another way you could have thought about it, you could have said, okay, this is going to be a linear equation because she's ascending at a constant rate."}, {"video_title": "Constructing linear equation from context.mp3", "Sentence": "So it's going to be this rate, 125 meters per hour, times the number of hours she has been hiking. So the number of hours is x times x. So this right over here is an equation for the relationship between the elevation and the number of hours. Another way you could have thought about it, you could have said, okay, this is going to be a linear equation because she's ascending at a constant rate. You could say the slope-intercept form for a linear equation is y is equal to mx plus b, where b is your y-intercept. What is the value of y when x is equal to zero? And you say, all right, when x is equal to zero, she's at an elevation of 1,200."}, {"video_title": "Constructing linear equation from context.mp3", "Sentence": "Another way you could have thought about it, you could have said, okay, this is going to be a linear equation because she's ascending at a constant rate. You could say the slope-intercept form for a linear equation is y is equal to mx plus b, where b is your y-intercept. What is the value of y when x is equal to zero? And you say, all right, when x is equal to zero, she's at an elevation of 1,200. And then m is our slope. So that's the rate at which our elevation is increasing. And that's what we calculated right over here."}, {"video_title": "Constructing linear equation from context.mp3", "Sentence": "And you say, all right, when x is equal to zero, she's at an elevation of 1,200. And then m is our slope. So that's the rate at which our elevation is increasing. And that's what we calculated right over here. Our slope is 125 meters per hour. So notice, these are equivalent. I just have these two terms are swapped."}, {"video_title": "Constructing linear equation from context.mp3", "Sentence": "And that's what we calculated right over here. Our slope is 125 meters per hour. So notice, these are equivalent. I just have these two terms are swapped. So we could either write y is equal to 1,200 plus 125x, or you could write it the other way around. You could write 125x plus 1,200. They are equivalent."}, {"video_title": "How to evaluate expressions in two variables with decimals and fractions   6th grade   Khan Academy.mp3", "Sentence": "So why don't you try to pause the video and evaluate this first before we work through it together. All right, so if we want to evaluate this thing, everywhere we see a j, we want to replace it with a 0.5, and everywhere we see a k, we want to replace it with a 0.25. So let's do that. This is going to be seven times, and instead of j, I'm going to put a 0.5 in there, and then we have plus five minus eight times k, and k, we're saying, is 0.25. So what is this going to be equal to? So if I were to take, and I can color code this, seven times 0.5, half of seven, that's going to be 3.5, then I have plus five, and then I'm going to subtract, I am subtracting eight times 0.25. 0.25, this is 1.4."}, {"video_title": "How to evaluate expressions in two variables with decimals and fractions   6th grade   Khan Academy.mp3", "Sentence": "This is going to be seven times, and instead of j, I'm going to put a 0.5 in there, and then we have plus five minus eight times k, and k, we're saying, is 0.25. So what is this going to be equal to? So if I were to take, and I can color code this, seven times 0.5, half of seven, that's going to be 3.5, then I have plus five, and then I'm going to subtract, I am subtracting eight times 0.25. 0.25, this is 1.4. I could rewrite this if I want. 0.25, that's the same thing as 1.4. Eight times 1.4, or another way to think about it is eight divided by four is going to be equal to two."}, {"video_title": "How to evaluate expressions in two variables with decimals and fractions   6th grade   Khan Academy.mp3", "Sentence": "0.25, this is 1.4. I could rewrite this if I want. 0.25, that's the same thing as 1.4. Eight times 1.4, or another way to think about it is eight divided by four is going to be equal to two. So this whole thing over here is going to be equal to two. So it's going to be minus, we have this minus out here, so minus two. And what is this going to be?"}, {"video_title": "How to evaluate expressions in two variables with decimals and fractions   6th grade   Khan Academy.mp3", "Sentence": "Eight times 1.4, or another way to think about it is eight divided by four is going to be equal to two. So this whole thing over here is going to be equal to two. So it's going to be minus, we have this minus out here, so minus two. And what is this going to be? Well, let's think about it. 3.5 plus five is 8.5, minus two is going to be 6.5. So this is equal to this, is equal to 6.5."}, {"video_title": "How to evaluate expressions in two variables with decimals and fractions   6th grade   Khan Academy.mp3", "Sentence": "And what is this going to be? Well, let's think about it. 3.5 plus five is 8.5, minus two is going to be 6.5. So this is equal to this, is equal to 6.5. Let's do another one of these. All right. And just like before, try to work through it on your own before we do it together."}, {"video_title": "How to evaluate expressions in two variables with decimals and fractions   6th grade   Khan Academy.mp3", "Sentence": "So this is equal to this, is equal to 6.5. Let's do another one of these. All right. And just like before, try to work through it on your own before we do it together. All right, now let's do it together. So over here I have this expression, 0.1m plus eight minus 12n, when m is equal to 30 and n is equal to 1.4. All right, so everywhere I see an m, I want to replace with a 30, and everywhere I see an n, I want to replace with a 1.4."}, {"video_title": "How to evaluate expressions in two variables with decimals and fractions   6th grade   Khan Academy.mp3", "Sentence": "And just like before, try to work through it on your own before we do it together. All right, now let's do it together. So over here I have this expression, 0.1m plus eight minus 12n, when m is equal to 30 and n is equal to 1.4. All right, so everywhere I see an m, I want to replace with a 30, and everywhere I see an n, I want to replace with a 1.4. So this is going to be equal to 0.1 times m, m is 30, times 30, plus eight minus 12 times n, where n is 1.4. So what is 1.10? This right over here, 0.1, that's the same thing as 1.10 of 30."}, {"video_title": "How to evaluate expressions in two variables with decimals and fractions   6th grade   Khan Academy.mp3", "Sentence": "All right, so everywhere I see an m, I want to replace with a 30, and everywhere I see an n, I want to replace with a 1.4. So this is going to be equal to 0.1 times m, m is 30, times 30, plus eight minus 12 times n, where n is 1.4. So what is 1.10? This right over here, 0.1, that's the same thing as 1.10 of 30. Well, 1.10 of 30, that's going to be three, so this part is three, three plus eight, and then we're going to have minus, well, what is 12 times 1.4? That's going to be 12 fourths, or 12 divided by four, which is going to be equal to three. And now when we evaluate this, so that is equal to this, we have a three plus eight minus three, well, threes are going to, you have a positive three, and then you're going to subtract three, and you're just going to be left with, you're just going to be left with an eight, and you're done."}, {"video_title": "Worked example equivalent systems of equations   High School Math   Khan Academy.mp3", "Sentence": "They each took a few steps that led to the systems shown in the table below. So we have the teacher's original system, what Mivek got after doing some operations, what Kamala got after doing some operations, which of them obtained a system that is equivalent to the teacher's system? So the first question we should ask ourselves is what does it mean to even have an equivalent system? For the sake of this question, or for our purposes, an equivalent system is a system that has the same solution. So if there's some xy pair that satisfies the teacher's system, that is the solution to the teacher's system, well, Mivek's system, we're gonna call it equivalent if it has the same solution. Similarly, if Kamala's system has the same solution, then we're going to call it equivalent to the teacher's system. So let's make some comparisons here."}, {"video_title": "Worked example equivalent systems of equations   High School Math   Khan Academy.mp3", "Sentence": "For the sake of this question, or for our purposes, an equivalent system is a system that has the same solution. So if there's some xy pair that satisfies the teacher's system, that is the solution to the teacher's system, well, Mivek's system, we're gonna call it equivalent if it has the same solution. Similarly, if Kamala's system has the same solution, then we're going to call it equivalent to the teacher's system. So let's make some comparisons here. So first, let's look at Mivek. So his first equation is actually unchanged from the teacher's equation. It's unchanged from the teacher's equation."}, {"video_title": "Worked example equivalent systems of equations   High School Math   Khan Academy.mp3", "Sentence": "So let's make some comparisons here. So first, let's look at Mivek. So his first equation is actually unchanged from the teacher's equation. It's unchanged from the teacher's equation. So any solution that meets both of these equations is for sure gonna meet this top equation, because it's literally the same as the top equation of the teacher, so that works out. Now let's look at the second one. The second one is definitely, this is a different, this second equation is definitely a different equation over here."}, {"video_title": "Worked example equivalent systems of equations   High School Math   Khan Academy.mp3", "Sentence": "It's unchanged from the teacher's equation. So any solution that meets both of these equations is for sure gonna meet this top equation, because it's literally the same as the top equation of the teacher, so that works out. Now let's look at the second one. The second one is definitely, this is a different, this second equation is definitely a different equation over here. We can check that it's not just being multiplied by some number on both sides. If we, to go from one to zero, if you were multiplying, you'd have to multiply one times zero, and then if you wanted, in order to maintain the equality, you would have to do that on both sides, but zero times this left-hand side would have been zero. You would have gotten zero equals zero."}, {"video_title": "Worked example equivalent systems of equations   High School Math   Khan Academy.mp3", "Sentence": "The second one is definitely, this is a different, this second equation is definitely a different equation over here. We can check that it's not just being multiplied by some number on both sides. If we, to go from one to zero, if you were multiplying, you'd have to multiply one times zero, and then if you wanted, in order to maintain the equality, you would have to do that on both sides, but zero times this left-hand side would have been zero. You would have gotten zero equals zero. So he didn't just scale both sides by some number. Looks like he did another operation. He probably, he probably looks like he's adding or subtracting something to both sides."}, {"video_title": "Worked example equivalent systems of equations   High School Math   Khan Academy.mp3", "Sentence": "You would have gotten zero equals zero. So he didn't just scale both sides by some number. Looks like he did another operation. He probably, he probably looks like he's adding or subtracting something to both sides. Let's see how he could have gotten this right over here. So he took negative four x plus five y is equal to one, and it looks like from that, he was able to get negative three x plus seven y is equal to zero. So let's see what he might have, what he had to do to do that."}, {"video_title": "Worked example equivalent systems of equations   High School Math   Khan Academy.mp3", "Sentence": "He probably, he probably looks like he's adding or subtracting something to both sides. Let's see how he could have gotten this right over here. So he took negative four x plus five y is equal to one, and it looks like from that, he was able to get negative three x plus seven y is equal to zero. So let's see what he might have, what he had to do to do that. Let's see, he would have had to, to go from negative four x to negative three x, he would have had to add an x, so I can just write an x right over there. To go from five y to seven y, he would have had to add two y. So on the left-hand side, he would have had to add x plus two y."}, {"video_title": "Worked example equivalent systems of equations   High School Math   Khan Academy.mp3", "Sentence": "So let's see what he might have, what he had to do to do that. Let's see, he would have had to, to go from negative four x to negative three x, he would have had to add an x, so I can just write an x right over there. To go from five y to seven y, he would have had to add two y. So on the left-hand side, he would have had to add x plus two y. Notice we have an x plus two y right over there. And on the right-hand side, he would have had to add or subtract a one or add a negative one. Notice we see a negative one right over there."}, {"video_title": "Worked example equivalent systems of equations   High School Math   Khan Academy.mp3", "Sentence": "So on the left-hand side, he would have had to add x plus two y. Notice we have an x plus two y right over there. And on the right-hand side, he would have had to add or subtract a one or add a negative one. Notice we see a negative one right over there. So what he essentially did is he added the left-hand sides of these two equations to get this new left-hand side right over here, and he added the right-hand sides to get this new right-hand side. And that is a legitimate operation. This new equation that you got, this new linear equation, it's going to represent a different line than this one right over here, but the resulting system is going to have the same solution."}, {"video_title": "Worked example equivalent systems of equations   High School Math   Khan Academy.mp3", "Sentence": "Notice we see a negative one right over there. So what he essentially did is he added the left-hand sides of these two equations to get this new left-hand side right over here, and he added the right-hand sides to get this new right-hand side. And that is a legitimate operation. This new equation that you got, this new linear equation, it's going to represent a different line than this one right over here, but the resulting system is going to have the same solution. Why do we feel confident that the resulting system is going to have the same solution? Well, for an xy pair that satisfies both of these equations, that's what a solution would be, for that xy pair, x plus two y is equal to negative one. So for that solution, we're adding the same thing to both sides."}, {"video_title": "Worked example equivalent systems of equations   High School Math   Khan Academy.mp3", "Sentence": "This new equation that you got, this new linear equation, it's going to represent a different line than this one right over here, but the resulting system is going to have the same solution. Why do we feel confident that the resulting system is going to have the same solution? Well, for an xy pair that satisfies both of these equations, that's what a solution would be, for that xy pair, x plus two y is equal to negative one. So for that solution, we're adding the same thing to both sides. We're saying, look, I'm going to add x plus two y to the left-hand side. Well, I need to add, if I don't want to change the solution, I have to add the same thing to the right-hand side. Well, they're telling us, for the solution to this equation, x plus two y is equal to negative one."}, {"video_title": "Worked example equivalent systems of equations   High School Math   Khan Academy.mp3", "Sentence": "So for that solution, we're adding the same thing to both sides. We're saying, look, I'm going to add x plus two y to the left-hand side. Well, I need to add, if I don't want to change the solution, I have to add the same thing to the right-hand side. Well, they're telling us, for the solution to this equation, x plus two y is equal to negative one. So negative one is the same thing as x plus two y for that solution. So we're not going to change the resulting solution of the system. So it's a completely legitimate operation."}, {"video_title": "Worked example equivalent systems of equations   High School Math   Khan Academy.mp3", "Sentence": "Well, they're telling us, for the solution to this equation, x plus two y is equal to negative one. So negative one is the same thing as x plus two y for that solution. So we're not going to change the resulting solution of the system. So it's a completely legitimate operation. What Vivek did is adding the left-hand sides and adding the right-hand sides to get this new second equation. That's not going to change the solution of the system. In fact, that's a technique we often use to eventually find the solution of a system."}, {"video_title": "Worked example equivalent systems of equations   High School Math   Khan Academy.mp3", "Sentence": "So it's a completely legitimate operation. What Vivek did is adding the left-hand sides and adding the right-hand sides to get this new second equation. That's not going to change the solution of the system. In fact, that's a technique we often use to eventually find the solution of a system. So now let's look at Camila. So her first equation is actually the exact same equation as the teacher's second equation. Now, let's see, her second equation, how does it relate possibly to the first equation?"}, {"video_title": "Worked example equivalent systems of equations   High School Math   Khan Academy.mp3", "Sentence": "In fact, that's a technique we often use to eventually find the solution of a system. So now let's look at Camila. So her first equation is actually the exact same equation as the teacher's second equation. Now, let's see, her second equation, how does it relate possibly to the first equation? So just looking at it offhand, it looks like it might just be, it looks like she just multiplied both sides times a number. And it looks like that number, she clearly multiplied the right-hand side times negative eight. So times negative eight."}, {"video_title": "Worked example equivalent systems of equations   High School Math   Khan Academy.mp3", "Sentence": "Now, let's see, her second equation, how does it relate possibly to the first equation? So just looking at it offhand, it looks like it might just be, it looks like she just multiplied both sides times a number. And it looks like that number, she clearly multiplied the right-hand side times negative eight. So times negative eight. Negative one times negative eight is positive eight. And it looks like she also multiplied the left-hand side by negative eight. Negative eight times x is negative eight x."}, {"video_title": "Worked example equivalent systems of equations   High School Math   Khan Academy.mp3", "Sentence": "So times negative eight. Negative one times negative eight is positive eight. And it looks like she also multiplied the left-hand side by negative eight. Negative eight times x is negative eight x. Negative eight times two y is negative 16 y. So she just multiplied both sides by the same value, which actually doesn't change the equation. This actually is going to be, it changes the way it looks, but it actually represents the same line."}, {"video_title": "Worked example equivalent systems of equations   High School Math   Khan Academy.mp3", "Sentence": "Negative eight times x is negative eight x. Negative eight times two y is negative 16 y. So she just multiplied both sides by the same value, which actually doesn't change the equation. This actually is going to be, it changes the way it looks, but it actually represents the same line. So this is definitely still an equivalent system. These are still the same constraints. You're going to have the same solution."}, {"video_title": "Worked example equivalent systems of equations   High School Math   Khan Academy.mp3", "Sentence": "This actually is going to be, it changes the way it looks, but it actually represents the same line. So this is definitely still an equivalent system. These are still the same constraints. You're going to have the same solution. Whenever you're dealing with systems, you're not going to change the solution of the system as long as you either multiply both sides of an equation by a scalar, or you're adding and subtracting the equations. When I say add or subtract equations, you're adding the left-hand side to the left-hand side, adding the right-hand side to the right-hand side like we had here, or subtracting the one from the other on the left-hand side. We subtract the bottom from the top on the left-hand, and we subtract the bottom from the top on the right-hand."}, {"video_title": "Converting recursive & explicit forms of arithmetic sequences   High School Math   Khan Academy.mp3", "Sentence": "So I have a function here, h of n, and let's say that it explicitly defines the terms of a sequence. So let me make a little, let me make a quick, a quick table here. So we have n, and then we have h of n. When n is equal to one, h of n is negative 31 minus seven times one minus one, which is going to be, this is just going to be zero, so it's going to be negative 31. When n is equal to two, it's going to be negative 31 minus seven times two minus one, so two minus one. Well, this is just going to be one, so it's negative 31 minus seven, which is equal to negative 38. When n is equal to three, it's going to be negative 31 minus seven times three minus one, which is just two. So we're going to subtract seven twice."}, {"video_title": "Converting recursive & explicit forms of arithmetic sequences   High School Math   Khan Academy.mp3", "Sentence": "When n is equal to two, it's going to be negative 31 minus seven times two minus one, so two minus one. Well, this is just going to be one, so it's negative 31 minus seven, which is equal to negative 38. When n is equal to three, it's going to be negative 31 minus seven times three minus one, which is just two. So we're going to subtract seven twice. So it's going to be negative 31 minus 14, which is equal to negative 45. So what do we see happening here? Well, we're starting at negative 31, and then we keep subtracting, we keep subtracting negative seven."}, {"video_title": "Converting recursive & explicit forms of arithmetic sequences   High School Math   Khan Academy.mp3", "Sentence": "So we're going to subtract seven twice. So it's going to be negative 31 minus 14, which is equal to negative 45. So what do we see happening here? Well, we're starting at negative 31, and then we keep subtracting, we keep subtracting negative seven. We keep subtracting negative seven from that. In fact, we subtract negative seven one less than the term, we subtract negative seven one less times than the term we're dealing with. If we're dealing with the third term, we subtract negative seven twice."}, {"video_title": "Converting recursive & explicit forms of arithmetic sequences   High School Math   Khan Academy.mp3", "Sentence": "Well, we're starting at negative 31, and then we keep subtracting, we keep subtracting negative seven. We keep subtracting negative seven from that. In fact, we subtract negative seven one less than the term, we subtract negative seven one less times than the term we're dealing with. If we're dealing with the third term, we subtract negative seven twice. If we're dealing with the second term, we subtract negative seven once. So this is all nice, but what I want you to do now is pause the video and see if you can define this exact same sequence. So the sequence here is you start at negative 31, and you keep subtracting negative seven, so negative 38, negative 45, the next one is going to be negative 52, and you go on and on and on, you keep subtracting negative seven."}, {"video_title": "Converting recursive & explicit forms of arithmetic sequences   High School Math   Khan Academy.mp3", "Sentence": "If we're dealing with the third term, we subtract negative seven twice. If we're dealing with the second term, we subtract negative seven once. So this is all nice, but what I want you to do now is pause the video and see if you can define this exact same sequence. So the sequence here is you start at negative 31, and you keep subtracting negative seven, so negative 38, negative 45, the next one is going to be negative 52, and you go on and on and on, you keep subtracting negative seven. Can we define this sequence in terms of a recursive function? So why don't you have a go at that? All right, let's try to define it in terms of a recursive function."}, {"video_title": "Converting recursive & explicit forms of arithmetic sequences   High School Math   Khan Academy.mp3", "Sentence": "So the sequence here is you start at negative 31, and you keep subtracting negative seven, so negative 38, negative 45, the next one is going to be negative 52, and you go on and on and on, you keep subtracting negative seven. Can we define this sequence in terms of a recursive function? So why don't you have a go at that? All right, let's try to define it in terms of a recursive function. Let's just call that g of n. So g of n, and in some ways a recursive function is easier, because you could say, okay, look, the first term, when n is equal to one, if n is equal to one, let me just write it, if n is equal to one, if n is equal to one, what's g of n going to be? It's going to be negative 31, negative 31, and if n is greater than one, and a whole number, so this is going to be defined for all positive integers, and whole number, well, it's just going to be the previous term, so g of n minus one, minus seven, minus seven. So we're saying, hey, if we're just picking an arbitrary term, we just have to look at the previous term, and then subtract, and then subtract seven."}, {"video_title": "Converting recursive & explicit forms of arithmetic sequences   High School Math   Khan Academy.mp3", "Sentence": "All right, let's try to define it in terms of a recursive function. Let's just call that g of n. So g of n, and in some ways a recursive function is easier, because you could say, okay, look, the first term, when n is equal to one, if n is equal to one, let me just write it, if n is equal to one, if n is equal to one, what's g of n going to be? It's going to be negative 31, negative 31, and if n is greater than one, and a whole number, so this is going to be defined for all positive integers, and whole number, well, it's just going to be the previous term, so g of n minus one, minus seven, minus seven. So we're saying, hey, if we're just picking an arbitrary term, we just have to look at the previous term, and then subtract, and then subtract seven. And it all works out nice and easy, because you keep looking at previous, previous, previous terms, all the way until you get to the base case, which is when n is equal to one, and then you can build up back from that, and you get this exact same sequence. Let's do another example, or let's go the other way around. So here we have a sequence defined recursively, and I want to create a function that defines a sequence explicitly."}, {"video_title": "Converting recursive & explicit forms of arithmetic sequences   High School Math   Khan Academy.mp3", "Sentence": "So we're saying, hey, if we're just picking an arbitrary term, we just have to look at the previous term, and then subtract, and then subtract seven. And it all works out nice and easy, because you keep looking at previous, previous, previous terms, all the way until you get to the base case, which is when n is equal to one, and then you can build up back from that, and you get this exact same sequence. Let's do another example, or let's go the other way around. So here we have a sequence defined recursively, and I want to create a function that defines a sequence explicitly. So let's think about this. So one way to think about it, this sequence, when n is equal to one, it starts at 9.6, and then every term is the previous term, minus 0.1. So the second term is going to be the previous term, minus 0.1, so it's going to be 9.5."}, {"video_title": "Converting recursive & explicit forms of arithmetic sequences   High School Math   Khan Academy.mp3", "Sentence": "So here we have a sequence defined recursively, and I want to create a function that defines a sequence explicitly. So let's think about this. So one way to think about it, this sequence, when n is equal to one, it starts at 9.6, and then every term is the previous term, minus 0.1. So the second term is going to be the previous term, minus 0.1, so it's going to be 9.5. Then you're going to go to 9.4. Then you're going to go to 9.3. We could keep going on, and on, and on."}, {"video_title": "Converting recursive & explicit forms of arithmetic sequences   High School Math   Khan Academy.mp3", "Sentence": "So the second term is going to be the previous term, minus 0.1, so it's going to be 9.5. Then you're going to go to 9.4. Then you're going to go to 9.3. We could keep going on, and on, and on. Or if we want, we can make a little table here, and we could say, this is n, this is h of n, and you see when n is equal to one, h of n is 9.6. When n is equal to two, we're now in this case over here, it's going to be h of two minus one, so it's going to be h of one minus 0.1. Well, it's just going to be this minus 0.1, which is going to be 9.5."}, {"video_title": "Converting recursive & explicit forms of arithmetic sequences   High School Math   Khan Academy.mp3", "Sentence": "We could keep going on, and on, and on. Or if we want, we can make a little table here, and we could say, this is n, this is h of n, and you see when n is equal to one, h of n is 9.6. When n is equal to two, we're now in this case over here, it's going to be h of two minus one, so it's going to be h of one minus 0.1. Well, it's just going to be this minus 0.1, which is going to be 9.5. When h is three, it's going to be h of two, h of two minus 0.1, minus 0.1. Well, h of two is right over here. You subtract a tenth, you're going to get 9.4, exactly what we saw over here."}, {"video_title": "Converting recursive & explicit forms of arithmetic sequences   High School Math   Khan Academy.mp3", "Sentence": "Well, it's just going to be this minus 0.1, which is going to be 9.5. When h is three, it's going to be h of two, h of two minus 0.1, minus 0.1. Well, h of two is right over here. You subtract a tenth, you're going to get 9.4, exactly what we saw over here. Let's see if we can pause the video now and define this, create a function that constructs or defines this arithmetic sequence explicitly. Here it was recursively, we want to define it explicitly. All right, so let's just call it, I don't know, let's just call it f of n. We could say, look, it's going to be 9.6, but we're going to subtract 0.1 a certain number of times, depending on what term we're talking about."}, {"video_title": "Converting recursive & explicit forms of arithmetic sequences   High School Math   Khan Academy.mp3", "Sentence": "You subtract a tenth, you're going to get 9.4, exactly what we saw over here. Let's see if we can pause the video now and define this, create a function that constructs or defines this arithmetic sequence explicitly. Here it was recursively, we want to define it explicitly. All right, so let's just call it, I don't know, let's just call it f of n. We could say, look, it's going to be 9.6, but we're going to subtract 0.1 a certain number of times, depending on what term we're talking about. We're going to subtract 0.1, but how many times are we going to subtract it as a function of n? Let's see, if we're talking about the first term, we subtract zero times. The second term, we subtract one time."}, {"video_title": "Converting recursive & explicit forms of arithmetic sequences   High School Math   Khan Academy.mp3", "Sentence": "All right, so let's just call it, I don't know, let's just call it f of n. We could say, look, it's going to be 9.6, but we're going to subtract 0.1 a certain number of times, depending on what term we're talking about. We're going to subtract 0.1, but how many times are we going to subtract it as a function of n? Let's see, if we're talking about the first term, we subtract zero times. The second term, we subtract one time. The third term, we subtract two times. The fourth term, we subtract three times. Whatever term we're talking about, we subtract that term minus one times."}, {"video_title": "Converting recursive & explicit forms of arithmetic sequences   High School Math   Khan Academy.mp3", "Sentence": "The second term, we subtract one time. The third term, we subtract two times. The fourth term, we subtract three times. Whatever term we're talking about, we subtract that term minus one times. If we're talking about the nth term, we subtract this value n minus one times. You can verify that this is going to work. When n is equal to one, this term here is going to be zero, so this whole thing's going to be zero, you get 9.6."}, {"video_title": "Converting recursive & explicit forms of arithmetic sequences   High School Math   Khan Academy.mp3", "Sentence": "Whatever term we're talking about, we subtract that term minus one times. If we're talking about the nth term, we subtract this value n minus one times. You can verify that this is going to work. When n is equal to one, this term here is going to be zero, so this whole thing's going to be zero, you get 9.6. When n is equal to two, two minus one, you subtract 0.1 one time. 9.6 minus 0.1 is 9.5. You could keep doing that, you could draw a table and evaluate these if you want to, but the key thing is, you're starting at 9.6 and you're subtracting 0.1 one fewer times than the term you're looking at."}, {"video_title": "Solving systems of two linear equations example   Algebra I   Khan Academy.mp3", "Sentence": "You are solving a system of two linear equations and two variables. You have found more than one solution that satisfies the system. Which of the following statements is true? So before even reading these statements, let's just think about what's going on. So let me draw my axes here. Let's draw my axes. So this is going to be my vertical axis."}, {"video_title": "Solving systems of two linear equations example   Algebra I   Khan Academy.mp3", "Sentence": "So before even reading these statements, let's just think about what's going on. So let me draw my axes here. Let's draw my axes. So this is going to be my vertical axis. That could be one of the variables. And then this is my horizontal axis. That's one of the other variables."}, {"video_title": "Solving systems of two linear equations example   Algebra I   Khan Academy.mp3", "Sentence": "So this is going to be my vertical axis. That could be one of the variables. And then this is my horizontal axis. That's one of the other variables. And maybe for the sake of convention, this could be x and this could be y. But they're whatever our two variables are. So it's a system of two linear equations."}, {"video_title": "Solving systems of two linear equations example   Algebra I   Khan Academy.mp3", "Sentence": "That's one of the other variables. And maybe for the sake of convention, this could be x and this could be y. But they're whatever our two variables are. So it's a system of two linear equations. So if we're graphing them, each of the linear equations in two variables can be represented by a line. Now, there's only three scenarios here. One scenario is where the lines don't intersect at all."}, {"video_title": "Solving systems of two linear equations example   Algebra I   Khan Academy.mp3", "Sentence": "So it's a system of two linear equations. So if we're graphing them, each of the linear equations in two variables can be represented by a line. Now, there's only three scenarios here. One scenario is where the lines don't intersect at all. So the only way that you're going to have two lines in two dimensions that don't intersect is if they have the same slope and they have different y-intercepts. So that's one scenario. But that's not the scenario that's being described here."}, {"video_title": "Solving systems of two linear equations example   Algebra I   Khan Academy.mp3", "Sentence": "One scenario is where the lines don't intersect at all. So the only way that you're going to have two lines in two dimensions that don't intersect is if they have the same slope and they have different y-intercepts. So that's one scenario. But that's not the scenario that's being described here. They say you have found more than one solution that satisfies the system. Here there are no solutions. So that's not the scenario that we're talking about."}, {"video_title": "Solving systems of two linear equations example   Algebra I   Khan Academy.mp3", "Sentence": "But that's not the scenario that's being described here. They say you have found more than one solution that satisfies the system. Here there are no solutions. So that's not the scenario that we're talking about. There's another scenario where they intersect in exactly one place. So they intersect in exactly one place. There's one point, one xy coordinate right over there that satisfies both of these constraints."}, {"video_title": "Solving systems of two linear equations example   Algebra I   Khan Academy.mp3", "Sentence": "So that's not the scenario that we're talking about. There's another scenario where they intersect in exactly one place. So they intersect in exactly one place. There's one point, one xy coordinate right over there that satisfies both of these constraints. But this also is not the scenario they're talking about. They're telling us that you have found more than one solution that satisfies the system. So this isn't the scenario either."}, {"video_title": "Solving systems of two linear equations example   Algebra I   Khan Academy.mp3", "Sentence": "There's one point, one xy coordinate right over there that satisfies both of these constraints. But this also is not the scenario they're talking about. They're telling us that you have found more than one solution that satisfies the system. So this isn't the scenario either. So the only other scenario that we can have, we don't have parallel lines that don't intersect. We don't have lines that only intersect in one place. The only other scenario is that we're dealing with a situation where both linear equations are essentially the same constraint."}, {"video_title": "Solving systems of two linear equations example   Algebra I   Khan Academy.mp3", "Sentence": "So this isn't the scenario either. So the only other scenario that we can have, we don't have parallel lines that don't intersect. We don't have lines that only intersect in one place. The only other scenario is that we're dealing with a situation where both linear equations are essentially the same constraint. They both are essentially representing the same xy relationship. That's the only way that I can have two lines. And this only applies to linear relationship and lines."}, {"video_title": "Solving systems of two linear equations example   Algebra I   Khan Academy.mp3", "Sentence": "The only other scenario is that we're dealing with a situation where both linear equations are essentially the same constraint. They both are essentially representing the same xy relationship. That's the only way that I can have two lines. And this only applies to linear relationship and lines. But the only way that two lines can intersect more than one place is if they intersect everywhere. So in this situation we know that we must have an infinite number of solutions. So which of these choices say that?"}, {"video_title": "Solving systems of two linear equations example   Algebra I   Khan Academy.mp3", "Sentence": "And this only applies to linear relationship and lines. But the only way that two lines can intersect more than one place is if they intersect everywhere. So in this situation we know that we must have an infinite number of solutions. So which of these choices say that? This one right here. There are infinitely many more solutions to the system. Right over there."}, {"video_title": "Area of triangles intuition   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "In another video, we saw that if we're looking at the area of a parallelogram, and we also know the length of a base, and we know its height, that the area is still going to be base times height. Now, it's not as obvious when you look at the parallelogram, but in that video, we did a little manipulation of the area. We said, hey, let's take this little section right over here. So, we took that little section right over there, and then we move it over to the right-hand side, and just like that, you see that as long as the base and the height is the same as this rectangle here, I'm able to construct the same rectangle by moving that area over. And that's why the area of this parallelogram is base times height. I didn't add or take away area, I just shifted area from the left-hand side to the right-hand side to show you that the area of that parallelogram was the same as this area of the rectangle. It's still going to be base times height."}, {"video_title": "Area of triangles intuition   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "So, we took that little section right over there, and then we move it over to the right-hand side, and just like that, you see that as long as the base and the height is the same as this rectangle here, I'm able to construct the same rectangle by moving that area over. And that's why the area of this parallelogram is base times height. I didn't add or take away area, I just shifted area from the left-hand side to the right-hand side to show you that the area of that parallelogram was the same as this area of the rectangle. It's still going to be base times height. So, hopefully that convinces you, that convinces you that the area of a parallelogram is base times height, because we're now going to use that to get the intuition for the area of a triangle. So, let's look at some triangles here. So, that is a triangle, and we're given the base and the height."}, {"video_title": "Area of triangles intuition   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "It's still going to be base times height. So, hopefully that convinces you, that convinces you that the area of a parallelogram is base times height, because we're now going to use that to get the intuition for the area of a triangle. So, let's look at some triangles here. So, that is a triangle, and we're given the base and the height. Now, we're gonna try to think about, well, what's this area, what's the area of this triangle going to be? And you can imagine it's going to be dependent on base and height. Well, to think about that, let me copy and paste this triangle."}, {"video_title": "Area of triangles intuition   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "So, that is a triangle, and we're given the base and the height. Now, we're gonna try to think about, well, what's this area, what's the area of this triangle going to be? And you can imagine it's going to be dependent on base and height. Well, to think about that, let me copy and paste this triangle. So, let me copy, and then let me paste it. And what I'm gonna do is, so now I have two of the triangles, so this is now going to be twice the area, and I'm gonna rotate it around, I'm gonna rotate it around like that, rotate it around like that, and then add it to the original area, and you see something very interesting is happening. I have now constructed a parallelogram, I have now constructed a parallelogram that has twice the area of our original triangle, because I have two of our original triangles right over here, you saw me do it."}, {"video_title": "Area of triangles intuition   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "Well, to think about that, let me copy and paste this triangle. So, let me copy, and then let me paste it. And what I'm gonna do is, so now I have two of the triangles, so this is now going to be twice the area, and I'm gonna rotate it around, I'm gonna rotate it around like that, rotate it around like that, and then add it to the original area, and you see something very interesting is happening. I have now constructed a parallelogram, I have now constructed a parallelogram that has twice the area of our original triangle, because I have two of our original triangles right over here, you saw me do it. I copied and pasted it, and then I flipped it over, and I constructed the parallelogram. Now, why is this interesting? Well, the area of the entire parallelogram, the area of the entire parallelogram is going to be the length of this base, base times this height, you also have the height written on the H upside down over here, it's going to be base times height."}, {"video_title": "Area of triangles intuition   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "I have now constructed a parallelogram, I have now constructed a parallelogram that has twice the area of our original triangle, because I have two of our original triangles right over here, you saw me do it. I copied and pasted it, and then I flipped it over, and I constructed the parallelogram. Now, why is this interesting? Well, the area of the entire parallelogram, the area of the entire parallelogram is going to be the length of this base, base times this height, you also have the height written on the H upside down over here, it's going to be base times height. That's going to be for the parallelogram, for the entire, let me draw a parallelogram right over here. That's going to be the area of the entire parallelogram. So, what would be the area of our original triangle?"}, {"video_title": "Area of triangles intuition   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "Well, the area of the entire parallelogram, the area of the entire parallelogram is going to be the length of this base, base times this height, you also have the height written on the H upside down over here, it's going to be base times height. That's going to be for the parallelogram, for the entire, let me draw a parallelogram right over here. That's going to be the area of the entire parallelogram. So, what would be the area of our original triangle? What would be the area of our original triangle? Well, we already saw that this area of the parallelogram, it's twice the area of our original triangle. So, our original triangle is just going to have half the area."}, {"video_title": "Area of triangles intuition   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "So, what would be the area of our original triangle? What would be the area of our original triangle? Well, we already saw that this area of the parallelogram, it's twice the area of our original triangle. So, our original triangle is just going to have half the area. So, this area right over here is going to be 1 1 2, the area of the parallelogram. 1 1 2, base, let me do those same colors. 1 1 2, base times height."}, {"video_title": "Area of triangles intuition   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "So, our original triangle is just going to have half the area. So, this area right over here is going to be 1 1 2, the area of the parallelogram. 1 1 2, base, let me do those same colors. 1 1 2, base times height. 1 1 2, base times height. And you might say, okay, maybe it worked for this triangle, but I want to see it work for more triangles. And so, to help you there, I've added another triangle right over here."}, {"video_title": "Area of triangles intuition   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "1 1 2, base times height. 1 1 2, base times height. And you might say, okay, maybe it worked for this triangle, but I want to see it work for more triangles. And so, to help you there, I've added another triangle right over here. You could view this as an obtuse triangle. This angle right over here is greater than 90 degrees. But I'm going to do the same trick."}, {"video_title": "Area of triangles intuition   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "And so, to help you there, I've added another triangle right over here. You could view this as an obtuse triangle. This angle right over here is greater than 90 degrees. But I'm going to do the same trick. We have the base, and then we have the height. Here, you can think of, if you start at this point right over here, and if you drop a ball, the length that the ball goes, or if you had a string here, to kind of get to the ground level, you could view this as the ground level right over there, that that's going to be the height. It's not sitting in the triangle like we saw last time, but it's still the height of the triangle."}, {"video_title": "Area of triangles intuition   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "But I'm going to do the same trick. We have the base, and then we have the height. Here, you can think of, if you start at this point right over here, and if you drop a ball, the length that the ball goes, or if you had a string here, to kind of get to the ground level, you could view this as the ground level right over there, that that's going to be the height. It's not sitting in the triangle like we saw last time, but it's still the height of the triangle. If this was a building of some kind, you'd say, well, this is the height. How far off the ground is it? Well, what's the area of this going to be?"}, {"video_title": "Area of triangles intuition   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "It's not sitting in the triangle like we saw last time, but it's still the height of the triangle. If this was a building of some kind, you'd say, well, this is the height. How far off the ground is it? Well, what's the area of this going to be? Well, you can imagine. It's going to be 1 1 2 base times height. How do we feel good about that?"}, {"video_title": "Area of triangles intuition   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "Well, what's the area of this going to be? Well, you can imagine. It's going to be 1 1 2 base times height. How do we feel good about that? Well, let's do the same magic here. So, let me copy and paste this. I'm going to copy and then paste it."}, {"video_title": "Area of triangles intuition   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "How do we feel good about that? Well, let's do the same magic here. So, let me copy and paste this. I'm going to copy and then paste it. Whoops, that didn't work. Let me copy and then paste it. And so, I have two of these triangles now, but I'm going to flip this one over so that I can construct a parallelogram."}, {"video_title": "Area of triangles intuition   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "I'm going to copy and then paste it. Whoops, that didn't work. Let me copy and then paste it. And so, I have two of these triangles now, but I'm going to flip this one over so that I can construct a parallelogram. I'm going to flip it over and move it over here. I'm going to have to rotate it a little bit more. So, I think you get the general idea."}, {"video_title": "Area of triangles intuition   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "And so, I have two of these triangles now, but I'm going to flip this one over so that I can construct a parallelogram. I'm going to flip it over and move it over here. I'm going to have to rotate it a little bit more. So, I think you get the general idea. So, now I've constructed a parallelogram that has twice the area of our original triangle. It has twice the area of our original triangle. And so, if I talked about the area of the entire parallelogram, it would be base times the height of the parallelogram."}, {"video_title": "Area of triangles intuition   Algebra I   High School Math   Khan Academy.mp3", "Sentence": "So, I think you get the general idea. So, now I've constructed a parallelogram that has twice the area of our original triangle. It has twice the area of our original triangle. And so, if I talked about the area of the entire parallelogram, it would be base times the height of the parallelogram. Base times the height of the parallelogram. But if we're only talking about the area of, if we're only talking about this area right over here, which is our original triangle, it's going to be half the area of the parallelogram. So, it's going to be 1 1 2 of that."}, {"video_title": "Slope from two ordered pairs example 1   Algebra I   Khan Academy.mp3", "Sentence": "Find the slope of the line that goes through the ordered pairs 4, 2 and negative 3, 16. So just as a reminder, slope is defined as rise over run, or you could view that rise as just change in y and run as just change in x. So you can see the triangles here, that's the delta symbol, it literally means change in. Or another way, and you might see this formula and it tends to be really complicated, but just remember it's just these two things over here. Sometimes slope will be specified with the variable m, and they'll say that m is the same thing, and this is really the same thing as change in y. They'll write y2 minus y1 over x2 minus x1. So the notation tends to be kind of complicated, but all this means is you take the y value of your end point and subtract from it the y value of your starting point."}, {"video_title": "Slope from two ordered pairs example 1   Algebra I   Khan Academy.mp3", "Sentence": "Or another way, and you might see this formula and it tends to be really complicated, but just remember it's just these two things over here. Sometimes slope will be specified with the variable m, and they'll say that m is the same thing, and this is really the same thing as change in y. They'll write y2 minus y1 over x2 minus x1. So the notation tends to be kind of complicated, but all this means is you take the y value of your end point and subtract from it the y value of your starting point. That'll essentially give you your change in y. And it says take the x value of your end point and subtract from that the x value of your starting point, and that'll give you change in x. So whatever of these work for you, let's actually figure out the slope of the line that goes through these two points."}, {"video_title": "Slope from two ordered pairs example 1   Algebra I   Khan Academy.mp3", "Sentence": "So the notation tends to be kind of complicated, but all this means is you take the y value of your end point and subtract from it the y value of your starting point. That'll essentially give you your change in y. And it says take the x value of your end point and subtract from that the x value of your starting point, and that'll give you change in x. So whatever of these work for you, let's actually figure out the slope of the line that goes through these two points. So we're starting at, and actually we could do it both ways. We could start at this point and go to that point and calculate the slope, or we could start at this point and go to that point and calculate the slope. So let's do it both ways."}, {"video_title": "Slope from two ordered pairs example 1   Algebra I   Khan Academy.mp3", "Sentence": "So whatever of these work for you, let's actually figure out the slope of the line that goes through these two points. So we're starting at, and actually we could do it both ways. We could start at this point and go to that point and calculate the slope, or we could start at this point and go to that point and calculate the slope. So let's do it both ways. So let's say that our starting point is the point 4, 2, and let's say that our end point is negative 3, 16. So what is the change in x over here? What is the change in x in this scenario?"}, {"video_title": "Slope from two ordered pairs example 1   Algebra I   Khan Academy.mp3", "Sentence": "So let's do it both ways. So let's say that our starting point is the point 4, 2, and let's say that our end point is negative 3, 16. So what is the change in x over here? What is the change in x in this scenario? So we're going from 4 to negative 3. If something goes from 4 to negative 3, what was its change? Well, you have to go down 4 to get to 0, and then you have to go down another 3 to get to negative 3."}, {"video_title": "Slope from two ordered pairs example 1   Algebra I   Khan Academy.mp3", "Sentence": "What is the change in x in this scenario? So we're going from 4 to negative 3. If something goes from 4 to negative 3, what was its change? Well, you have to go down 4 to get to 0, and then you have to go down another 3 to get to negative 3. So our change in x here is negative 7. Actually, let me write it this way. Our change in x is equal to negative 3 minus 4, which is equal to negative 7."}, {"video_title": "Slope from two ordered pairs example 1   Algebra I   Khan Academy.mp3", "Sentence": "Well, you have to go down 4 to get to 0, and then you have to go down another 3 to get to negative 3. So our change in x here is negative 7. Actually, let me write it this way. Our change in x is equal to negative 3 minus 4, which is equal to negative 7. If I'm going from 4 to negative 3, I went down by 7. Our change in x is negative 7. Let's do the same thing for the change in y."}, {"video_title": "Slope from two ordered pairs example 1   Algebra I   Khan Academy.mp3", "Sentence": "Our change in x is equal to negative 3 minus 4, which is equal to negative 7. If I'm going from 4 to negative 3, I went down by 7. Our change in x is negative 7. Let's do the same thing for the change in y. Notice, I implicitly used this formula over here. Our change in x was this value, our end point, our end x value, minus our starting x value. Let's do the same thing for our change in y."}, {"video_title": "Slope from two ordered pairs example 1   Algebra I   Khan Academy.mp3", "Sentence": "Let's do the same thing for the change in y. Notice, I implicitly used this formula over here. Our change in x was this value, our end point, our end x value, minus our starting x value. Let's do the same thing for our change in y. Our change in y, if we're starting at 2 and we go to 16, that means we moved up 14. Or another way you could say it, you could take your ending y value and subtract from that your starting y value, and you get 14. So what is the slope over here?"}, {"video_title": "Slope from two ordered pairs example 1   Algebra I   Khan Academy.mp3", "Sentence": "Let's do the same thing for our change in y. Our change in y, if we're starting at 2 and we go to 16, that means we moved up 14. Or another way you could say it, you could take your ending y value and subtract from that your starting y value, and you get 14. So what is the slope over here? The slope is just change in y over change in x. The slope over here is change in y over change in x, which is our change in y is 14, and our change in x is negative 7. If we want to simplify this, 14 divided by negative 7 is negative 2."}, {"video_title": "Slope from two ordered pairs example 1   Algebra I   Khan Academy.mp3", "Sentence": "So what is the slope over here? The slope is just change in y over change in x. The slope over here is change in y over change in x, which is our change in y is 14, and our change in x is negative 7. If we want to simplify this, 14 divided by negative 7 is negative 2. What I want to show you is that we could have done it the other way around. We could have made this the starting point and this the end point. What we would have gotten is the negative values of each of these, but then they would have cancelled out and we would still get negative 2."}, {"video_title": "Slope from two ordered pairs example 1   Algebra I   Khan Academy.mp3", "Sentence": "If we want to simplify this, 14 divided by negative 7 is negative 2. What I want to show you is that we could have done it the other way around. We could have made this the starting point and this the end point. What we would have gotten is the negative values of each of these, but then they would have cancelled out and we would still get negative 2. Let's try it out. Let's say that our start point was negative 3 comma 16, and let's say that our end point is the 4 comma 2. In this situation, what is our change in x?"}, {"video_title": "Slope from two ordered pairs example 1   Algebra I   Khan Academy.mp3", "Sentence": "What we would have gotten is the negative values of each of these, but then they would have cancelled out and we would still get negative 2. Let's try it out. Let's say that our start point was negative 3 comma 16, and let's say that our end point is the 4 comma 2. In this situation, what is our change in x? If I start at negative 3 and I go to 4, that means I went up 7. Or if you want to just calculate that, you would do 4 minus negative 3. Needless to say, we just went up 7."}, {"video_title": "Slope from two ordered pairs example 1   Algebra I   Khan Academy.mp3", "Sentence": "In this situation, what is our change in x? If I start at negative 3 and I go to 4, that means I went up 7. Or if you want to just calculate that, you would do 4 minus negative 3. Needless to say, we just went up 7. What is our change in y? Our change in y over here, or we could say our rise. If we start at 16 and we end at 2, that means we went down 14."}, {"video_title": "Slope from two ordered pairs example 1   Algebra I   Khan Academy.mp3", "Sentence": "Needless to say, we just went up 7. What is our change in y? Our change in y over here, or we could say our rise. If we start at 16 and we end at 2, that means we went down 14. Or you could just say 2 minus 16 is negative 14. We went down by 14. This was our run."}, {"video_title": "Slope from two ordered pairs example 1   Algebra I   Khan Academy.mp3", "Sentence": "If we start at 16 and we end at 2, that means we went down 14. Or you could just say 2 minus 16 is negative 14. We went down by 14. This was our run. If you say rise over run, which is the same thing as change in y over change in x, our rise is negative 14 and our run here is 7. Notice, these are just the negatives of these values from when we swapped them. Once again, this is equal to negative 2."}, {"video_title": "Slope from two ordered pairs example 1   Algebra I   Khan Academy.mp3", "Sentence": "This was our run. If you say rise over run, which is the same thing as change in y over change in x, our rise is negative 14 and our run here is 7. Notice, these are just the negatives of these values from when we swapped them. Once again, this is equal to negative 2. Let's just visualize this. Let me do a quick graph here just to show you what a downward slope would look like. Let me draw our two points."}, {"video_title": "Slope from two ordered pairs example 1   Algebra I   Khan Academy.mp3", "Sentence": "Once again, this is equal to negative 2. Let's just visualize this. Let me do a quick graph here just to show you what a downward slope would look like. Let me draw our two points. This is my x-axis. That is my y-axis. This point over here, 4, 2."}, {"video_title": "Slope from two ordered pairs example 1   Algebra I   Khan Academy.mp3", "Sentence": "Let me draw our two points. This is my x-axis. That is my y-axis. This point over here, 4, 2. Let me graph it. We are going to go all the way up to 16. Let me save some space here."}, {"video_title": "Slope from two ordered pairs example 1   Algebra I   Khan Academy.mp3", "Sentence": "This point over here, 4, 2. Let me graph it. We are going to go all the way up to 16. Let me save some space here. We have 1, 2, 3, 4. It is 4, 1, 2. 4, 2 is right over here."}, {"video_title": "Slope from two ordered pairs example 1   Algebra I   Khan Academy.mp3", "Sentence": "Let me save some space here. We have 1, 2, 3, 4. It is 4, 1, 2. 4, 2 is right over here. Then we have the point negative 3, 16. Let me draw that over here. We have negative 1, 2, 3."}, {"video_title": "Slope from two ordered pairs example 1   Algebra I   Khan Academy.mp3", "Sentence": "4, 2 is right over here. Then we have the point negative 3, 16. Let me draw that over here. We have negative 1, 2, 3. We have to go up to 16. This is 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16. It goes right over here."}, {"video_title": "Slope from two ordered pairs example 1   Algebra I   Khan Academy.mp3", "Sentence": "We have negative 1, 2, 3. We have to go up to 16. This is 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16. It goes right over here. This is negative 3, 16. The line that goes between them is going to look something like this. I will try my best to draw a relatively straight line."}, {"video_title": "Slope from two ordered pairs example 1   Algebra I   Khan Academy.mp3", "Sentence": "It goes right over here. This is negative 3, 16. The line that goes between them is going to look something like this. I will try my best to draw a relatively straight line. That line will keep going. The line will keep going. That is my best attempt."}, {"video_title": "Slope from two ordered pairs example 1   Algebra I   Khan Academy.mp3", "Sentence": "I will try my best to draw a relatively straight line. That line will keep going. The line will keep going. That is my best attempt. Notice it is downward sloping. As you increase an x value, the line goes down. It is going from the top left to the bottom right."}, {"video_title": "Slope from two ordered pairs example 1   Algebra I   Khan Academy.mp3", "Sentence": "That is my best attempt. Notice it is downward sloping. As you increase an x value, the line goes down. It is going from the top left to the bottom right. As x gets bigger, y gets smaller. That is what a downward sloping line looks like. Just to visualize our change in x's and our change in y's that we dealt with here."}, {"video_title": "Slope from two ordered pairs example 1   Algebra I   Khan Academy.mp3", "Sentence": "It is going from the top left to the bottom right. As x gets bigger, y gets smaller. That is what a downward sloping line looks like. Just to visualize our change in x's and our change in y's that we dealt with here. When we started at 4, 2 and ended at negative 3, 16, that was analogous to starting here and ending over there. We said our change in x was negative 7. We had to move back."}, {"video_title": "Slope from two ordered pairs example 1   Algebra I   Khan Academy.mp3", "Sentence": "Just to visualize our change in x's and our change in y's that we dealt with here. When we started at 4, 2 and ended at negative 3, 16, that was analogous to starting here and ending over there. We said our change in x was negative 7. We had to move back. Our run, we had to move in the left direction by 7. That is why it was negative 7. Then we had to move in the y direction positive 14."}, {"video_title": "Slope from two ordered pairs example 1   Algebra I   Khan Academy.mp3", "Sentence": "We had to move back. Our run, we had to move in the left direction by 7. That is why it was negative 7. Then we had to move in the y direction positive 14. That is why our rise was positive. It was 14 over negative 7 or negative 2. When we did it the other way, we started at this point and ended at this point."}, {"video_title": "Slope from two ordered pairs example 1   Algebra I   Khan Academy.mp3", "Sentence": "Then we had to move in the y direction positive 14. That is why our rise was positive. It was 14 over negative 7 or negative 2. When we did it the other way, we started at this point and ended at this point. We started at negative 3, 16 and ended at that point. In that situation, our run was positive 7. Now we had to go down in the y direction since we switched the starting and the end point."}, {"video_title": "Slope from two ordered pairs example 1   Algebra I   Khan Academy.mp3", "Sentence": "When we did it the other way, we started at this point and ended at this point. We started at negative 3, 16 and ended at that point. In that situation, our run was positive 7. Now we had to go down in the y direction since we switched the starting and the end point. Now we had to go down negative 14. Our run is now positive 7 and our rise is now negative 14. Either way, we got the same slope."}, {"video_title": "Solving a proportion with an unknown variable (example)   7th grade   Khan Academy.mp3", "Sentence": "Or 8 to 36, or the ratio of 8 over 36, is equal to the ratio of 10 to what? And there's a bunch of different ways to solve this. And I'll explore really all of them. Or not all of them, or a good selection of them. So one way to think about it is these two need to be equivalent ratios, or really equivalent fractions. So whatever happened to the numerator also has to happen to the denominator. So what do we have to multiply 8 by to get 10?"}, {"video_title": "Solving a proportion with an unknown variable (example)   7th grade   Khan Academy.mp3", "Sentence": "Or not all of them, or a good selection of them. So one way to think about it is these two need to be equivalent ratios, or really equivalent fractions. So whatever happened to the numerator also has to happen to the denominator. So what do we have to multiply 8 by to get 10? Well, you could multiply 8 times 10 over 8 will definitely give you 10. So we're multiplying by 10 over 8 over here. Or another way to write 10 over 8, 10 over 8 is the same thing as 5 over 4."}, {"video_title": "Solving a proportion with an unknown variable (example)   7th grade   Khan Academy.mp3", "Sentence": "So what do we have to multiply 8 by to get 10? Well, you could multiply 8 times 10 over 8 will definitely give you 10. So we're multiplying by 10 over 8 over here. Or another way to write 10 over 8, 10 over 8 is the same thing as 5 over 4. So we're multiplying by 5 over 4 to get to 10, from 8 to 10. Well, if we did that to the numerator, in order to have an equivalent fraction, you have to do the same thing to the denominator. You have to multiply it."}, {"video_title": "Solving a proportion with an unknown variable (example)   7th grade   Khan Academy.mp3", "Sentence": "Or another way to write 10 over 8, 10 over 8 is the same thing as 5 over 4. So we're multiplying by 5 over 4 to get to 10, from 8 to 10. Well, if we did that to the numerator, in order to have an equivalent fraction, you have to do the same thing to the denominator. You have to multiply it. You have to multiply it times 5 over 4. And so we could say this n, this thing that we just solved for, this n is going to be equal to 36 times 5 divided by 4. Or you could say that this is going to be equal to 36 times 5 divided by 4."}, {"video_title": "Solving a proportion with an unknown variable (example)   7th grade   Khan Academy.mp3", "Sentence": "You have to multiply it. You have to multiply it times 5 over 4. And so we could say this n, this thing that we just solved for, this n is going to be equal to 36 times 5 divided by 4. Or you could say that this is going to be equal to 36 times 5 divided by 4. And now, 36 divided by 4, we know what that is. We can divide both the numerator and the denominator by 4. You divide the numerator by 4, you get 9."}, {"video_title": "Solving a proportion with an unknown variable (example)   7th grade   Khan Academy.mp3", "Sentence": "Or you could say that this is going to be equal to 36 times 5 divided by 4. And now, 36 divided by 4, we know what that is. We can divide both the numerator and the denominator by 4. You divide the numerator by 4, you get 9. Divide the denominator by 4, you get 1. You get 45. So that's one way to think about it."}, {"video_title": "Solving a proportion with an unknown variable (example)   7th grade   Khan Academy.mp3", "Sentence": "You divide the numerator by 4, you get 9. Divide the denominator by 4, you get 1. You get 45. So that's one way to think about it. 8 over 36 is equal to 10 over 45. Another way to think about it is, what do we have to multiply 8 by to get its denominator? How much larger is the denominator 36 than 8?"}, {"video_title": "Solving a proportion with an unknown variable (example)   7th grade   Khan Academy.mp3", "Sentence": "So that's one way to think about it. 8 over 36 is equal to 10 over 45. Another way to think about it is, what do we have to multiply 8 by to get its denominator? How much larger is the denominator 36 than 8? Well, let's just divide 36 over 8. So 36 over 8 is the same thing as, so we can simplify dividing the numerator and the denominator by 4. That's their greatest common divisor."}, {"video_title": "Solving a proportion with an unknown variable (example)   7th grade   Khan Academy.mp3", "Sentence": "How much larger is the denominator 36 than 8? Well, let's just divide 36 over 8. So 36 over 8 is the same thing as, so we can simplify dividing the numerator and the denominator by 4. That's their greatest common divisor. That's the same thing as 9 halves. So if you multiply the numerator by 9 halves, you get the denominator. So we're multiplying by 9 halves to get the denominator over here."}, {"video_title": "Solving a proportion with an unknown variable (example)   7th grade   Khan Academy.mp3", "Sentence": "That's their greatest common divisor. That's the same thing as 9 halves. So if you multiply the numerator by 9 halves, you get the denominator. So we're multiplying by 9 halves to get the denominator over here. Well, then we have to do the same thing over here. If 36 is 9 halves times 8, let me write this. 36, or let me write 8 times 9 halves is equal to 36."}, {"video_title": "Solving a proportion with an unknown variable (example)   7th grade   Khan Academy.mp3", "Sentence": "So we're multiplying by 9 halves to get the denominator over here. Well, then we have to do the same thing over here. If 36 is 9 halves times 8, let me write this. 36, or let me write 8 times 9 halves is equal to 36. That's how we go from the numerator to denominator. Then to figure out what the denominator here is, if we want the same fraction, we have to multiply by 9 halves again. So then we'll get 10 times 9 halves is going to be equal to n, is going to be equal to this denominator."}, {"video_title": "Solving a proportion with an unknown variable (example)   7th grade   Khan Academy.mp3", "Sentence": "36, or let me write 8 times 9 halves is equal to 36. That's how we go from the numerator to denominator. Then to figure out what the denominator here is, if we want the same fraction, we have to multiply by 9 halves again. So then we'll get 10 times 9 halves is going to be equal to n, is going to be equal to this denominator. And so this is the same thing as saying 10 times 9 over 2. Divide the numerator and the denominator by 2, you get 5 over 1, which is 45. So 45 is equal to n. Once again, we got the same way."}, {"video_title": "Solving a proportion with an unknown variable (example)   7th grade   Khan Academy.mp3", "Sentence": "So then we'll get 10 times 9 halves is going to be equal to n, is going to be equal to this denominator. And so this is the same thing as saying 10 times 9 over 2. Divide the numerator and the denominator by 2, you get 5 over 1, which is 45. So 45 is equal to n. Once again, we got the same way. Completely legitimate way to solve it. Now sometimes when you see a proportion like this, sometimes people say, oh, you can cross multiply. And you can cross multiply, and I'll teach you how to do that."}, {"video_title": "Solving a proportion with an unknown variable (example)   7th grade   Khan Academy.mp3", "Sentence": "So 45 is equal to n. Once again, we got the same way. Completely legitimate way to solve it. Now sometimes when you see a proportion like this, sometimes people say, oh, you can cross multiply. And you can cross multiply, and I'll teach you how to do that. And that's sometimes a quick way to do it. But I don't like teaching it the first time you look at proportions, because it's really just something mechanical. You really don't understand what you're doing."}, {"video_title": "Solving a proportion with an unknown variable (example)   7th grade   Khan Academy.mp3", "Sentence": "And you can cross multiply, and I'll teach you how to do that. And that's sometimes a quick way to do it. But I don't like teaching it the first time you look at proportions, because it's really just something mechanical. You really don't understand what you're doing. And it really comes out of a little bit of algebra. And I'll show you the algebra as well. But if you don't understand it, if it doesn't make as much sense to you at this point, don't worry too much about it."}, {"video_title": "Solving a proportion with an unknown variable (example)   7th grade   Khan Academy.mp3", "Sentence": "You really don't understand what you're doing. And it really comes out of a little bit of algebra. And I'll show you the algebra as well. But if you don't understand it, if it doesn't make as much sense to you at this point, don't worry too much about it. So we have 8 over 36 is equal to 10 over n. When you cross multiply, you're saying that the numerator here times the denominator over here is going to be equal to, so 8 times n is going to be equal to the denominator over here. Let me do this in a different color. The denominator over here times the numerator over here."}, {"video_title": "Solving a proportion with an unknown variable (example)   7th grade   Khan Academy.mp3", "Sentence": "But if you don't understand it, if it doesn't make as much sense to you at this point, don't worry too much about it. So we have 8 over 36 is equal to 10 over n. When you cross multiply, you're saying that the numerator here times the denominator over here is going to be equal to, so 8 times n is going to be equal to the denominator over here. Let me do this in a different color. The denominator over here times the numerator over here. This is what it means to cross multiply. So this is going to be equal to 36 times 10. Or you could say, let me do this in a neutral color now."}, {"video_title": "Solving a proportion with an unknown variable (example)   7th grade   Khan Academy.mp3", "Sentence": "The denominator over here times the numerator over here. This is what it means to cross multiply. So this is going to be equal to 36 times 10. Or you could say, let me do this in a neutral color now. You could say that 8n is equal to 360. And so you're saying 8 times what is equal to 360? Or to figure out what that times what is, you divide 360 divided by 8."}, {"video_title": "Solving a proportion with an unknown variable (example)   7th grade   Khan Academy.mp3", "Sentence": "Or you could say, let me do this in a neutral color now. You could say that 8n is equal to 360. And so you're saying 8 times what is equal to 360? Or to figure out what that times what is, you divide 360 divided by 8. So we could divide, and this is a little bit of algebra here. We're dividing both sides of the equation by 8. And we're getting n is equal to 360 divided by 8."}, {"video_title": "Solving a proportion with an unknown variable (example)   7th grade   Khan Academy.mp3", "Sentence": "Or to figure out what that times what is, you divide 360 divided by 8. So we could divide, and this is a little bit of algebra here. We're dividing both sides of the equation by 8. And we're getting n is equal to 360 divided by 8. And you can do that without thinking in strict algebraic terms. You can say 8 times what is 360? Well, 8 times 360 over 8."}, {"video_title": "Solving a proportion with an unknown variable (example)   7th grade   Khan Academy.mp3", "Sentence": "And we're getting n is equal to 360 divided by 8. And you can do that without thinking in strict algebraic terms. You can say 8 times what is 360? Well, 8 times 360 over 8. If I write 8 times question mark is equal to 360, well, question mark could definitely be 360 over 8. If I multiply these out, this guy and that guy cancel out, and it's definitely 360. And that's why it's 360 over 8."}, {"video_title": "Solving a proportion with an unknown variable (example)   7th grade   Khan Academy.mp3", "Sentence": "Well, 8 times 360 over 8. If I write 8 times question mark is equal to 360, well, question mark could definitely be 360 over 8. If I multiply these out, this guy and that guy cancel out, and it's definitely 360. And that's why it's 360 over 8. But now we want to actually divide this to actually get our right answer or a simplified answer. 8 goes into 360. 8 goes into 36 four times."}, {"video_title": "Solving a proportion with an unknown variable (example)   7th grade   Khan Academy.mp3", "Sentence": "And that's why it's 360 over 8. But now we want to actually divide this to actually get our right answer or a simplified answer. 8 goes into 360. 8 goes into 36 four times. 4 times 8 is 32. You have a remainder of 4. Bring down the 0."}, {"video_title": "Solving a proportion with an unknown variable (example)   7th grade   Khan Academy.mp3", "Sentence": "8 goes into 36 four times. 4 times 8 is 32. You have a remainder of 4. Bring down the 0. 8 goes into 45 times. 5 times 8 is 40. And then you have no remainder."}, {"video_title": "Solving a proportion with an unknown variable (example)   7th grade   Khan Academy.mp3", "Sentence": "Bring down the 0. 8 goes into 45 times. 5 times 8 is 40. And then you have no remainder. And you're done. Once again, we got n is equal to 45. Now, the last thing I'm going to show you involves a little bit of algebra."}, {"video_title": "Solving a proportion with an unknown variable (example)   7th grade   Khan Academy.mp3", "Sentence": "And then you have no remainder. And you're done. Once again, we got n is equal to 45. Now, the last thing I'm going to show you involves a little bit of algebra. If any of the ways before this work, that's fine. And where this is sitting in the playlist, you're not expected to know the algebra. But I want to show you the algebra just because I wanted to show you that this cross multiplication isn't some magic, that using algebra, we will get this exact same thing."}, {"video_title": "Solving a proportion with an unknown variable (example)   7th grade   Khan Academy.mp3", "Sentence": "Now, the last thing I'm going to show you involves a little bit of algebra. If any of the ways before this work, that's fine. And where this is sitting in the playlist, you're not expected to know the algebra. But I want to show you the algebra just because I wanted to show you that this cross multiplication isn't some magic, that using algebra, we will get this exact same thing. But you could stop watching this if you find this part confusing. So let's rewrite our proportion. 8 over 36 is equal to 10 over n. And we want to solve for n. Well, the easiest way to solve for n is maybe multiply both."}, {"video_title": "Solving a proportion with an unknown variable (example)   7th grade   Khan Academy.mp3", "Sentence": "But I want to show you the algebra just because I wanted to show you that this cross multiplication isn't some magic, that using algebra, we will get this exact same thing. But you could stop watching this if you find this part confusing. So let's rewrite our proportion. 8 over 36 is equal to 10 over n. And we want to solve for n. Well, the easiest way to solve for n is maybe multiply both. This thing on the left is equal to this thing on the right. So we can multiply them both by the same thing. And the equality will still hold."}, {"video_title": "Solving a proportion with an unknown variable (example)   7th grade   Khan Academy.mp3", "Sentence": "8 over 36 is equal to 10 over n. And we want to solve for n. Well, the easiest way to solve for n is maybe multiply both. This thing on the left is equal to this thing on the right. So we can multiply them both by the same thing. And the equality will still hold. So we can multiply both of them by n. On the right-hand side, the n's cancel out. On the left-hand side, we have 836 times n is equal to 10. Now if we want to solve for n, we can literally multiply."}, {"video_title": "Solving a proportion with an unknown variable (example)   7th grade   Khan Academy.mp3", "Sentence": "And the equality will still hold. So we can multiply both of them by n. On the right-hand side, the n's cancel out. On the left-hand side, we have 836 times n is equal to 10. Now if we want to solve for n, we can literally multiply. If we want just an n here, we'd want to multiply this side times 36. Do that in a different color. We'd want to multiply this side times 36 times 8."}, {"video_title": "Solving a proportion with an unknown variable (example)   7th grade   Khan Academy.mp3", "Sentence": "Now if we want to solve for n, we can literally multiply. If we want just an n here, we'd want to multiply this side times 36. Do that in a different color. We'd want to multiply this side times 36 times 8. Because if you multiply these guys out, you get 1. And you just have an n. But since we're doing it to the left-hand side, we also have to do it to the right-hand side. So 10 times 36 over 8."}, {"video_title": "Solving a proportion with an unknown variable (example)   7th grade   Khan Academy.mp3", "Sentence": "We'd want to multiply this side times 36 times 8. Because if you multiply these guys out, you get 1. And you just have an n. But since we're doing it to the left-hand side, we also have to do it to the right-hand side. So 10 times 36 over 8. These guys cancel out. And we're left with n is equal to 10 times 36 is 360 over 8. And notice, we're getting the exact same value that we got with cross-multiplying."}, {"video_title": "Solving a proportion with an unknown variable (example)   7th grade   Khan Academy.mp3", "Sentence": "So 10 times 36 over 8. These guys cancel out. And we're left with n is equal to 10 times 36 is 360 over 8. And notice, we're getting the exact same value that we got with cross-multiplying. And with cross-multiplying, you're actually doing two steps. Actually, you're doing an extra step here. You're multiplying both sides by n so that you get your 8n."}, {"video_title": "Solving a proportion with an unknown variable (example)   7th grade   Khan Academy.mp3", "Sentence": "And notice, we're getting the exact same value that we got with cross-multiplying. And with cross-multiplying, you're actually doing two steps. Actually, you're doing an extra step here. You're multiplying both sides by n so that you get your 8n. And then you're multiplying both sides by 36 so that you get your 36 on both sides. And you get this value here. But at the end, when you simplify it, you'll get the exact same answer."}, {"video_title": "How to solve equations of the form ax = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Let's say we have the equation 7 times x is equal to 14. Now before even trying to solve this equation, what I want to do is think a little bit about what this actually means. 7x equals 14, this is the exact same thing as saying 7 times x, 7 times x is equal to 14. Now you might be able to do this in your head, you could literally go through the 7 times tables and say 7 times 1 is equal to 7, so that won't work, 7 times 2 is equal to 14, so 2 works here. So you would immediately be able to solve it, you would be able to just by trying different numbers out say hey, that's going to be a 2. But what we're going to do in this video is think about how to solve this systematically, because what we're going to find is as these equations get more and more complicated, you're not going to be able to just think about it and do it in your head. So it's really important that 1, you understand how to manipulate these equations, but even more important, understand what they actually represent."}, {"video_title": "How to solve equations of the form ax = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Now you might be able to do this in your head, you could literally go through the 7 times tables and say 7 times 1 is equal to 7, so that won't work, 7 times 2 is equal to 14, so 2 works here. So you would immediately be able to solve it, you would be able to just by trying different numbers out say hey, that's going to be a 2. But what we're going to do in this video is think about how to solve this systematically, because what we're going to find is as these equations get more and more complicated, you're not going to be able to just think about it and do it in your head. So it's really important that 1, you understand how to manipulate these equations, but even more important, understand what they actually represent. This literally just says 7 times x is equal to 14. In algebra, we don't write the times there, when you write two numbers next to each other, or a number next to a variable like this, it just means that you are multiplying, it's just a shorthand, a shorthand notation. And in general we don't use the multiplication sign, because it's confusing, because x is the most common variable used in algebra, and if I were to write 7 times x is equal to 14, and I write my times sign or my x a little bit strange, it might look like xx or times times, so in general when you're dealing with equations, especially when one of the variables is an x, you wouldn't use the traditional multiplication sign."}, {"video_title": "How to solve equations of the form ax = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So it's really important that 1, you understand how to manipulate these equations, but even more important, understand what they actually represent. This literally just says 7 times x is equal to 14. In algebra, we don't write the times there, when you write two numbers next to each other, or a number next to a variable like this, it just means that you are multiplying, it's just a shorthand, a shorthand notation. And in general we don't use the multiplication sign, because it's confusing, because x is the most common variable used in algebra, and if I were to write 7 times x is equal to 14, and I write my times sign or my x a little bit strange, it might look like xx or times times, so in general when you're dealing with equations, especially when one of the variables is an x, you wouldn't use the traditional multiplication sign. You might use something like this. You might use dot to represent multiplication, so you might have 7 times x is equal to 14, but this is still a little unusual. If you have something multiplying by a variable, you'll just write 7x."}, {"video_title": "How to solve equations of the form ax = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "And in general we don't use the multiplication sign, because it's confusing, because x is the most common variable used in algebra, and if I were to write 7 times x is equal to 14, and I write my times sign or my x a little bit strange, it might look like xx or times times, so in general when you're dealing with equations, especially when one of the variables is an x, you wouldn't use the traditional multiplication sign. You might use something like this. You might use dot to represent multiplication, so you might have 7 times x is equal to 14, but this is still a little unusual. If you have something multiplying by a variable, you'll just write 7x. That literally means 7 times x. Now, to understand how you can manipulate this equation to solve it, let's visualize this. So 7 times x, what is that?"}, {"video_title": "How to solve equations of the form ax = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "If you have something multiplying by a variable, you'll just write 7x. That literally means 7 times x. Now, to understand how you can manipulate this equation to solve it, let's visualize this. So 7 times x, what is that? That's the same thing. So I'm just going to rewrite this equation, but I'm going to rewrite it in visual form. So 7 times x."}, {"video_title": "How to solve equations of the form ax = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So 7 times x, what is that? That's the same thing. So I'm just going to rewrite this equation, but I'm going to rewrite it in visual form. So 7 times x. So that literally means x added to itself 7 times. That's the definition of multiplication. So it's literally x plus x plus x plus x plus x."}, {"video_title": "How to solve equations of the form ax = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So 7 times x. So that literally means x added to itself 7 times. That's the definition of multiplication. So it's literally x plus x plus x plus x plus x. Let's see, that's 5x's plus x plus x. So that right there is literally 7x's. This is 7x right there."}, {"video_title": "How to solve equations of the form ax = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So it's literally x plus x plus x plus x plus x. Let's see, that's 5x's plus x plus x. So that right there is literally 7x's. This is 7x right there. Let me rewrite it down. This right here is 7x. Now, this equation tells us that 7x is equal to 14."}, {"video_title": "How to solve equations of the form ax = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "This is 7x right there. Let me rewrite it down. This right here is 7x. Now, this equation tells us that 7x is equal to 14. So it's saying that this is equal to 14. Now let me just draw 14 objects here. So let's say I have 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14."}, {"video_title": "How to solve equations of the form ax = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Now, this equation tells us that 7x is equal to 14. So it's saying that this is equal to 14. Now let me just draw 14 objects here. So let's say I have 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14. So literally we're saying 7x is equal to 14 things. These are equivalent statements. Now, the reason why I drew it out this way is so that you really understand what we're going to do when we divide both sides by 7."}, {"video_title": "How to solve equations of the form ax = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So let's say I have 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14. So literally we're saying 7x is equal to 14 things. These are equivalent statements. Now, the reason why I drew it out this way is so that you really understand what we're going to do when we divide both sides by 7. So let me erase this right here. So the standard step whenever... Oh, I didn't want to do that. Let me do this."}, {"video_title": "How to solve equations of the form ax = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Now, the reason why I drew it out this way is so that you really understand what we're going to do when we divide both sides by 7. So let me erase this right here. So the standard step whenever... Oh, I didn't want to do that. Let me do this. Let me draw that last circle. So in general, whenever you simplify an equation down to a coefficient is just the number multiplying the variable. So some number multiplying the variable, or we could call that a coefficient times a variable, equal to something else."}, {"video_title": "How to solve equations of the form ax = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Let me do this. Let me draw that last circle. So in general, whenever you simplify an equation down to a coefficient is just the number multiplying the variable. So some number multiplying the variable, or we could call that a coefficient times a variable, equal to something else. What you want to do is just divide both sides by 7 in this case, or divide both sides by the coefficient. So if you divide both sides by 7, what do you get? 7 times something divided by 7 is just going to be that original something."}, {"video_title": "How to solve equations of the form ax = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So some number multiplying the variable, or we could call that a coefficient times a variable, equal to something else. What you want to do is just divide both sides by 7 in this case, or divide both sides by the coefficient. So if you divide both sides by 7, what do you get? 7 times something divided by 7 is just going to be that original something. 7's cancelled out, and 14 divided by 7 is 2. So your solution is going to be x is equal to 2. But just to make it very tangible in your head, what's going on here is when we're dividing both sides of the equation by 7, we're literally dividing both sides by 7."}, {"video_title": "How to solve equations of the form ax = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "7 times something divided by 7 is just going to be that original something. 7's cancelled out, and 14 divided by 7 is 2. So your solution is going to be x is equal to 2. But just to make it very tangible in your head, what's going on here is when we're dividing both sides of the equation by 7, we're literally dividing both sides by 7. This is an equation. It's saying that this is equal to that. Anything I do to the left-hand side, I have to do to the right."}, {"video_title": "How to solve equations of the form ax = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "But just to make it very tangible in your head, what's going on here is when we're dividing both sides of the equation by 7, we're literally dividing both sides by 7. This is an equation. It's saying that this is equal to that. Anything I do to the left-hand side, I have to do to the right. If they start off being equal, I can't just do an operation to one side and have it still be equal. They were the same thing. So if I divide the left-hand side by 7, so let me divide it into 7 groups."}, {"video_title": "How to solve equations of the form ax = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Anything I do to the left-hand side, I have to do to the right. If they start off being equal, I can't just do an operation to one side and have it still be equal. They were the same thing. So if I divide the left-hand side by 7, so let me divide it into 7 groups. So there are 7 x's here, so that's 1, 2, 3, 4, 5, 6, 7. So it's 1, 2, 3, 4, 5, 6, 7 groups. Now if I divide that into 7 groups, I'll also want to divide the right-hand side into 7 groups."}, {"video_title": "How to solve equations of the form ax = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So if I divide the left-hand side by 7, so let me divide it into 7 groups. So there are 7 x's here, so that's 1, 2, 3, 4, 5, 6, 7. So it's 1, 2, 3, 4, 5, 6, 7 groups. Now if I divide that into 7 groups, I'll also want to divide the right-hand side into 7 groups. 1, 2, 3, 4, 5, 6, 7. So if this whole thing is equal to this whole thing, then each of these little chunks that we broke into, these 7 chunks, are going to be equivalent. So this chunk, you could say, is equal to that chunk."}, {"video_title": "How to solve equations of the form ax = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Now if I divide that into 7 groups, I'll also want to divide the right-hand side into 7 groups. 1, 2, 3, 4, 5, 6, 7. So if this whole thing is equal to this whole thing, then each of these little chunks that we broke into, these 7 chunks, are going to be equivalent. So this chunk, you could say, is equal to that chunk. This chunk is equal to this chunk. They're all equivalent chunks. There are 7 chunks here, 7 chunks here."}, {"video_title": "How to solve equations of the form ax = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So this chunk, you could say, is equal to that chunk. This chunk is equal to this chunk. They're all equivalent chunks. There are 7 chunks here, 7 chunks here. So each x must be equal to 2 of these objects. So we get x is equal to, in this case, we had the objects drawn out, where there's 2 of them. x is equal to 2."}, {"video_title": "How to solve equations of the form ax = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "There are 7 chunks here, 7 chunks here. So each x must be equal to 2 of these objects. So we get x is equal to, in this case, we had the objects drawn out, where there's 2 of them. x is equal to 2. Now, let's just do a couple more examples here, just so it really gets in your mind that we're dealing with an equation, and any operation that you do to one side of the equation, you should do to the other. So let me scroll down a little bit. So let's say I have 3x is equal to 15."}, {"video_title": "How to solve equations of the form ax = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "x is equal to 2. Now, let's just do a couple more examples here, just so it really gets in your mind that we're dealing with an equation, and any operation that you do to one side of the equation, you should do to the other. So let me scroll down a little bit. So let's say I have 3x is equal to 15. Now, once again, you might be able to do this in your head. This is saying 3 times some number is equal to 15. You could go through your 3 times tables and figure it out."}, {"video_title": "How to solve equations of the form ax = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So let's say I have 3x is equal to 15. Now, once again, you might be able to do this in your head. This is saying 3 times some number is equal to 15. You could go through your 3 times tables and figure it out. But if you just wanted to do this systematically, and it is good to understand it systematically, you say, OK, this thing on the left is equal to this thing on the right. What do I have to do to this thing on the left to have just an x there? Well, to have just an x there, I want to divide it by 3."}, {"video_title": "How to solve equations of the form ax = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "You could go through your 3 times tables and figure it out. But if you just wanted to do this systematically, and it is good to understand it systematically, you say, OK, this thing on the left is equal to this thing on the right. What do I have to do to this thing on the left to have just an x there? Well, to have just an x there, I want to divide it by 3. And my whole motivation for doing that is to have 3 times something divided by 3. The 3's will cancel out, and I'm just going to be left with an x. Now, 3x was equal to 15."}, {"video_title": "How to solve equations of the form ax = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Well, to have just an x there, I want to divide it by 3. And my whole motivation for doing that is to have 3 times something divided by 3. The 3's will cancel out, and I'm just going to be left with an x. Now, 3x was equal to 15. If I'm dividing the left side by 3, in order for the equality to still hold, I also have to divide the right side by 3. And what does that give us? Well, the left-hand side, we're just going to be left with an x."}, {"video_title": "How to solve equations of the form ax = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Now, 3x was equal to 15. If I'm dividing the left side by 3, in order for the equality to still hold, I also have to divide the right side by 3. And what does that give us? Well, the left-hand side, we're just going to be left with an x. So it's just going to be an x. And then the right-hand side, what is 15 divided by 3? Well, it is just 5."}, {"video_title": "How to solve equations of the form ax = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Well, the left-hand side, we're just going to be left with an x. So it's just going to be an x. And then the right-hand side, what is 15 divided by 3? Well, it is just 5. Now, you could have also done this equation in a slightly different way, although they are really equivalent. If I start with 3x is equal to 15, you might say, hey, Sal, instead of dividing by 3, I could also get rid of this 3. I could just be left with an x."}, {"video_title": "How to solve equations of the form ax = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Well, it is just 5. Now, you could have also done this equation in a slightly different way, although they are really equivalent. If I start with 3x is equal to 15, you might say, hey, Sal, instead of dividing by 3, I could also get rid of this 3. I could just be left with an x. If I multiply both sides of this equation by 1 3rd. So if I multiply both sides of this equation by 1 3rd, that should also work. You say, look, 1 3rd of 3 is 1."}, {"video_title": "How to solve equations of the form ax = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "I could just be left with an x. If I multiply both sides of this equation by 1 3rd. So if I multiply both sides of this equation by 1 3rd, that should also work. You say, look, 1 3rd of 3 is 1. When you just multiply this part right here, 1 3rd times 3, that is just 1. 1x is equal to 15 times 1 3rd is equal to 5. And 1 times x is the same thing as just x."}, {"video_title": "How to solve equations of the form ax = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "You say, look, 1 3rd of 3 is 1. When you just multiply this part right here, 1 3rd times 3, that is just 1. 1x is equal to 15 times 1 3rd is equal to 5. And 1 times x is the same thing as just x. So this is the same thing as x is equal to 5. And these are actually equivalent ways of doing it. If you divide both sides by 3, that is equivalent to multiplying both sides of the equation by 1 3rd."}, {"video_title": "How to solve equations of the form ax = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "And 1 times x is the same thing as just x. So this is the same thing as x is equal to 5. And these are actually equivalent ways of doing it. If you divide both sides by 3, that is equivalent to multiplying both sides of the equation by 1 3rd. Now, let's do one more, and I'm going to make it a little bit more complicated. So let's say, and I'm going to change the variable a little bit. So let's say I have 2y plus 4y is equal to 18."}, {"video_title": "How to solve equations of the form ax = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "If you divide both sides by 3, that is equivalent to multiplying both sides of the equation by 1 3rd. Now, let's do one more, and I'm going to make it a little bit more complicated. So let's say, and I'm going to change the variable a little bit. So let's say I have 2y plus 4y is equal to 18. Now all of a sudden, it's a little harder to do in your head. We're saying 2 times something plus 4 times that same something is going to be equal to 18. So here it's harder to think about what number that is."}, {"video_title": "How to solve equations of the form ax = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So let's say I have 2y plus 4y is equal to 18. Now all of a sudden, it's a little harder to do in your head. We're saying 2 times something plus 4 times that same something is going to be equal to 18. So here it's harder to think about what number that is. You could try it. You might say, well, if y was 1, it would be 2 times 1 plus 4 times 1. Well, that doesn't work."}, {"video_title": "How to solve equations of the form ax = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So here it's harder to think about what number that is. You could try it. You might say, well, if y was 1, it would be 2 times 1 plus 4 times 1. Well, that doesn't work. But let's think about how to do it systematically. You could keep guessing, and you might eventually get the answer. But how do you do this systematically?"}, {"video_title": "How to solve equations of the form ax = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Well, that doesn't work. But let's think about how to do it systematically. You could keep guessing, and you might eventually get the answer. But how do you do this systematically? Let's visualize it. So if I have 2 y's, what does that mean? It literally means I have 2 y's added to each other."}, {"video_title": "How to solve equations of the form ax = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "But how do you do this systematically? Let's visualize it. So if I have 2 y's, what does that mean? It literally means I have 2 y's added to each other. So it's literally y plus y. And then to that, I'm adding 4 y's, which are literally 4 y's added to each other. So it's y plus y plus y plus y."}, {"video_title": "How to solve equations of the form ax = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "It literally means I have 2 y's added to each other. So it's literally y plus y. And then to that, I'm adding 4 y's, which are literally 4 y's added to each other. So it's y plus y plus y plus y. And that has got to be equal to 18. So that is equal to 18. Now, how many y's do I have here on the left-hand side?"}, {"video_title": "How to solve equations of the form ax = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So it's y plus y plus y plus y. And that has got to be equal to 18. So that is equal to 18. Now, how many y's do I have here on the left-hand side? How many y's do I have? I have 1, 2, 3, 4, 5, 6 y's. So you could simplify this as 6 y's equal to 18."}, {"video_title": "How to solve equations of the form ax = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Now, how many y's do I have here on the left-hand side? How many y's do I have? I have 1, 2, 3, 4, 5, 6 y's. So you could simplify this as 6 y's equal to 18. And if you think about it, it makes complete sense. So this is the whole thing right here, the 2 y plus the 4 y is 6 y. So 2 y plus 4 y is 6 y, which makes sense."}, {"video_title": "How to solve equations of the form ax = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So you could simplify this as 6 y's equal to 18. And if you think about it, it makes complete sense. So this is the whole thing right here, the 2 y plus the 4 y is 6 y. So 2 y plus 4 y is 6 y, which makes sense. If I have 2 apples plus 4 apples, I'm going to have 6 apples. If I have 2 y's plus 4 y's, I'm going to have 6 y's. Now, that's going to be equal to 18."}, {"video_title": "How to solve equations of the form ax = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So 2 y plus 4 y is 6 y, which makes sense. If I have 2 apples plus 4 apples, I'm going to have 6 apples. If I have 2 y's plus 4 y's, I'm going to have 6 y's. Now, that's going to be equal to 18. That is going to be equal to 18. And now, hopefully, we understand how to do this. If I have 6 times something is equal to 18, if I divide both sides of this equation by 6, I'll solve for the something."}, {"video_title": "How to solve equations of the form ax = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Now, that's going to be equal to 18. That is going to be equal to 18. And now, hopefully, we understand how to do this. If I have 6 times something is equal to 18, if I divide both sides of this equation by 6, I'll solve for the something. So divide the left-hand side by 6. And divide the right-hand side by 6. And we are left with y is equal to 3."}, {"video_title": "How to solve equations of the form ax = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "If I have 6 times something is equal to 18, if I divide both sides of this equation by 6, I'll solve for the something. So divide the left-hand side by 6. And divide the right-hand side by 6. And we are left with y is equal to 3. And you could try it out. That's what's cool about an equation. You can always check to see if you got the right answer."}, {"video_title": "How to solve equations of the form ax = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "And we are left with y is equal to 3. And you could try it out. That's what's cool about an equation. You can always check to see if you got the right answer. Let's see if that works. 2 times 3 plus 4 times 3 is equal to what? 2 times 3, this right here, is 6."}, {"video_title": "How to solve equations of the form ax = b   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "You can always check to see if you got the right answer. Let's see if that works. 2 times 3 plus 4 times 3 is equal to what? 2 times 3, this right here, is 6. And then 4 times 3 is 12. 6 plus 12 is indeed equal to 18. So it works out."}, {"video_title": "Standard form for linear equations   Algebra I   Khan Academy.mp3", "Sentence": "m is the coefficient on this mx term right over here, and m would represent the slope. And then from b, you're able to figure out the y-intercept. The y, you're able to figure out the y-intercept from this. It's literally the graph that represents the xy pairs that satisfy this equation. It would intersect the y-axis at the point x equals zero, y is equal to b, and its slope would be m. We've already seen that multiple times. We've also seen that you can also express things in point-slope form. So let me actually make it clear."}, {"video_title": "Standard form for linear equations   Algebra I   Khan Academy.mp3", "Sentence": "It's literally the graph that represents the xy pairs that satisfy this equation. It would intersect the y-axis at the point x equals zero, y is equal to b, and its slope would be m. We've already seen that multiple times. We've also seen that you can also express things in point-slope form. So let me actually make it clear. This is slope-intercept. Slope-intercept. And these are just different ways of writing the same equations."}, {"video_title": "Standard form for linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So let me actually make it clear. This is slope-intercept. Slope-intercept. And these are just different ways of writing the same equations. You can algebraically manipulate from one to the other. Another way is point-slope. Point-slope form."}, {"video_title": "Standard form for linear equations   Algebra I   Khan Academy.mp3", "Sentence": "And these are just different ways of writing the same equations. You can algebraically manipulate from one to the other. Another way is point-slope. Point-slope form. And in point-slope form, if you know that some, if you know that there's an equation where the line that represents the solutions of that equation has a slope m, slope is equal to m, and if you know that x equals a, y equals b satisfies that equation, then in point-slope form, you can express the equation as y minus b is equal to m over, or m times x minus a. This is point-slope form, and we do videos on that. But what I really want to get into in this video is another form, and it's a form that you might have already seen."}, {"video_title": "Standard form for linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Point-slope form. And in point-slope form, if you know that some, if you know that there's an equation where the line that represents the solutions of that equation has a slope m, slope is equal to m, and if you know that x equals a, y equals b satisfies that equation, then in point-slope form, you can express the equation as y minus b is equal to m over, or m times x minus a. This is point-slope form, and we do videos on that. But what I really want to get into in this video is another form, and it's a form that you might have already seen. And that is standard form. Standard, standard form. And standard form takes the shape of ax plus by is equal to c, where a, b, and c are integers."}, {"video_title": "Standard form for linear equations   Algebra I   Khan Academy.mp3", "Sentence": "But what I really want to get into in this video is another form, and it's a form that you might have already seen. And that is standard form. Standard, standard form. And standard form takes the shape of ax plus by is equal to c, where a, b, and c are integers. And what I want to do in this video, like we've done in the ones on point-slope and slope-intercept, is gain an appreciation for what is standard form good at, and what is standard form less good at. So let's give a tangible example here. So let's say I have the linear equation."}, {"video_title": "Standard form for linear equations   Algebra I   Khan Academy.mp3", "Sentence": "And standard form takes the shape of ax plus by is equal to c, where a, b, and c are integers. And what I want to do in this video, like we've done in the ones on point-slope and slope-intercept, is gain an appreciation for what is standard form good at, and what is standard form less good at. So let's give a tangible example here. So let's say I have the linear equation. It's in standard form, nine x plus 16 y is equal to 72. And we wanted to graph this. So the thing that standard form is really good for is figuring out not just the y-intercept, y-intercept is pretty good if you're using slope-intercept form, but we can figure out the y-intercept pretty clearly from standard form, and the x-intercept."}, {"video_title": "Standard form for linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So let's say I have the linear equation. It's in standard form, nine x plus 16 y is equal to 72. And we wanted to graph this. So the thing that standard form is really good for is figuring out not just the y-intercept, y-intercept is pretty good if you're using slope-intercept form, but we can figure out the y-intercept pretty clearly from standard form, and the x-intercept. The x-intercept isn't so easy to figure out from these other forms right over here. So how do we do that? Well, to figure out the x and y-intercepts, let's just set up a little table here, x comma y."}, {"video_title": "Standard form for linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So the thing that standard form is really good for is figuring out not just the y-intercept, y-intercept is pretty good if you're using slope-intercept form, but we can figure out the y-intercept pretty clearly from standard form, and the x-intercept. The x-intercept isn't so easy to figure out from these other forms right over here. So how do we do that? Well, to figure out the x and y-intercepts, let's just set up a little table here, x comma y. And so the x-intercept is going to happen when y is equal to zero, and the y-intercept is going to happen when x is equal to zero. So when y is zero, what is x? So when y is zero, 16 times zero is zero, that term disappears, and you're left with nine x is equal to 72."}, {"video_title": "Standard form for linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Well, to figure out the x and y-intercepts, let's just set up a little table here, x comma y. And so the x-intercept is going to happen when y is equal to zero, and the y-intercept is going to happen when x is equal to zero. So when y is zero, what is x? So when y is zero, 16 times zero is zero, that term disappears, and you're left with nine x is equal to 72. So if nine times x is 72, 72 divided by nine is eight. So x would be equal to eight. So once again, that was pretty easy to figure out."}, {"video_title": "Standard form for linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So when y is zero, 16 times zero is zero, that term disappears, and you're left with nine x is equal to 72. So if nine times x is 72, 72 divided by nine is eight. So x would be equal to eight. So once again, that was pretty easy to figure out. This term goes away, and you just have to say, hey, nine times x is 72, x would be eight. When y is equal to zero, x is eight. So the point, let's see, y is zero, x is one, two, three, four, five, six, seven, eight, that's this point, that right over here, this point right over here, is the x-intercept."}, {"video_title": "Standard form for linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So once again, that was pretty easy to figure out. This term goes away, and you just have to say, hey, nine times x is 72, x would be eight. When y is equal to zero, x is eight. So the point, let's see, y is zero, x is one, two, three, four, five, six, seven, eight, that's this point, that right over here, this point right over here, is the x-intercept. When we talk about x-intercept, we're referring to the point where the line actually intersects the x-axis. Now what about the y-intercept? Well, we said x equals zero, this disappears, and we're left with 16y is equal to 72."}, {"video_title": "Standard form for linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So the point, let's see, y is zero, x is one, two, three, four, five, six, seven, eight, that's this point, that right over here, this point right over here, is the x-intercept. When we talk about x-intercept, we're referring to the point where the line actually intersects the x-axis. Now what about the y-intercept? Well, we said x equals zero, this disappears, and we're left with 16y is equal to 72. And so we could solve, we could solve that. So we could say, all right, 16y is equal to 72, and then divide both sides by 16, we get y is equal to 72 over 16, and let's see, what is that equal to? That is equal to, let's see, they're both divisible by eight."}, {"video_title": "Standard form for linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Well, we said x equals zero, this disappears, and we're left with 16y is equal to 72. And so we could solve, we could solve that. So we could say, all right, 16y is equal to 72, and then divide both sides by 16, we get y is equal to 72 over 16, and let's see, what is that equal to? That is equal to, let's see, they're both divisible by eight. So that's nine over two, or we could say it's 4.5. So when x is zero, y is 4.5. And so we could plot that point as well."}, {"video_title": "Standard form for linear equations   Algebra I   Khan Academy.mp3", "Sentence": "That is equal to, let's see, they're both divisible by eight. So that's nine over two, or we could say it's 4.5. So when x is zero, y is 4.5. And so we could plot that point as well. X is zero, y is one, two, three, four, point five. And just with these two points, two points are enough to graph a line, we can now graph it. So let's do that."}, {"video_title": "Standard form for linear equations   Algebra I   Khan Academy.mp3", "Sentence": "And so we could plot that point as well. X is zero, y is one, two, three, four, point five. And just with these two points, two points are enough to graph a line, we can now graph it. So let's do that. So let me, whoops, I thought I was using the tool that would draw a straight line. Let me see if I can, so the line will look something like that. There you have it, I've just graphed, I've just graphed, this is the line that represents all the x and y pairs that satisfy the equation, 9x plus 16y is equal to 72."}, {"video_title": "Standard form for linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So let's do that. So let me, whoops, I thought I was using the tool that would draw a straight line. Let me see if I can, so the line will look something like that. There you have it, I've just graphed, I've just graphed, this is the line that represents all the x and y pairs that satisfy the equation, 9x plus 16y is equal to 72. Now, I mentioned standard form's good at certain things, and the good thing that standard form is, where it's maybe somewhat unique relative to the other forms we looked at, is it's very easy to figure out the x-intercept. It was very easy to figure out the x-intercept from standard form. And it wasn't too hard to figure out the y-intercept either."}, {"video_title": "Standard form for linear equations   Algebra I   Khan Academy.mp3", "Sentence": "There you have it, I've just graphed, I've just graphed, this is the line that represents all the x and y pairs that satisfy the equation, 9x plus 16y is equal to 72. Now, I mentioned standard form's good at certain things, and the good thing that standard form is, where it's maybe somewhat unique relative to the other forms we looked at, is it's very easy to figure out the x-intercept. It was very easy to figure out the x-intercept from standard form. And it wasn't too hard to figure out the y-intercept either. If we looked at slope-intercept form, the y-intercept just kind of jumps out at you. At point-slope form, neither the x nor the y-intercept kind of jump out at you. The place where slope-intercept or point-slope form are frankly better is that it's pretty easy to pick out the slope here, while in standard form, you would have to do a little bit of work."}, {"video_title": "Standard form for linear equations   Algebra I   Khan Academy.mp3", "Sentence": "And it wasn't too hard to figure out the y-intercept either. If we looked at slope-intercept form, the y-intercept just kind of jumps out at you. At point-slope form, neither the x nor the y-intercept kind of jump out at you. The place where slope-intercept or point-slope form are frankly better is that it's pretty easy to pick out the slope here, while in standard form, you would have to do a little bit of work. You could use these two points, you could use the x and y-intercept as two points and figure out the slope from there, so you could literally say, okay, if I'm going from this point to this point, my change in x, to go from eight to zero is negative eight, and to go from zero to 4.5, or the little delta there unnecessarily, let me... So when you go from eight to zero, your change in x is equal to negative eight, and to go from zero to 4.5, your change in y is going to be 4.5. So your slope, once you've figured this out, you could say, okay, this is going to be change in y, 4.5, over change in x, over negative eight."}, {"video_title": "Standard form for linear equations   Algebra I   Khan Academy.mp3", "Sentence": "The place where slope-intercept or point-slope form are frankly better is that it's pretty easy to pick out the slope here, while in standard form, you would have to do a little bit of work. You could use these two points, you could use the x and y-intercept as two points and figure out the slope from there, so you could literally say, okay, if I'm going from this point to this point, my change in x, to go from eight to zero is negative eight, and to go from zero to 4.5, or the little delta there unnecessarily, let me... So when you go from eight to zero, your change in x is equal to negative eight, and to go from zero to 4.5, your change in y is going to be 4.5. So your slope, once you've figured this out, you could say, okay, this is going to be change in y, 4.5, over change in x, over negative eight. And since I, at least, I don't like a decimal up here, let's multiply the numerator and the denominator by two, you get negative nine over 16. Now once again, we have to do a little bit of work here. We either use these two points, it didn't just jump immediately out of this, although you might see a little bit of a pattern of what's going on here, but you still have to think about, is it negative, is it positive?"}, {"video_title": "Standard form for linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So your slope, once you've figured this out, you could say, okay, this is going to be change in y, 4.5, over change in x, over negative eight. And since I, at least, I don't like a decimal up here, let's multiply the numerator and the denominator by two, you get negative nine over 16. Now once again, we have to do a little bit of work here. We either use these two points, it didn't just jump immediately out of this, although you might see a little bit of a pattern of what's going on here, but you still have to think about, is it negative, is it positive? You have to do a little bit of algebraic manipulation. Or what I typically do if I'm looking for the slope, I actually might put this into one of the other forms, especially slope-intercept form. But standard form by itself, great for figuring out both the x and y intercepts, and it's frankly not that hard to convert it to slope-intercept form."}, {"video_title": "Standard form for linear equations   Algebra I   Khan Academy.mp3", "Sentence": "We either use these two points, it didn't just jump immediately out of this, although you might see a little bit of a pattern of what's going on here, but you still have to think about, is it negative, is it positive? You have to do a little bit of algebraic manipulation. Or what I typically do if I'm looking for the slope, I actually might put this into one of the other forms, especially slope-intercept form. But standard form by itself, great for figuring out both the x and y intercepts, and it's frankly not that hard to convert it to slope-intercept form. Let's do that. Just to make it clear. So if you start with nine x, let me do that in yellow."}, {"video_title": "Standard form for linear equations   Algebra I   Khan Academy.mp3", "Sentence": "But standard form by itself, great for figuring out both the x and y intercepts, and it's frankly not that hard to convert it to slope-intercept form. Let's do that. Just to make it clear. So if you start with nine x, let me do that in yellow. If we start with nine x plus 16 y is equal to 72, and we want to put it in slope-intercept form, we can subtract nine x from both sides, you get 16 y is equal to negative nine x plus 72, and then divide both sides by 16. So divide everything by 16, and you'll be left with y is equal to negative nine 16 x, that's the slope, you see it right there, plus 72 over 16, we already figured out that's nine halves, or 4.5. So I could write, well, I'll just write that as 4.5."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "And they tell us to complete the table so each row represents a solution of the following equation. And they give us the equation, and then they want us to figure out what does y equal when x is equal to negative five, and what does x equal when y is equal to eight? And to figure this out, I've actually copied and pasted this part of the problem onto my scratch pad, so let me get that out. And so this is the exact same problem. And there's a couple of ways that we could try to tackle it. One way is you could try to simplify this more, get all your x's on one side and all your y's on the other side. Or we could just literally substitute when x equals negative five, what must y equal?"}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "And so this is the exact same problem. And there's a couple of ways that we could try to tackle it. One way is you could try to simplify this more, get all your x's on one side and all your y's on the other side. Or we could just literally substitute when x equals negative five, what must y equal? Actually, let me do it the second way first. So if we take this equation and we substitute x with negative five, what do we get? We get negative three times, well, we're gonna say x is negative five, times negative five plus seven y is equal to five times, x is once again, it's gonna be negative five, x is negative five, five times negative five plus two y."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "Or we could just literally substitute when x equals negative five, what must y equal? Actually, let me do it the second way first. So if we take this equation and we substitute x with negative five, what do we get? We get negative three times, well, we're gonna say x is negative five, times negative five plus seven y is equal to five times, x is once again, it's gonna be negative five, x is negative five, five times negative five plus two y. See, negative three times negative five is positive 15, plus seven y is equal to negative 25 plus two y. And now to solve for y, let's see, I could subtract two y from both sides so that I get rid of the two y here on the right. So let me subtract two y, subtract two y from both sides."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "We get negative three times, well, we're gonna say x is negative five, times negative five plus seven y is equal to five times, x is once again, it's gonna be negative five, x is negative five, five times negative five plus two y. See, negative three times negative five is positive 15, plus seven y is equal to negative 25 plus two y. And now to solve for y, let's see, I could subtract two y from both sides so that I get rid of the two y here on the right. So let me subtract two y, subtract two y from both sides. And then if I want all my constants on the right-hand side, I can subtract 15 from both sides. So let me subtract 15 from both sides. And I am going to be left with 15 minus 15, that's zero."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "So let me subtract two y, subtract two y from both sides. And then if I want all my constants on the right-hand side, I can subtract 15 from both sides. So let me subtract 15 from both sides. And I am going to be left with 15 minus 15, that's zero. That's the whole point of subtracting 15 from both sides, so I get rid of this 15 here. Seven y minus two y, seven of something minus two of that same something is gonna be five of that something. It's gonna be equal to five y, is equal to negative 25 minus 15."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "And I am going to be left with 15 minus 15, that's zero. That's the whole point of subtracting 15 from both sides, so I get rid of this 15 here. Seven y minus two y, seven of something minus two of that same something is gonna be five of that something. It's gonna be equal to five y, is equal to negative 25 minus 15. Well, that's gonna be negative 40. And then two y minus two y, well, that's just gonna be zero. That was the whole point of subtracting two y from both sides."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "It's gonna be equal to five y, is equal to negative 25 minus 15. Well, that's gonna be negative 40. And then two y minus two y, well, that's just gonna be zero. That was the whole point of subtracting two y from both sides. So you have five times y is equal to negative 40, or if we divide both sides by five, we divide both sides by five, we would get y is equal to negative eight. So when x is equal to negative five, y is equal to negative eight. Y is equal to negative eight."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "That was the whole point of subtracting two y from both sides. So you have five times y is equal to negative 40, or if we divide both sides by five, we divide both sides by five, we would get y is equal to negative eight. So when x is equal to negative five, y is equal to negative eight. Y is equal to negative eight. And actually, we can fill that in. So this y is going to be equal to negative eight. And now we gotta figure this out."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "Y is equal to negative eight. And actually, we can fill that in. So this y is going to be equal to negative eight. And now we gotta figure this out. What does x equal when y is positive eight? Well, we can go back to our scratch pad here. And now let's take the same equation, but let's make y equal to positive eight."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "And now we gotta figure this out. What does x equal when y is positive eight? Well, we can go back to our scratch pad here. And now let's take the same equation, but let's make y equal to positive eight. So you have negative three x plus seven. Now y is going to be eight. Y is eight."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "And now let's take the same equation, but let's make y equal to positive eight. So you have negative three x plus seven. Now y is going to be eight. Y is eight. Seven times eight is equal to five times x plus two times, once again, y is eight. Two times eight. So we get negative three x plus 56, that's 56, is equal to five x plus 16."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "Y is eight. Seven times eight is equal to five times x plus two times, once again, y is eight. Two times eight. So we get negative three x plus 56, that's 56, is equal to five x plus 16. Now if we want to get all of our constants on one side and of all of our x terms on the other side, well, what could we do? Let's see, we could add three x to both sides. That would get rid of all of the x's on this side and put them all on this side."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "So we get negative three x plus 56, that's 56, is equal to five x plus 16. Now if we want to get all of our constants on one side and of all of our x terms on the other side, well, what could we do? Let's see, we could add three x to both sides. That would get rid of all of the x's on this side and put them all on this side. So we're gonna add three x to both sides. And let's see, if we want to get all the constants on the left-hand side, we'd want to get rid of the 16, so we could subtract 16 from the right-hand side. If we do it from the right, we're gonna have to do it from the left as well."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "That would get rid of all of the x's on this side and put them all on this side. So we're gonna add three x to both sides. And let's see, if we want to get all the constants on the left-hand side, we'd want to get rid of the 16, so we could subtract 16 from the right-hand side. If we do it from the right, we're gonna have to do it from the left as well. And we're going to be left with, these cancel out, 56 minus 16 is positive 40. And then, let's see, 16 minus 16 is zero. Five x plus three x is equal to eight x."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "If we do it from the right, we're gonna have to do it from the left as well. And we're going to be left with, these cancel out, 56 minus 16 is positive 40. And then, let's see, 16 minus 16 is zero. Five x plus three x is equal to eight x. We get eight x is equal to 40. We could divide both sides by eight. And we get five is equal to x."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "Five x plus three x is equal to eight x. We get eight x is equal to 40. We could divide both sides by eight. And we get five is equal to x. So this right over here is going to be equal to five. So let's go back. Let's go back now."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "And we get five is equal to x. So this right over here is going to be equal to five. So let's go back. Let's go back now. So when y is positive eight, x is positive five. Now they ask us, use your two solutions to graph the equation. So let's see if we can do, whoops, let me use my mouse now."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "Let's go back now. So when y is positive eight, x is positive five. Now they ask us, use your two solutions to graph the equation. So let's see if we can do, whoops, let me use my mouse now. So to graph the equation. So when x is negative five, y is negative eight. So the point negative five comma negative eight."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "So let's see if we can do, whoops, let me use my mouse now. So to graph the equation. So when x is negative five, y is negative eight. So the point negative five comma negative eight. So that's right over there. Actually, let me move my browser up so you can see that. Negative five, when x is negative five, y is negative eight."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "So the point negative five comma negative eight. So that's right over there. Actually, let me move my browser up so you can see that. Negative five, when x is negative five, y is negative eight. And when x is positive five, we see that up here, when x is positive five, y is positive eight. When x is positive five, y is positive eight. And we're done."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "Negative five, when x is negative five, y is negative eight. And when x is positive five, we see that up here, when x is positive five, y is positive eight. When x is positive five, y is positive eight. And we're done. We can check our answer if we like. We got it right. Now I said there was two ways to tackle it."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "And we're done. We can check our answer if we like. We got it right. Now I said there was two ways to tackle it. I kind of just did it, I guess you could say the naive way. I just substituted negative five directly into this and solved for y. And then I substituted y equals positive eight directly into this and then solved for x."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "Now I said there was two ways to tackle it. I kind of just did it, I guess you could say the naive way. I just substituted negative five directly into this and solved for y. And then I substituted y equals positive eight directly into this and then solved for x. Another way that I could have done it that actually probably would have been, or for sure would have been the easier way to do it, is ahead of time to try to simplify this expression. So what I could have done right from the get-go is said, hey, let's put all my x's on one side and all my y's on the other side. So this is negative three x plus seven y is equal to five x plus two y."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "And then I substituted y equals positive eight directly into this and then solved for x. Another way that I could have done it that actually probably would have been, or for sure would have been the easier way to do it, is ahead of time to try to simplify this expression. So what I could have done right from the get-go is said, hey, let's put all my x's on one side and all my y's on the other side. So this is negative three x plus seven y is equal to five x plus two y. And let's say I want to get all my y's on the left and all my x's on the right. So I don't want this negative three x on the left, so I'd want to add three x. Adding three x would cancel this out, but I can't just do it on the left-hand side, I'd have to do it on the right-hand side as well."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "So this is negative three x plus seven y is equal to five x plus two y. And let's say I want to get all my y's on the left and all my x's on the right. So I don't want this negative three x on the left, so I'd want to add three x. Adding three x would cancel this out, but I can't just do it on the left-hand side, I'd have to do it on the right-hand side as well. And then if I want to get rid of this two y on the right, I could subtract two y from the right, but of course I'd also want to do it from the left. And then what am I left with? So negative three x plus three x is zero, seven y minus two y is five y."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "Adding three x would cancel this out, but I can't just do it on the left-hand side, I'd have to do it on the right-hand side as well. And then if I want to get rid of this two y on the right, I could subtract two y from the right, but of course I'd also want to do it from the left. And then what am I left with? So negative three x plus three x is zero, seven y minus two y is five y. And then I have five x plus three x is eight x. Two y minus two y is zero. And then if I wanted to, I could solve for y, I could divide both sides by five and I'd get y is equal to 8 5ths x."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "So negative three x plus three x is zero, seven y minus two y is five y. And then I have five x plus three x is eight x. Two y minus two y is zero. And then if I wanted to, I could solve for y, I could divide both sides by five and I'd get y is equal to 8 5ths x. So this right over here represents the same exact equation as this over here, it's just written in a different way. All of the xy pairs that satisfy this would satisfy this and vice versa. And this is much easier, because if x is now negative five, if x is negative five, y would be 8 5ths times negative five."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "And then if I wanted to, I could solve for y, I could divide both sides by five and I'd get y is equal to 8 5ths x. So this right over here represents the same exact equation as this over here, it's just written in a different way. All of the xy pairs that satisfy this would satisfy this and vice versa. And this is much easier, because if x is now negative five, if x is negative five, y would be 8 5ths times negative five. Well that's going to be negative eight. And when y is equal to eight, well you actually could even do this up here. You could say five times eight is equal to eight x."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "And this is much easier, because if x is now negative five, if x is negative five, y would be 8 5ths times negative five. Well that's going to be negative eight. And when y is equal to eight, well you actually could even do this up here. You could say five times eight is equal to eight x. And then you could see, well five times eight is the same thing as eight times five. So x would be equal to five. So I think this would actually have been a simpler way to do it."}, {"video_title": "Simplifying square roots   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "So 117 doesn't jump out at me as some type of a perfect square. So let's actually take its prime factorization and see if any of those prime factors show up more than once. So it's clearly an odd number. It's clearly not divisible by 2. To test whether it's divisible by 3, we can add up all of the digits. And we explained why this works in another place on Khan Academy. But if you add up all the digits, you get a 9."}, {"video_title": "Simplifying square roots   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "It's clearly not divisible by 2. To test whether it's divisible by 3, we can add up all of the digits. And we explained why this works in another place on Khan Academy. But if you add up all the digits, you get a 9. And 9 is divisible by 3. So 117 is going to be divisible by 3. Now let's do a little aside here and figure out what 117 divided by 3 actually is."}, {"video_title": "Simplifying square roots   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "But if you add up all the digits, you get a 9. And 9 is divisible by 3. So 117 is going to be divisible by 3. Now let's do a little aside here and figure out what 117 divided by 3 actually is. So 3 doesn't go into 1. It does go into 11 3 times. 3 times 3 is 9."}, {"video_title": "Simplifying square roots   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "Now let's do a little aside here and figure out what 117 divided by 3 actually is. So 3 doesn't go into 1. It does go into 11 3 times. 3 times 3 is 9. Subtract. You got a remainder of 2. Bring down a 7."}, {"video_title": "Simplifying square roots   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "3 times 3 is 9. Subtract. You got a remainder of 2. Bring down a 7. 3 goes into 27 9 times. 9 times 3 is 27. Subtract, and you're done."}, {"video_title": "Simplifying square roots   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "Bring down a 7. 3 goes into 27 9 times. 9 times 3 is 27. Subtract, and you're done. It goes in perfectly. And so we can factor 117 as 3 times 39. Now 39 we can factor as."}, {"video_title": "Simplifying square roots   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "Subtract, and you're done. It goes in perfectly. And so we can factor 117 as 3 times 39. Now 39 we can factor as. That jumps out more at us. That's divisible by 3. That's equivalent to 3 times 13."}, {"video_title": "Simplifying square roots   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "Now 39 we can factor as. That jumps out more at us. That's divisible by 3. That's equivalent to 3 times 13. And then all of these are now prime numbers. So we could say that this thing is the same as 5 times the square root of 3 times 3 times 13. And this is going to be the same thing as, and we know this from our exponent properties, as 5 times the square root of 3 times 3 times the square root of 13."}, {"video_title": "Simplifying square roots   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "That's equivalent to 3 times 13. And then all of these are now prime numbers. So we could say that this thing is the same as 5 times the square root of 3 times 3 times 13. And this is going to be the same thing as, and we know this from our exponent properties, as 5 times the square root of 3 times 3 times the square root of 13. Now what's the square root of 3 times 3? Well, that's the square root of 9. That's the square root of 3 squared."}, {"video_title": "Simplifying square roots   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "And this is going to be the same thing as, and we know this from our exponent properties, as 5 times the square root of 3 times 3 times the square root of 13. Now what's the square root of 3 times 3? Well, that's the square root of 9. That's the square root of 3 squared. Any of those, well, that's just going to give you 3. So this is just going to simplify to 3. So this whole thing is 5 times 3 times the square root of 13."}, {"video_title": "Simplifying square roots   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "That's the square root of 3 squared. Any of those, well, that's just going to give you 3. So this is just going to simplify to 3. So this whole thing is 5 times 3 times the square root of 13. So this part right over here would give us 15 times the square root of 13. Let's do one more example here. So let's try to simplify 3 times the square root of 26."}, {"video_title": "Simplifying square roots   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "So this whole thing is 5 times 3 times the square root of 13. So this part right over here would give us 15 times the square root of 13. Let's do one more example here. So let's try to simplify 3 times the square root of 26. Actually, I'm going to put 26 in yellow like I did in the previous problem. 26. Well, 26 is clearly an even number."}, {"video_title": "Simplifying square roots   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "So let's try to simplify 3 times the square root of 26. Actually, I'm going to put 26 in yellow like I did in the previous problem. 26. Well, 26 is clearly an even number. So it's going to be divisible by 2. We can rewrite it as 2 times 13. And then we're done."}, {"video_title": "Simplifying square roots   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "Well, 26 is clearly an even number. So it's going to be divisible by 2. We can rewrite it as 2 times 13. And then we're done. 13 is a prime number. We can't factor this anymore. And so 26 doesn't have any perfect squares in it."}, {"video_title": "Simplifying square roots   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "And then we're done. 13 is a prime number. We can't factor this anymore. And so 26 doesn't have any perfect squares in it. It's not like we can factor it out as a factor of some other numbers and some perfect squares like we did here. 117 is 13 times 9. It's the product of a perfect square and 13."}, {"video_title": "Simplifying square roots   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "And so 26 doesn't have any perfect squares in it. It's not like we can factor it out as a factor of some other numbers and some perfect squares like we did here. 117 is 13 times 9. It's the product of a perfect square and 13. 26 isn't. So we've simplified this about as much as we can. We would just leave this as 3 times the square root of 26."}, {"video_title": "Solving the troll riddle visually   Algebra II   Khan Academy.mp3", "Sentence": "And if we don't solve his riddle, he's gonna push us into the water. So we are under pressure. And at least we made some headway in the last video. We were able to represent his clues mathematically as a system of equations. What I wanna do in this video is think about whether we can solve for the system of equations. And you will see that there are many ways of solving a system of equations. But this time I wanna do it visually, because at least in my mind, it helps really kind of get the intuition of what these things are saying."}, {"video_title": "Solving the troll riddle visually   Algebra II   Khan Academy.mp3", "Sentence": "We were able to represent his clues mathematically as a system of equations. What I wanna do in this video is think about whether we can solve for the system of equations. And you will see that there are many ways of solving a system of equations. But this time I wanna do it visually, because at least in my mind, it helps really kind of get the intuition of what these things are saying. So let's draw some axes over here. Let's draw an F axis. That's the number of fives that I have."}, {"video_title": "Solving the troll riddle visually   Algebra II   Khan Academy.mp3", "Sentence": "But this time I wanna do it visually, because at least in my mind, it helps really kind of get the intuition of what these things are saying. So let's draw some axes over here. Let's draw an F axis. That's the number of fives that I have. And let's draw a T axis. That is the number of tens I have. And let's say that this right over here, this right over here is 500 tens."}, {"video_title": "Solving the troll riddle visually   Algebra II   Khan Academy.mp3", "Sentence": "That's the number of fives that I have. And let's draw a T axis. That is the number of tens I have. And let's say that this right over here, this right over here is 500 tens. That is 1,000 tens. And let's say that this is, sorry, that's 500 fives. That's 1,000 fives."}, {"video_title": "Solving the troll riddle visually   Algebra II   Khan Academy.mp3", "Sentence": "And let's say that this right over here, this right over here is 500 tens. That is 1,000 tens. And let's say that this is, sorry, that's 500 fives. That's 1,000 fives. This is 500 tens. 500 tens. And this is 1,000 tens."}, {"video_title": "Solving the troll riddle visually   Algebra II   Khan Academy.mp3", "Sentence": "That's 1,000 fives. This is 500 tens. 500 tens. And this is 1,000 tens. So let's think about all of the combinations of Fs and Ts that satisfy this first equation. If T, if we have no tens, then we're gonna have 900 fives. So if we have no tens, we're gonna have 900 fives."}, {"video_title": "Solving the troll riddle visually   Algebra II   Khan Academy.mp3", "Sentence": "And this is 1,000 tens. So let's think about all of the combinations of Fs and Ts that satisfy this first equation. If T, if we have no tens, then we're gonna have 900 fives. So if we have no tens, we're gonna have 900 fives. So that looks like it's right about there. So that's the point zero tens, 900, 900 fives. But what if it went the other way?"}, {"video_title": "Solving the troll riddle visually   Algebra II   Khan Academy.mp3", "Sentence": "So if we have no tens, we're gonna have 900 fives. So that looks like it's right about there. So that's the point zero tens, 900, 900 fives. But what if it went the other way? If we have no fives, we're gonna have 900 tens. So that's gonna be the point 900 tens, zero fives. So all the combinations of Fs and Ts that satisfy this are going to be on this line, are going to be on that line right over there."}, {"video_title": "Solving the troll riddle visually   Algebra II   Khan Academy.mp3", "Sentence": "But what if it went the other way? If we have no fives, we're gonna have 900 tens. So that's gonna be the point 900 tens, zero fives. So all the combinations of Fs and Ts that satisfy this are going to be on this line, are going to be on that line right over there. And I just draw a dotted line just because it's easier for me to draw it straight. So that represents all the Fs and Ts that satisfy the first constraint. Obviously, there's a bunch of them."}, {"video_title": "Solving the troll riddle visually   Algebra II   Khan Academy.mp3", "Sentence": "So all the combinations of Fs and Ts that satisfy this are going to be on this line, are going to be on that line right over there. And I just draw a dotted line just because it's easier for me to draw it straight. So that represents all the Fs and Ts that satisfy the first constraint. Obviously, there's a bunch of them. So we don't know which is the one that is actually what the troll has. But lucky for us, we have a second constraint, this one right over here. So let's do the same thing."}, {"video_title": "Solving the troll riddle visually   Algebra II   Khan Academy.mp3", "Sentence": "Obviously, there's a bunch of them. So we don't know which is the one that is actually what the troll has. But lucky for us, we have a second constraint, this one right over here. So let's do the same thing. In this constraint, what happens if we have no tens? If tens are zero, then we have five F is equal to 5,500. Let me do a little table here because this is a little bit more involved."}, {"video_title": "Solving the troll riddle visually   Algebra II   Khan Academy.mp3", "Sentence": "So let's do the same thing. In this constraint, what happens if we have no tens? If tens are zero, then we have five F is equal to 5,500. Let me do a little table here because this is a little bit more involved. It's a little bit more involved. So for the second equation, tens and fives. If I have no tens, if I have no tens, I have five F is equal to 5,500, F will be 1,100."}, {"video_title": "Solving the troll riddle visually   Algebra II   Khan Academy.mp3", "Sentence": "Let me do a little table here because this is a little bit more involved. It's a little bit more involved. So for the second equation, tens and fives. If I have no tens, if I have no tens, I have five F is equal to 5,500, F will be 1,100. F is, I must have 1,100 fives. If I have no fives, if I have no fives, and this is zero, and I have 10 T is equal to 5,500, that means I have 550, 550 tens. So let's plot both of those points."}, {"video_title": "Solving the troll riddle visually   Algebra II   Khan Academy.mp3", "Sentence": "If I have no tens, if I have no tens, I have five F is equal to 5,500, F will be 1,100. F is, I must have 1,100 fives. If I have no fives, if I have no fives, and this is zero, and I have 10 T is equal to 5,500, that means I have 550, 550 tens. So let's plot both of those points. T equals zero, F is 1,100, that's right about there. So that is zero, 1,100 is on the line that represents this equation. And that when F is zero, T is 550."}, {"video_title": "Solving the troll riddle visually   Algebra II   Khan Academy.mp3", "Sentence": "So let's plot both of those points. T equals zero, F is 1,100, that's right about there. So that is zero, 1,100 is on the line that represents this equation. And that when F is zero, T is 550. So let's see, this is about, see this would be six, seven, eight, nine. So 550 is gonna be right over here. So that is the point 550, 550 comma zero."}, {"video_title": "Solving the troll riddle visually   Algebra II   Khan Academy.mp3", "Sentence": "And that when F is zero, T is 550. So let's see, this is about, see this would be six, seven, eight, nine. So 550 is gonna be right over here. So that is the point 550, 550 comma zero. And all of these points, let me try to draw a straight line again. I can do a better job than that. I can do a better job than that."}, {"video_title": "Solving the troll riddle visually   Algebra II   Khan Academy.mp3", "Sentence": "So that is the point 550, 550 comma zero. And all of these points, let me try to draw a straight line again. I can do a better job than that. I can do a better job than that. So all of these points are the points, let me try one more time, one more time. We wanna get this right. We don't wanna get pushed into the water by the troll."}, {"video_title": "Solving the troll riddle visually   Algebra II   Khan Academy.mp3", "Sentence": "I can do a better job than that. So all of these points are the points, let me try one more time, one more time. We wanna get this right. We don't wanna get pushed into the water by the troll. So there you go, that looks pretty good. So every point on this blue line represents an F, T combination that satisfies the second constraint. So what is an F and T, or number of fives and number of tens that satisfy both constraints?"}, {"video_title": "Solving the troll riddle visually   Algebra II   Khan Academy.mp3", "Sentence": "We don't wanna get pushed into the water by the troll. So there you go, that looks pretty good. So every point on this blue line represents an F, T combination that satisfies the second constraint. So what is an F and T, or number of fives and number of tens that satisfy both constraints? Well, it would be a point that is sitting on both of the lines. And what is a point that is sitting on both of the lines? Well, that's where they intersect."}, {"video_title": "Solving the troll riddle visually   Algebra II   Khan Academy.mp3", "Sentence": "So what is an F and T, or number of fives and number of tens that satisfy both constraints? Well, it would be a point that is sitting on both of the lines. And what is a point that is sitting on both of the lines? Well, that's where they intersect. This point right over here is clearly on the blue line and it is clearly on the yellow line. And what we can do is if we drew this graph really, really, really precisely, we could see how many fives that is and how many tens that is. And if you look at it, if you look at it very precisely, and actually I encourage you to graph it very precisely and come up with how many fives and how many tens that is."}, {"video_title": "Solving the troll riddle visually   Algebra II   Khan Academy.mp3", "Sentence": "Well, that's where they intersect. This point right over here is clearly on the blue line and it is clearly on the yellow line. And what we can do is if we drew this graph really, really, really precisely, we could see how many fives that is and how many tens that is. And if you look at it, if you look at it very precisely, and actually I encourage you to graph it very precisely and come up with how many fives and how many tens that is. Well, when we do it right over here, I'm going to eyeball it. If we look at it right over here, it looks like we have about 700 fives, 700 fives, and it looks like we have about 200 tens. And this is based on my really rough graph, but let's see if that worked."}, {"video_title": "Solving the troll riddle visually   Algebra II   Khan Academy.mp3", "Sentence": "And if you look at it, if you look at it very precisely, and actually I encourage you to graph it very precisely and come up with how many fives and how many tens that is. Well, when we do it right over here, I'm going to eyeball it. If we look at it right over here, it looks like we have about 700 fives, 700 fives, and it looks like we have about 200 tens. And this is based on my really rough graph, but let's see if that worked. 700 plus 200, 700 plus 200 is equal to 900. And if I have 700 fives, 700, let me write this down. Five times 700 is going to be the value of the fives, which is $3,500."}, {"video_title": "Solving the troll riddle visually   Algebra II   Khan Academy.mp3", "Sentence": "And this is based on my really rough graph, but let's see if that worked. 700 plus 200, 700 plus 200 is equal to 900. And if I have 700 fives, 700, let me write this down. Five times 700 is going to be the value of the fives, which is $3,500. And then 10 plus 10 times 200, 10 times 200, which is $2,000, $2,000 is the value of the tens. And if you add up the two values, you indeed get to $5,500. So this looks right."}, {"video_title": "Solving the troll riddle visually   Algebra II   Khan Academy.mp3", "Sentence": "Five times 700 is going to be the value of the fives, which is $3,500. And then 10 plus 10 times 200, 10 times 200, which is $2,000, $2,000 is the value of the tens. And if you add up the two values, you indeed get to $5,500. So this looks right. And so we can tell the troll, troll, I know, I know how many five and $10 bills you have. You have 700 $5 bills and you have 200 $10 bills. The troll is impressed and he lets you cross the bridge to be the hero or heroine of this fantasy adventure."}, {"video_title": "Interpret quadratic models Factored form   Algebra I   Khan Academy.mp3", "Sentence": "So pause this video and see if you can figure that out. All right, now let's think about this together. So let's just imagine actually what the graph of this function looks like. And it'll also help us imagine what's going on with the helicopter. So our horizontal axis, this is t, time in minutes, and then our vertical axis is height, so height as a function of time, and maybe I just write it like this. I'll just write height, and this is given in meters above the ground. Now, I don't know exactly what the graph looks like, but given that I have a negative coefficient on my quadratic term, I know that it is a downward-opening parabola like that."}, {"video_title": "Interpret quadratic models Factored form   Algebra I   Khan Academy.mp3", "Sentence": "And it'll also help us imagine what's going on with the helicopter. So our horizontal axis, this is t, time in minutes, and then our vertical axis is height, so height as a function of time, and maybe I just write it like this. I'll just write height, and this is given in meters above the ground. Now, I don't know exactly what the graph looks like, but given that I have a negative coefficient on my quadratic term, I know that it is a downward-opening parabola like that. And it says that the helicopter takes off of a platform. So however high the platform is, then it takes off, and it's going to do something like this. I don't know exactly what the graph looks like, but probably something like this."}, {"video_title": "Interpret quadratic models Factored form   Algebra I   Khan Academy.mp3", "Sentence": "Now, I don't know exactly what the graph looks like, but given that I have a negative coefficient on my quadratic term, I know that it is a downward-opening parabola like that. And it says that the helicopter takes off of a platform. So however high the platform is, then it takes off, and it's going to do something like this. I don't know exactly what the graph looks like, but probably something like this. Now, if they asked us what is the highest point of the helicopter and at what time does it happen, then we'd wanna figure out what the vertex is of this parabola. But that's not what they're asking. They're asking when does a helicopter land on the ground?"}, {"video_title": "Interpret quadratic models Factored form   Algebra I   Khan Academy.mp3", "Sentence": "I don't know exactly what the graph looks like, but probably something like this. Now, if they asked us what is the highest point of the helicopter and at what time does it happen, then we'd wanna figure out what the vertex is of this parabola. But that's not what they're asking. They're asking when does a helicopter land on the ground? That's this time right over here. So if we wanted to find the vertex, we would wanna put this into vertex form. But here, we wanna figure out when does that function equal zero?"}, {"video_title": "Interpret quadratic models Factored form   Algebra I   Khan Academy.mp3", "Sentence": "They're asking when does a helicopter land on the ground? That's this time right over here. So if we wanted to find the vertex, we would wanna put this into vertex form. But here, we wanna figure out when does that function equal zero? We want to find a zero of this quadratic right over here. So the best way that I can think about doing it is try to factor it. Try to set this thing equal to zero, and then factor it, and then see what t values make that equal to zero."}, {"video_title": "Interpret quadratic models Factored form   Algebra I   Khan Academy.mp3", "Sentence": "But here, we wanna figure out when does that function equal zero? We want to find a zero of this quadratic right over here. So the best way that I can think about doing it is try to factor it. Try to set this thing equal to zero, and then factor it, and then see what t values make that equal to zero. So let me do that. So I say negative three t squared plus 24t plus 60. Remember, we care when our height is equal to zero, equals zero."}, {"video_title": "Interpret quadratic models Factored form   Algebra I   Khan Academy.mp3", "Sentence": "Try to set this thing equal to zero, and then factor it, and then see what t values make that equal to zero. So let me do that. So I say negative three t squared plus 24t plus 60. Remember, we care when our height is equal to zero, equals zero. So let's see. Maybe the first thing I would do, just to simplify this second degree term a little bit, let's just divide both sides by negative three. If we did that, this would become t squared, 24 divided by negative three is negative eight, negative eight t. 60 divided by negative three is negative 20."}, {"video_title": "Interpret quadratic models Factored form   Algebra I   Khan Academy.mp3", "Sentence": "Remember, we care when our height is equal to zero, equals zero. So let's see. Maybe the first thing I would do, just to simplify this second degree term a little bit, let's just divide both sides by negative three. If we did that, this would become t squared, 24 divided by negative three is negative eight, negative eight t. 60 divided by negative three is negative 20. And then zero divided by negative three is of course still zero. And now can I think of two numbers whose product is negative 20, so they would have to have different signs in order to get a negative product, and whose sum is negative eight? So let's see."}, {"video_title": "Interpret quadratic models Factored form   Algebra I   Khan Academy.mp3", "Sentence": "If we did that, this would become t squared, 24 divided by negative three is negative eight, negative eight t. 60 divided by negative three is negative 20. And then zero divided by negative three is of course still zero. And now can I think of two numbers whose product is negative 20, so they would have to have different signs in order to get a negative product, and whose sum is negative eight? So let's see. What about negative 10 and two? That seems to work. So I could write this as t minus 10 times t plus two is equal to zero."}, {"video_title": "Interpret quadratic models Factored form   Algebra I   Khan Academy.mp3", "Sentence": "So let's see. What about negative 10 and two? That seems to work. So I could write this as t minus 10 times t plus two is equal to zero. And so in order to make this entire expression equal to zero, either one of these could be equal to zero. So either t minus 10 is equal to zero, or t plus two is equal to zero. And of course on the left here, I can add 10 to both sides."}, {"video_title": "Interpret quadratic models Factored form   Algebra I   Khan Academy.mp3", "Sentence": "So I could write this as t minus 10 times t plus two is equal to zero. And so in order to make this entire expression equal to zero, either one of these could be equal to zero. So either t minus 10 is equal to zero, or t plus two is equal to zero. And of course on the left here, I can add 10 to both sides. So either t equals 10, or I could subtract two from both sides here, t is equal to negative two. So there's two places where the function is equal to zero. One at time t equals negative two, and one at time t is equal to 10."}, {"video_title": "Interpret quadratic models Factored form   Algebra I   Khan Academy.mp3", "Sentence": "And of course on the left here, I can add 10 to both sides. So either t equals 10, or I could subtract two from both sides here, t is equal to negative two. So there's two places where the function is equal to zero. One at time t equals negative two, and one at time t is equal to 10. Now we're assuming we're dealing with positive time here. We don't know what the helicopter was doing before the takeoff. So we wouldn't really think about this."}, {"video_title": "Interpret quadratic models Factored form   Algebra I   Khan Academy.mp3", "Sentence": "One at time t equals negative two, and one at time t is equal to 10. Now we're assuming we're dealing with positive time here. We don't know what the helicopter was doing before the takeoff. So we wouldn't really think about this. So what we really care about is that t is equal to 10 minutes. That's when the helicopter is right over there. And actually we know at t equals zero, these two terms become zero."}, {"video_title": "Interpret quadratic models Factored form   Algebra I   Khan Academy.mp3", "Sentence": "So we wouldn't really think about this. So what we really care about is that t is equal to 10 minutes. That's when the helicopter is right over there. And actually we know at t equals zero, these two terms become zero. We know it takes off at 60 meters. It goes up. If we figured out the vertex, we would know how high it went, but then it starts going back down, and in 10 minutes after takeoff, it is back at zero, back on the ground."}, {"video_title": "Solving the troll riddle visually   Algebra II   Khan Academy.mp3", "Sentence": "And if we don't solve his riddle, he's gonna push us into the water. So we are under pressure. And at least we made some headway in the last video. We were able to represent his clues mathematically as a system of equations. What I wanna do in this video is think about whether we can solve for the system of equations. And you will see that there are many ways of solving a system of equations. But this time I wanna do it visually, because at least in my mind, it helps really kind of get the intuition of what these things are saying."}, {"video_title": "Solving the troll riddle visually   Algebra II   Khan Academy.mp3", "Sentence": "We were able to represent his clues mathematically as a system of equations. What I wanna do in this video is think about whether we can solve for the system of equations. And you will see that there are many ways of solving a system of equations. But this time I wanna do it visually, because at least in my mind, it helps really kind of get the intuition of what these things are saying. So let's draw some axes over here. Let's draw an F axis. That's the number of fives that I have."}, {"video_title": "Solving the troll riddle visually   Algebra II   Khan Academy.mp3", "Sentence": "But this time I wanna do it visually, because at least in my mind, it helps really kind of get the intuition of what these things are saying. So let's draw some axes over here. Let's draw an F axis. That's the number of fives that I have. And let's draw a T axis. That is the number of tens I have. And let's say that this right over here, this right over here is 500 tens."}, {"video_title": "Solving the troll riddle visually   Algebra II   Khan Academy.mp3", "Sentence": "That's the number of fives that I have. And let's draw a T axis. That is the number of tens I have. And let's say that this right over here, this right over here is 500 tens. That is 1,000 tens. And let's say that this is, sorry, that's 500 fives. That's 1,000 fives."}, {"video_title": "Solving the troll riddle visually   Algebra II   Khan Academy.mp3", "Sentence": "And let's say that this right over here, this right over here is 500 tens. That is 1,000 tens. And let's say that this is, sorry, that's 500 fives. That's 1,000 fives. This is 500 tens. 500 tens. And this is 1,000 tens."}, {"video_title": "Solving the troll riddle visually   Algebra II   Khan Academy.mp3", "Sentence": "That's 1,000 fives. This is 500 tens. 500 tens. And this is 1,000 tens. So let's think about all of the combinations of Fs and Ts that satisfy this first equation. If T, if we have no tens, then we're gonna have 900 fives. So if we have no tens, we're gonna have 900 fives."}, {"video_title": "Solving the troll riddle visually   Algebra II   Khan Academy.mp3", "Sentence": "And this is 1,000 tens. So let's think about all of the combinations of Fs and Ts that satisfy this first equation. If T, if we have no tens, then we're gonna have 900 fives. So if we have no tens, we're gonna have 900 fives. So that looks like it's right about there. So that's the point zero tens, 900, 900 fives. But what if it went the other way?"}, {"video_title": "Solving the troll riddle visually   Algebra II   Khan Academy.mp3", "Sentence": "So if we have no tens, we're gonna have 900 fives. So that looks like it's right about there. So that's the point zero tens, 900, 900 fives. But what if it went the other way? If we have no fives, we're gonna have 900 tens. So that's gonna be the point 900 tens, zero fives. So all the combinations of Fs and Ts that satisfy this are going to be on this line, are going to be on that line right over there."}, {"video_title": "Solving the troll riddle visually   Algebra II   Khan Academy.mp3", "Sentence": "But what if it went the other way? If we have no fives, we're gonna have 900 tens. So that's gonna be the point 900 tens, zero fives. So all the combinations of Fs and Ts that satisfy this are going to be on this line, are going to be on that line right over there. And I just draw a dotted line just because it's easier for me to draw it straight. So that represents all the Fs and Ts that satisfy the first constraint. Obviously, there's a bunch of them."}, {"video_title": "Solving the troll riddle visually   Algebra II   Khan Academy.mp3", "Sentence": "So all the combinations of Fs and Ts that satisfy this are going to be on this line, are going to be on that line right over there. And I just draw a dotted line just because it's easier for me to draw it straight. So that represents all the Fs and Ts that satisfy the first constraint. Obviously, there's a bunch of them. So we don't know which is the one that is actually what the troll has. But lucky for us, we have a second constraint, this one right over here. So let's do the same thing."}, {"video_title": "Solving the troll riddle visually   Algebra II   Khan Academy.mp3", "Sentence": "Obviously, there's a bunch of them. So we don't know which is the one that is actually what the troll has. But lucky for us, we have a second constraint, this one right over here. So let's do the same thing. In this constraint, what happens if we have no tens? If tens are zero, then we have five F is equal to 5,500. Let me do a little table here because this is a little bit more involved."}, {"video_title": "Solving the troll riddle visually   Algebra II   Khan Academy.mp3", "Sentence": "So let's do the same thing. In this constraint, what happens if we have no tens? If tens are zero, then we have five F is equal to 5,500. Let me do a little table here because this is a little bit more involved. It's a little bit more involved. So for the second equation, tens and fives. If I have no tens, if I have no tens, I have five F is equal to 5,500, F will be 1,100."}, {"video_title": "Solving the troll riddle visually   Algebra II   Khan Academy.mp3", "Sentence": "Let me do a little table here because this is a little bit more involved. It's a little bit more involved. So for the second equation, tens and fives. If I have no tens, if I have no tens, I have five F is equal to 5,500, F will be 1,100. F is, I must have 1,100 fives. If I have no fives, if I have no fives, and this is zero, and I have 10 T is equal to 5,500, that means I have 550, 550 tens. So let's plot both of those points."}, {"video_title": "Solving the troll riddle visually   Algebra II   Khan Academy.mp3", "Sentence": "If I have no tens, if I have no tens, I have five F is equal to 5,500, F will be 1,100. F is, I must have 1,100 fives. If I have no fives, if I have no fives, and this is zero, and I have 10 T is equal to 5,500, that means I have 550, 550 tens. So let's plot both of those points. T equals zero, F is 1,100, that's right about there. So that is zero, 1,100 is on the line that represents this equation. And that when F is zero, T is 550."}, {"video_title": "Solving the troll riddle visually   Algebra II   Khan Academy.mp3", "Sentence": "So let's plot both of those points. T equals zero, F is 1,100, that's right about there. So that is zero, 1,100 is on the line that represents this equation. And that when F is zero, T is 550. So let's see, this is about, see this would be six, seven, eight, nine. So 550 is gonna be right over here. So that is the point 550, 550 comma zero."}, {"video_title": "Solving the troll riddle visually   Algebra II   Khan Academy.mp3", "Sentence": "And that when F is zero, T is 550. So let's see, this is about, see this would be six, seven, eight, nine. So 550 is gonna be right over here. So that is the point 550, 550 comma zero. And all of these points, let me try to draw a straight line again. I can do a better job than that. I can do a better job than that."}, {"video_title": "Solving the troll riddle visually   Algebra II   Khan Academy.mp3", "Sentence": "So that is the point 550, 550 comma zero. And all of these points, let me try to draw a straight line again. I can do a better job than that. I can do a better job than that. So all of these points are the points, let me try one more time, one more time. We wanna get this right. We don't wanna get pushed into the water by the troll."}, {"video_title": "Solving the troll riddle visually   Algebra II   Khan Academy.mp3", "Sentence": "I can do a better job than that. So all of these points are the points, let me try one more time, one more time. We wanna get this right. We don't wanna get pushed into the water by the troll. So there you go, that looks pretty good. So every point on this blue line represents an F, T combination that satisfies the second constraint. So what is an F and T, or number of fives and number of tens that satisfy both constraints?"}, {"video_title": "Solving the troll riddle visually   Algebra II   Khan Academy.mp3", "Sentence": "We don't wanna get pushed into the water by the troll. So there you go, that looks pretty good. So every point on this blue line represents an F, T combination that satisfies the second constraint. So what is an F and T, or number of fives and number of tens that satisfy both constraints? Well, it would be a point that is sitting on both of the lines. And what is a point that is sitting on both of the lines? Well, that's where they intersect."}, {"video_title": "Solving the troll riddle visually   Algebra II   Khan Academy.mp3", "Sentence": "So what is an F and T, or number of fives and number of tens that satisfy both constraints? Well, it would be a point that is sitting on both of the lines. And what is a point that is sitting on both of the lines? Well, that's where they intersect. This point right over here is clearly on the blue line and it is clearly on the yellow line. And what we can do is if we drew this graph really, really, really precisely, we could see how many fives that is and how many tens that is. And if you look at it, if you look at it very precisely, and actually I encourage you to graph it very precisely and come up with how many fives and how many tens that is."}, {"video_title": "Solving the troll riddle visually   Algebra II   Khan Academy.mp3", "Sentence": "Well, that's where they intersect. This point right over here is clearly on the blue line and it is clearly on the yellow line. And what we can do is if we drew this graph really, really, really precisely, we could see how many fives that is and how many tens that is. And if you look at it, if you look at it very precisely, and actually I encourage you to graph it very precisely and come up with how many fives and how many tens that is. Well, when we do it right over here, I'm going to eyeball it. If we look at it right over here, it looks like we have about 700 fives, 700 fives, and it looks like we have about 200 tens. And this is based on my really rough graph, but let's see if that worked."}, {"video_title": "Solving the troll riddle visually   Algebra II   Khan Academy.mp3", "Sentence": "And if you look at it, if you look at it very precisely, and actually I encourage you to graph it very precisely and come up with how many fives and how many tens that is. Well, when we do it right over here, I'm going to eyeball it. If we look at it right over here, it looks like we have about 700 fives, 700 fives, and it looks like we have about 200 tens. And this is based on my really rough graph, but let's see if that worked. 700 plus 200, 700 plus 200 is equal to 900. And if I have 700 fives, 700, let me write this down. Five times 700 is going to be the value of the fives, which is $3,500."}, {"video_title": "Solving the troll riddle visually   Algebra II   Khan Academy.mp3", "Sentence": "And this is based on my really rough graph, but let's see if that worked. 700 plus 200, 700 plus 200 is equal to 900. And if I have 700 fives, 700, let me write this down. Five times 700 is going to be the value of the fives, which is $3,500. And then 10 plus 10 times 200, 10 times 200, which is $2,000, $2,000 is the value of the tens. And if you add up the two values, you indeed get to $5,500. So this looks right."}, {"video_title": "Solving the troll riddle visually   Algebra II   Khan Academy.mp3", "Sentence": "Five times 700 is going to be the value of the fives, which is $3,500. And then 10 plus 10 times 200, 10 times 200, which is $2,000, $2,000 is the value of the tens. And if you add up the two values, you indeed get to $5,500. So this looks right. And so we can tell the troll, troll, I know, I know how many five and $10 bills you have. You have 700 $5 bills and you have 200 $10 bills. The troll is impressed and he lets you cross the bridge to be the hero or heroine of this fantasy adventure."}, {"video_title": "Example of adding fractions with unlike denominators word problem   Pre-Algebra   Khan Academy.mp3", "Sentence": "Michael has got half a gallon of yellow paint. If they mix their paints together, will they have the one gallon they need? So let's think about that. We're going to add the 2 5ths of a gallon of red paint, and we're going to add that to half a gallon of yellow paint. And we want to see if this gets to being one whole gallon. So whenever we add fractions, right over here, we're not adding the same thing. Here we're adding 2 5ths."}, {"video_title": "Example of adding fractions with unlike denominators word problem   Pre-Algebra   Khan Academy.mp3", "Sentence": "We're going to add the 2 5ths of a gallon of red paint, and we're going to add that to half a gallon of yellow paint. And we want to see if this gets to being one whole gallon. So whenever we add fractions, right over here, we're not adding the same thing. Here we're adding 2 5ths. Here we're adding 1 halves. So in order to be able to add these two things, we need to get to a common denominator. And the common denominator, or the best common denominator to use, is the number that is the smallest multiple of both 5 and 2."}, {"video_title": "Example of adding fractions with unlike denominators word problem   Pre-Algebra   Khan Academy.mp3", "Sentence": "Here we're adding 2 5ths. Here we're adding 1 halves. So in order to be able to add these two things, we need to get to a common denominator. And the common denominator, or the best common denominator to use, is the number that is the smallest multiple of both 5 and 2. And since 5 and 2 are both prime numbers, the smallest number is just going to be their product. 10 is the smallest number that we can think of that is divisible by both 5 and 2. So let's rewrite each of these fractions with 10 as a denominator."}, {"video_title": "Example of adding fractions with unlike denominators word problem   Pre-Algebra   Khan Academy.mp3", "Sentence": "And the common denominator, or the best common denominator to use, is the number that is the smallest multiple of both 5 and 2. And since 5 and 2 are both prime numbers, the smallest number is just going to be their product. 10 is the smallest number that we can think of that is divisible by both 5 and 2. So let's rewrite each of these fractions with 10 as a denominator. So 2 5ths is going to be something over 10, and 1 half is going to be something over 10. And to help us visualize this, let me draw a grid with tenths in it. So that's that, and that's that right over here."}, {"video_title": "Example of adding fractions with unlike denominators word problem   Pre-Algebra   Khan Academy.mp3", "Sentence": "So let's rewrite each of these fractions with 10 as a denominator. So 2 5ths is going to be something over 10, and 1 half is going to be something over 10. And to help us visualize this, let me draw a grid with tenths in it. So that's that, and that's that right over here. So each of these are in tenths. These are 10 equal segments this bar is divided into. So let's try to visualize what 2 5ths looks like on this bar."}, {"video_title": "Example of adding fractions with unlike denominators word problem   Pre-Algebra   Khan Academy.mp3", "Sentence": "So that's that, and that's that right over here. So each of these are in tenths. These are 10 equal segments this bar is divided into. So let's try to visualize what 2 5ths looks like on this bar. Well, right now it's divided into tenths. If we were to divide this bar into fifths, then we're going to have 1. Actually, let me do it in that same color."}, {"video_title": "Example of adding fractions with unlike denominators word problem   Pre-Algebra   Khan Academy.mp3", "Sentence": "So let's try to visualize what 2 5ths looks like on this bar. Well, right now it's divided into tenths. If we were to divide this bar into fifths, then we're going to have 1. Actually, let me do it in that same color. So it's going to be 1 division, 2, 3, 4. So notice, if you go between the red marks, these are each a fifth of the bar. If you go between the red marks, and we have two of them."}, {"video_title": "Example of adding fractions with unlike denominators word problem   Pre-Algebra   Khan Academy.mp3", "Sentence": "Actually, let me do it in that same color. So it's going to be 1 division, 2, 3, 4. So notice, if you go between the red marks, these are each a fifth of the bar. If you go between the red marks, and we have two of them. So we're going to go 1 and 2. This right over here, this part of the bar represents 2 5ths of it. Now let's do the same thing for 1 half."}, {"video_title": "Example of adding fractions with unlike denominators word problem   Pre-Algebra   Khan Academy.mp3", "Sentence": "If you go between the red marks, and we have two of them. So we're going to go 1 and 2. This right over here, this part of the bar represents 2 5ths of it. Now let's do the same thing for 1 half. So let's divide this bar exactly in half. So let me do that. So I'm going to divide it exactly in half."}, {"video_title": "Example of adding fractions with unlike denominators word problem   Pre-Algebra   Khan Academy.mp3", "Sentence": "Now let's do the same thing for 1 half. So let's divide this bar exactly in half. So let me do that. So I'm going to divide it exactly in half. And 1 half literally represents one of the two equal sections. So this is 1 1 half. Now, to go from fifths to tenths, you're essentially taking each of the equal sections and you are multiplying by 2."}, {"video_title": "Example of adding fractions with unlike denominators word problem   Pre-Algebra   Khan Academy.mp3", "Sentence": "So I'm going to divide it exactly in half. And 1 half literally represents one of the two equal sections. So this is 1 1 half. Now, to go from fifths to tenths, you're essentially taking each of the equal sections and you are multiplying by 2. So to go from fifths to tenths, you're multiplying by 2. You have 5 equal sections. You split each of those into 2, so you have twice as many."}, {"video_title": "Example of adding fractions with unlike denominators word problem   Pre-Algebra   Khan Academy.mp3", "Sentence": "Now, to go from fifths to tenths, you're essentially taking each of the equal sections and you are multiplying by 2. So to go from fifths to tenths, you're multiplying by 2. You have 5 equal sections. You split each of those into 2, so you have twice as many. You now have 10 equal sections. So those two sections that were shaded in, well, you were going to multiply by 2 the same way. Those two are going to turn into 4 tenths."}, {"video_title": "Example of adding fractions with unlike denominators word problem   Pre-Algebra   Khan Academy.mp3", "Sentence": "You split each of those into 2, so you have twice as many. You now have 10 equal sections. So those two sections that were shaded in, well, you were going to multiply by 2 the same way. Those two are going to turn into 4 tenths. And you see it right over here when we shaded it initially. If you look at the tenths, you have 1 tenth, 2 tenths, 3 tenths, and 4 tenths. Let's do the same logic over here."}, {"video_title": "Example of adding fractions with unlike denominators word problem   Pre-Algebra   Khan Academy.mp3", "Sentence": "Those two are going to turn into 4 tenths. And you see it right over here when we shaded it initially. If you look at the tenths, you have 1 tenth, 2 tenths, 3 tenths, and 4 tenths. Let's do the same logic over here. If you have 2 halves and you want to make them into 10 tenths, you have to take each of the halves and split them into 5 sections. You're going to have 5 times as many sections. So to go from 2 to 10, we multiply by 5."}, {"video_title": "Example of adding fractions with unlike denominators word problem   Pre-Algebra   Khan Academy.mp3", "Sentence": "Let's do the same logic over here. If you have 2 halves and you want to make them into 10 tenths, you have to take each of the halves and split them into 5 sections. You're going to have 5 times as many sections. So to go from 2 to 10, we multiply by 5. So similarly, that one shaded in section, that in yellow, that's going to turn into 5. That 1 half is going to turn into 5 tenths. So we're going to multiply by 5."}, {"video_title": "Example of adding fractions with unlike denominators word problem   Pre-Algebra   Khan Academy.mp3", "Sentence": "So to go from 2 to 10, we multiply by 5. So similarly, that one shaded in section, that in yellow, that's going to turn into 5. That 1 half is going to turn into 5 tenths. So we're going to multiply by 5. Another way to think about it, whatever we did to the numerator or whatever we did to the denominator, we have to do to the numerator. Whatever we did to the denominator, we have to do to the numerator. Otherwise, we're changing the value of the fraction."}, {"video_title": "Example of adding fractions with unlike denominators word problem   Pre-Algebra   Khan Academy.mp3", "Sentence": "So we're going to multiply by 5. Another way to think about it, whatever we did to the numerator or whatever we did to the denominator, we have to do to the numerator. Whatever we did to the denominator, we have to do to the numerator. Otherwise, we're changing the value of the fraction. So 1 times 5 is going to get you to 5. And you see that over here when we shaded it in. That 1 half, if you look at the tenths, is equal to 1, 2, 3, 4, 5 tenths."}, {"video_title": "Example of adding fractions with unlike denominators word problem   Pre-Algebra   Khan Academy.mp3", "Sentence": "Otherwise, we're changing the value of the fraction. So 1 times 5 is going to get you to 5. And you see that over here when we shaded it in. That 1 half, if you look at the tenths, is equal to 1, 2, 3, 4, 5 tenths. And now we are ready to add these two things. 4 tenths plus 5 tenths, well, this is going to be equal to a certain number of tenths. It's going to be equal to 4 plus 5 tenths."}, {"video_title": "Example of adding fractions with unlike denominators word problem   Pre-Algebra   Khan Academy.mp3", "Sentence": "That 1 half, if you look at the tenths, is equal to 1, 2, 3, 4, 5 tenths. And now we are ready to add these two things. 4 tenths plus 5 tenths, well, this is going to be equal to a certain number of tenths. It's going to be equal to 4 plus 5 tenths. And we can once again visualize that. Let me draw our grid again. So 4 plus 5 tenths."}, {"video_title": "Example of adding fractions with unlike denominators word problem   Pre-Algebra   Khan Academy.mp3", "Sentence": "It's going to be equal to 4 plus 5 tenths. And we can once again visualize that. Let me draw our grid again. So 4 plus 5 tenths. I'll do it actually on top of the paint can right over here. So let me color in 4 tenths. So 1, 2, 3, 4."}, {"video_title": "Example of adding fractions with unlike denominators word problem   Pre-Algebra   Khan Academy.mp3", "Sentence": "So 4 plus 5 tenths. I'll do it actually on top of the paint can right over here. So let me color in 4 tenths. So 1, 2, 3, 4. And then let me color in the 5 tenths. And notice that was exactly the 4 tenths here, which is exactly the 2 fifths. Let me color in the 5 tenths."}, {"video_title": "Example of adding fractions with unlike denominators word problem   Pre-Algebra   Khan Academy.mp3", "Sentence": "So 1, 2, 3, 4. And then let me color in the 5 tenths. And notice that was exactly the 4 tenths here, which is exactly the 2 fifths. Let me color in the 5 tenths. 1, 2, 3, 4, and 5. And so how many total tenths do we have? We have a total of 1, 2, 3, 4, 5, 6, 7, 8, 9."}, {"video_title": "Example of adding fractions with unlike denominators word problem   Pre-Algebra   Khan Academy.mp3", "Sentence": "Let me color in the 5 tenths. 1, 2, 3, 4, and 5. And so how many total tenths do we have? We have a total of 1, 2, 3, 4, 5, 6, 7, 8, 9. 9 of the tenths are now shaded in. We had 9 tenths of a gallon of paint. So now to answer their question, will they have the gallon they need?"}, {"video_title": "Example of adding fractions with unlike denominators word problem   Pre-Algebra   Khan Academy.mp3", "Sentence": "We have a total of 1, 2, 3, 4, 5, 6, 7, 8, 9. 9 of the tenths are now shaded in. We had 9 tenths of a gallon of paint. So now to answer their question, will they have the gallon they need? No, they have less than a whole. A gallon would be 10 tenths. They only have 9 tenths."}, {"video_title": "Example of adding fractions with unlike denominators word problem   Pre-Algebra   Khan Academy.mp3", "Sentence": "So now to answer their question, will they have the gallon they need? No, they have less than a whole. A gallon would be 10 tenths. They only have 9 tenths. So no, they do not have enough of a gallon. Now another way you could have thought about this. You could have said, hey, look, 2 fifths is less than 1 half."}, {"video_title": "Example of adding fractions with unlike denominators word problem   Pre-Algebra   Khan Academy.mp3", "Sentence": "They only have 9 tenths. So no, they do not have enough of a gallon. Now another way you could have thought about this. You could have said, hey, look, 2 fifths is less than 1 half. And you could even visualize that right over here. So I have something less than 1 half plus 1 half. I'm not going to get a whole."}, {"video_title": "Percentage of a whole number   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "Let's see if we can figure out what 30% of 6 is. So one way of thinking about 30%, this literally means 30 per 100. So you could view this as 30 over 100 times 6 is the same thing as 30% of 6. Or you could view this as 30 hundredths times 6, so 0.30 times 6. And we could solve both of these, and you'll see that we'll get the same answer. If you do this multiplication right over here, 30 over 100, and you could view this times 6 over 1, this is equal to 180 over 100. And let's see, we can simplify."}, {"video_title": "Percentage of a whole number   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "Or you could view this as 30 hundredths times 6, so 0.30 times 6. And we could solve both of these, and you'll see that we'll get the same answer. If you do this multiplication right over here, 30 over 100, and you could view this times 6 over 1, this is equal to 180 over 100. And let's see, we can simplify. We can divide the numerator and the denominator by 10. And then we can divide the numerator and the denominator by 2. And we will get 9 fifths, which is the same thing as 1 and 4 fifths."}, {"video_title": "Percentage of a whole number   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "And let's see, we can simplify. We can divide the numerator and the denominator by 10. And then we can divide the numerator and the denominator by 2. And we will get 9 fifths, which is the same thing as 1 and 4 fifths. And then if we wanted to write this as a decimal, 4 fifths is 0.8. And if you want to verify that, you could verify that 5 goes into 4. And there's going to be a decimal, so let's throw some decimals in there."}, {"video_title": "Percentage of a whole number   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "And we will get 9 fifths, which is the same thing as 1 and 4 fifths. And then if we wanted to write this as a decimal, 4 fifths is 0.8. And if you want to verify that, you could verify that 5 goes into 4. And there's going to be a decimal, so let's throw some decimals in there. It goes into 4 zero times. We don't have to worry about that. It goes into 48 times."}, {"video_title": "Percentage of a whole number   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "And there's going to be a decimal, so let's throw some decimals in there. It goes into 4 zero times. We don't have to worry about that. It goes into 48 times. 8 times 5 is 40. Subtract, you have no remainder, and you just have zeros left here. So 4 fifths is 0.8."}, {"video_title": "Percentage of a whole number   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "It goes into 48 times. 8 times 5 is 40. Subtract, you have no remainder, and you just have zeros left here. So 4 fifths is 0.8. You got the 1 there. This is the same thing as 1.8, which you would have gotten if you divided 5 into 9. You would have gotten 1.8."}, {"video_title": "Percentage of a whole number   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "So 4 fifths is 0.8. You got the 1 there. This is the same thing as 1.8, which you would have gotten if you divided 5 into 9. You would have gotten 1.8. So 30% of 6 is equal to 1.8. And we can verify it doing this way as well. So if we were to multiply 0.30 times 6, let's do that."}, {"video_title": "Percentage of a whole number   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "You would have gotten 1.8. So 30% of 6 is equal to 1.8. And we can verify it doing this way as well. So if we were to multiply 0.30 times 6, let's do that. And I could just write that literally as 0.3 times 6. Well, 3 times 6 is 18. I have only one digit behind the decimal amongst both of these numbers that I'm multiplying."}, {"video_title": "Percentage of a whole number   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "So if we were to multiply 0.30 times 6, let's do that. And I could just write that literally as 0.3 times 6. Well, 3 times 6 is 18. I have only one digit behind the decimal amongst both of these numbers that I'm multiplying. I only have the 3 to the right of the decimal, so I'm only going to have one number to the right of the decimal here. So I'll just count one number. It's going to be 1.8."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So let's say we have the inequality 4x plus 3 is less than negative 1. So let's find all of the x's that satisfy this. So the first thing I'd like to do is get rid of this 3. So let's subtract 3 from both sides of this equation. So the left-hand side is just going to end up being 4x. These 3's cancel out. That just ends up with a 0."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So let's subtract 3 from both sides of this equation. So the left-hand side is just going to end up being 4x. These 3's cancel out. That just ends up with a 0. No reason to change the inequality just yet. We're just adding and subtracting from both sides. In this case, subtracting."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "That just ends up with a 0. No reason to change the inequality just yet. We're just adding and subtracting from both sides. In this case, subtracting. That doesn't change the inequality as long as we're subtracting the same value. We have negative 1 minus 3. That is negative 4."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "In this case, subtracting. That doesn't change the inequality as long as we're subtracting the same value. We have negative 1 minus 3. That is negative 4. Negative 1 minus 3 is negative 4. And then we'll want to divide both sides of this equation by 4. Let's divide both sides of this equation by 4."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "That is negative 4. Negative 1 minus 3 is negative 4. And then we'll want to divide both sides of this equation by 4. Let's divide both sides of this equation by 4. Once again, when you multiply or divide both sides of an inequality by a positive number, it doesn't change the inequality. So the left-hand side is just x. x is less than negative 4 divided by 4 is negative 1. x is less than negative 1. Or we could write this in interval notation."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Let's divide both sides of this equation by 4. Once again, when you multiply or divide both sides of an inequality by a positive number, it doesn't change the inequality. So the left-hand side is just x. x is less than negative 4 divided by 4 is negative 1. x is less than negative 1. Or we could write this in interval notation. All of the x's from negative infinity to negative 1, but not including negative 1. So we put a parenthesis right there. Let's do a slightly harder one."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Or we could write this in interval notation. All of the x's from negative infinity to negative 1, but not including negative 1. So we put a parenthesis right there. Let's do a slightly harder one. Let's say we have 5x is greater than 8x plus 27. Let's get all our x's on the left-hand side. The best way to do that is subtract 8x from both sides."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Let's do a slightly harder one. Let's say we have 5x is greater than 8x plus 27. Let's get all our x's on the left-hand side. The best way to do that is subtract 8x from both sides. You subtract 8x from both sides. The left-hand side becomes 5x minus 8x. That's negative 3x."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "The best way to do that is subtract 8x from both sides. You subtract 8x from both sides. The left-hand side becomes 5x minus 8x. That's negative 3x. We still have a greater than sign. We're just adding or subtracting the same quantity to both sides. These 8x's cancel out and you're just left with a 27."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "That's negative 3x. We still have a greater than sign. We're just adding or subtracting the same quantity to both sides. These 8x's cancel out and you're just left with a 27. So you have negative 3x is greater than 27. Now, to just turn this into an x, we want to divide both sides by negative 3. But remember, when you multiply or divide both sides of an inequality by a negative number, you swap the inequality."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "These 8x's cancel out and you're just left with a 27. So you have negative 3x is greater than 27. Now, to just turn this into an x, we want to divide both sides by negative 3. But remember, when you multiply or divide both sides of an inequality by a negative number, you swap the inequality. So if we divide both sides of this by negative 3, we have to swap this inequality. It will go from being a greater than sign to a less than sign. And this is a bit of a way that I remember greater than."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "But remember, when you multiply or divide both sides of an inequality by a negative number, you swap the inequality. So if we divide both sides of this by negative 3, we have to swap this inequality. It will go from being a greater than sign to a less than sign. And this is a bit of a way that I remember greater than. The left-hand side just looks bigger. This is greater than. If you just imagine this height, that height is greater than that height right there, which is just a point."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "And this is a bit of a way that I remember greater than. The left-hand side just looks bigger. This is greater than. If you just imagine this height, that height is greater than that height right there, which is just a point. I don't know if that confuses you or not. This is less than. This little point is less than the distance of that big opening."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "If you just imagine this height, that height is greater than that height right there, which is just a point. I don't know if that confuses you or not. This is less than. This little point is less than the distance of that big opening. That's how I remember it. But anyway, 3x over negative 3. So anyway, now that we divided both sides by a negative number, by negative 3, we swapped the inequality from greater than to less than."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "This little point is less than the distance of that big opening. That's how I remember it. But anyway, 3x over negative 3. So anyway, now that we divided both sides by a negative number, by negative 3, we swapped the inequality from greater than to less than. In the left-hand side, the negative 3's cancel out. You get x is less than 27 over negative 3, which is negative 9. Or in interval notation, it would be everything from negative infinity to negative 9, not including negative 9."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So anyway, now that we divided both sides by a negative number, by negative 3, we swapped the inequality from greater than to less than. In the left-hand side, the negative 3's cancel out. You get x is less than 27 over negative 3, which is negative 9. Or in interval notation, it would be everything from negative infinity to negative 9, not including negative 9. If you wanted to do it as a number line, it would look like this. This would be negative 9. Maybe this would be negative 8."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Or in interval notation, it would be everything from negative infinity to negative 9, not including negative 9. If you wanted to do it as a number line, it would look like this. This would be negative 9. Maybe this would be negative 8. Maybe this would be negative 10. You would start at negative 9, not include it because we don't have an equal sign here. You go all the way, everything less than that, all the way down as we see to negative infinity."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Maybe this would be negative 8. Maybe this would be negative 10. You would start at negative 9, not include it because we don't have an equal sign here. You go all the way, everything less than that, all the way down as we see to negative infinity. Let's do a nice, hairy problem. Let's say we have 8x minus 5 times 4x plus 1 is greater than or equal to negative 1 plus 2 times 4x minus 3. This might seem very daunting, but if we just simplify it step by step, you'll see it's no harder than any of the other problems we've tackled."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "You go all the way, everything less than that, all the way down as we see to negative infinity. Let's do a nice, hairy problem. Let's say we have 8x minus 5 times 4x plus 1 is greater than or equal to negative 1 plus 2 times 4x minus 3. This might seem very daunting, but if we just simplify it step by step, you'll see it's no harder than any of the other problems we've tackled. Let's just simplify this. You get 8x minus, let's distribute this negative 5. Let me say 8x and then distribute the negative 5."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "This might seem very daunting, but if we just simplify it step by step, you'll see it's no harder than any of the other problems we've tackled. Let's just simplify this. You get 8x minus, let's distribute this negative 5. Let me say 8x and then distribute the negative 5. Negative 5 times 4x is negative 20x. Negative 5, when I say negative 5, I'm talking about this whole thing. Negative 5 times 1 is negative 5."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Let me say 8x and then distribute the negative 5. Negative 5 times 4x is negative 20x. Negative 5, when I say negative 5, I'm talking about this whole thing. Negative 5 times 1 is negative 5. Then that's going to be greater than or equal to negative 1 plus 2 times 4x is 8x. 2 times negative 3 is negative 6. Now we can merge these two terms."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Negative 5 times 1 is negative 5. Then that's going to be greater than or equal to negative 1 plus 2 times 4x is 8x. 2 times negative 3 is negative 6. Now we can merge these two terms. 8x minus 20x is negative 12x minus 5 is greater than or equal to, we can merge these constant terms, negative 1 minus 6, that's negative 7. Then we have this plus 8x left over. I like to get all my x terms on the left-hand side, so let's subtract 8x from both sides of this equation."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Now we can merge these two terms. 8x minus 20x is negative 12x minus 5 is greater than or equal to, we can merge these constant terms, negative 1 minus 6, that's negative 7. Then we have this plus 8x left over. I like to get all my x terms on the left-hand side, so let's subtract 8x from both sides of this equation. Let's subtract 8x from both sides of this equation. From both sides, I'm subtracting 8x. This left-hand side, negative 12 minus 8, that's negative 20."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "I like to get all my x terms on the left-hand side, so let's subtract 8x from both sides of this equation. Let's subtract 8x from both sides of this equation. From both sides, I'm subtracting 8x. This left-hand side, negative 12 minus 8, that's negative 20. Negative 20x minus 5. Once again, no reason to change the inequality just yet. All we're doing is simplifying the sides or adding and subtracting from them."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "This left-hand side, negative 12 minus 8, that's negative 20. Negative 20x minus 5. Once again, no reason to change the inequality just yet. All we're doing is simplifying the sides or adding and subtracting from them. The right-hand side becomes, this thing cancels out, 8x minus 8x, that's 0. You're just left with a negative 7. Now I want to get rid of this negative 5, so let's add 5 to both sides of this equation."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "All we're doing is simplifying the sides or adding and subtracting from them. The right-hand side becomes, this thing cancels out, 8x minus 8x, that's 0. You're just left with a negative 7. Now I want to get rid of this negative 5, so let's add 5 to both sides of this equation. The left-hand side, you're just left with a negative 20x. The 5 and these 5s cancel out. No reason to change the inequality just yet."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Now I want to get rid of this negative 5, so let's add 5 to both sides of this equation. The left-hand side, you're just left with a negative 20x. The 5 and these 5s cancel out. No reason to change the inequality just yet. Negative 7 plus 5, that's negative 2. Now we're at an interesting point. We have negative 20x is greater than or equal to negative 2."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "No reason to change the inequality just yet. Negative 7 plus 5, that's negative 2. Now we're at an interesting point. We have negative 20x is greater than or equal to negative 2. If this was an equation or really any type of even inequality, we want to divide both sides by negative 20, but we have to remember, when you multiply or divide both sides of an inequality by a negative number, you have to swap the inequality, so let's remember that. If we divide this side by negative 20 and we divide this side by negative 20, all I did is took both of these sides, divided by negative 20. We have to swap the inequality."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "We have negative 20x is greater than or equal to negative 2. If this was an equation or really any type of even inequality, we want to divide both sides by negative 20, but we have to remember, when you multiply or divide both sides of an inequality by a negative number, you have to swap the inequality, so let's remember that. If we divide this side by negative 20 and we divide this side by negative 20, all I did is took both of these sides, divided by negative 20. We have to swap the inequality. The greater than or equal to has to become a less than or equal sign. Of course, these cancel out. You get x is less than or equal to, the negatives cancel out."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "We have to swap the inequality. The greater than or equal to has to become a less than or equal sign. Of course, these cancel out. You get x is less than or equal to, the negatives cancel out. 2 over 20 is 1 over 10. If we were writing it in interval notation, the upper bound would be 1 over 10. Notice we're including it because we have an equal sign, less than or equal, so we're including 1 over 10, and we're going to go all the way down to negative infinity, everything less than or equal to 1 over 10."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "You get x is less than or equal to, the negatives cancel out. 2 over 20 is 1 over 10. If we were writing it in interval notation, the upper bound would be 1 over 10. Notice we're including it because we have an equal sign, less than or equal, so we're including 1 over 10, and we're going to go all the way down to negative infinity, everything less than or equal to 1 over 10. This is just another way of writing that. Just for fun, let's draw the number line right here. This is maybe 0."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Notice we're including it because we have an equal sign, less than or equal, so we're including 1 over 10, and we're going to go all the way down to negative infinity, everything less than or equal to 1 over 10. This is just another way of writing that. Just for fun, let's draw the number line right here. This is maybe 0. That is 1. 1 over 10 might be over here. Everything less than or equal to 1 over 10."}, {"video_title": "Negative exponents   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "We already know that 2 to the fourth power can be viewed as, starting with a one, and then multiplying it by two four times, so let me do that. So times two, times two, times two, times two, and that will give us, let's see, two times two is four, eight, 16. So that will give us 16. Now I will ask you a more interesting question. What do you think two to the negative, negative four powers, and I encourage you to pause the video and think about that. Well you might be tempted to say, oh, maybe it's negative 16 or something like that, but remember what the exponent operation is trying to do. This is, one way of viewing it is, this is telling us how many times are we going to multiply two times negative one."}, {"video_title": "Negative exponents   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "Now I will ask you a more interesting question. What do you think two to the negative, negative four powers, and I encourage you to pause the video and think about that. Well you might be tempted to say, oh, maybe it's negative 16 or something like that, but remember what the exponent operation is trying to do. This is, one way of viewing it is, this is telling us how many times are we going to multiply two times negative one. But here we're gonna multiply negative four times. Well what does negative traditionally mean? Negative traditionally means the opposite."}, {"video_title": "Negative exponents   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "This is, one way of viewing it is, this is telling us how many times are we going to multiply two times negative one. But here we're gonna multiply negative four times. Well what does negative traditionally mean? Negative traditionally means the opposite. So here this is how many times you're going to multiply. Maybe when we make it negative, this says, how many times are we gonna, starting with a one, how many times are we going to divide by two? So let's think about that a little bit."}, {"video_title": "Negative exponents   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "Negative traditionally means the opposite. So here this is how many times you're going to multiply. Maybe when we make it negative, this says, how many times are we gonna, starting with a one, how many times are we going to divide by two? So let's think about that a little bit. So this could be viewed as one times, and we're gonna divide by two four times. Well dividing by two is the same thing as multiplying by 1 1\u20442. So we could say that this is one times 1 1\u20442, times 1 1\u20442, times, so let me just do it in one color, it's gonna take, so one times 1 1\u20442, times 1 1\u20442, times 1 1\u20442, times 1 1\u20442."}, {"video_title": "Negative exponents   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "So let's think about that a little bit. So this could be viewed as one times, and we're gonna divide by two four times. Well dividing by two is the same thing as multiplying by 1 1\u20442. So we could say that this is one times 1 1\u20442, times 1 1\u20442, times, so let me just do it in one color, it's gonna take, so one times 1 1\u20442, times 1 1\u20442, times 1 1\u20442, times 1 1\u20442. Notice, multiplying by 1 1\u20442 four times is the exact same thing as dividing by two four times. And in this situation, this would get you, well 1 1\u20442, well one times 1 1\u20442 is just 1 1\u20442, times 1 1\u20442 is 1 1\u20444, times 1 1\u20442 is 1 1\u20448, times 1 1\u20442 is one over 16. And so you probably see the relationship here."}, {"video_title": "Negative exponents   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "So we could say that this is one times 1 1\u20442, times 1 1\u20442, times, so let me just do it in one color, it's gonna take, so one times 1 1\u20442, times 1 1\u20442, times 1 1\u20442, times 1 1\u20442. Notice, multiplying by 1 1\u20442 four times is the exact same thing as dividing by two four times. And in this situation, this would get you, well 1 1\u20442, well one times 1 1\u20442 is just 1 1\u20442, times 1 1\u20442 is 1 1\u20444, times 1 1\u20442 is 1 1\u20448, times 1 1\u20442 is one over 16. And so you probably see the relationship here. If you're, this is essentially, you're starting with the one and you're dividing by two four times, you could also say, you could also say that two, two, I'm gonna do the same colors, two to the negative four, two to the negative four, is the same thing as one over two to the fourth power. One over two to the fourth power. Let me color code it nicely so you realize what the negative is doing."}, {"video_title": "Negative exponents   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "And so you probably see the relationship here. If you're, this is essentially, you're starting with the one and you're dividing by two four times, you could also say, you could also say that two, two, I'm gonna do the same colors, two to the negative four, two to the negative four, is the same thing as one over two to the fourth power. One over two to the fourth power. Let me color code it nicely so you realize what the negative is doing. So this negative right over here, let me do that in a better color, I'll do it in magenta, something that jumps out. So this negative right over here, this is what's causing us to go one over. So two to the negative four is the same thing based on the way we've defined it just up right here, as one over, or the reciprocal of two to the fourth, or one over two to the fourth."}, {"video_title": "Negative exponents   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "Let me color code it nicely so you realize what the negative is doing. So this negative right over here, let me do that in a better color, I'll do it in magenta, something that jumps out. So this negative right over here, this is what's causing us to go one over. So two to the negative four is the same thing based on the way we've defined it just up right here, as one over, or the reciprocal of two to the fourth, or one over two to the fourth. And so you could view this as being one over two times, so two times two times two times two, if you just view two to the fourth as taking four twos and multiplying them, or if you use this idea right over here, you could view it as starting with a one and multiplying it by two four times, either way, you are going to get one over, one over 16, one over 16. So let's do a few more examples of this just so that we make sure things are clear to us. So let's try three to the negative third power."}, {"video_title": "Negative exponents   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "So two to the negative four is the same thing based on the way we've defined it just up right here, as one over, or the reciprocal of two to the fourth, or one over two to the fourth. And so you could view this as being one over two times, so two times two times two times two, if you just view two to the fourth as taking four twos and multiplying them, or if you use this idea right over here, you could view it as starting with a one and multiplying it by two four times, either way, you are going to get one over, one over 16, one over 16. So let's do a few more examples of this just so that we make sure things are clear to us. So let's try three to the negative third power. So remember, whenever you see that negative, what my brain always does is say, I need to take the reciprocal here. So this is going to be equal to, and I'm gonna highlight the negative again, this is going to be one over three to the third power, one over three to the third power, which would be equal to, well, one over three times three, or you could say one over three times three times three, or one times three times three times three, is going to be 27. So this is going to be 1 27th."}, {"video_title": "Negative exponents   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "So let's try three to the negative third power. So remember, whenever you see that negative, what my brain always does is say, I need to take the reciprocal here. So this is going to be equal to, and I'm gonna highlight the negative again, this is going to be one over three to the third power, one over three to the third power, which would be equal to, well, one over three times three, or you could say one over three times three times three, or one times three times three times three, is going to be 27. So this is going to be 1 27th. Let's try another example. I'll do two or three more. So let's take a negative number to a negative exponent just to see if we can confuse ourselves."}, {"video_title": "Negative exponents   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "So this is going to be 1 27th. Let's try another example. I'll do two or three more. So let's take a negative number to a negative exponent just to see if we can confuse ourselves. So let's take the number negative four, negative four, and let's take it, I don't want my numbers to get too big too fast, so let's take, let's just take negative two, let's take negative two, and let's take it to the negative three power. Negative, I wanna make my negatives in magenta. Negative three power."}, {"video_title": "Negative exponents   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "So let's take a negative number to a negative exponent just to see if we can confuse ourselves. So let's take the number negative four, negative four, and let's take it, I don't want my numbers to get too big too fast, so let's take, let's just take negative two, let's take negative two, and let's take it to the negative three power. Negative, I wanna make my negatives in magenta. Negative three power. Negative, negative three power. So at first this might be daunting, do the negatives cancel, and that'll just be the remnants in your brain that you're trying to think of multiplying negatives, do not apply that here. Remember, you see a negative exponent, that just means the reciprocal of the positive exponent."}, {"video_title": "Negative exponents   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "Negative three power. Negative, negative three power. So at first this might be daunting, do the negatives cancel, and that'll just be the remnants in your brain that you're trying to think of multiplying negatives, do not apply that here. Remember, you see a negative exponent, that just means the reciprocal of the positive exponent. So one over negative two, negative two to the third power, to the positive third power. And this is equal to, this is equal to one over negative two, negative two times negative two times negative two, times negative two, or you could do it as one times negative two times negative two times negative two, which is going to give you 1 over negative 8 or negative 1 eighth. Let me scroll over a little bit."}, {"video_title": "Negative exponents   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "Remember, you see a negative exponent, that just means the reciprocal of the positive exponent. So one over negative two, negative two to the third power, to the positive third power. And this is equal to, this is equal to one over negative two, negative two times negative two times negative two, times negative two, or you could do it as one times negative two times negative two times negative two, which is going to give you 1 over negative 8 or negative 1 eighth. Let me scroll over a little bit. I don't want to have to start squinching things. So this is equal to negative 1 eighth. Let's do one more example, just in an attempt to confuse ourselves."}, {"video_title": "Negative exponents   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "Let me scroll over a little bit. I don't want to have to start squinching things. So this is equal to negative 1 eighth. Let's do one more example, just in an attempt to confuse ourselves. Let's take 5 eighths and raise this to the negative 2 power. So once again, this negative, oh, I got a fraction. This is a negative here."}, {"video_title": "Negative exponents   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "Let's do one more example, just in an attempt to confuse ourselves. Let's take 5 eighths and raise this to the negative 2 power. So once again, this negative, oh, I got a fraction. This is a negative here. Remember, this just means 1 over 5 eighths to the second power. So this is just going to be the same thing as 1 over 5 eighths squared, which is going to be the same thing. So this is going to be equal to 1 over 5 eighths times 5 eighths, which is 25 over 64."}, {"video_title": "Negative exponents   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "This is a negative here. Remember, this just means 1 over 5 eighths to the second power. So this is just going to be the same thing as 1 over 5 eighths squared, which is going to be the same thing. So this is going to be equal to 1 over 5 eighths times 5 eighths, which is 25 over 64. 1 over 25 over 64 is just going to be 64 over 25. So another way to think about it is you're going to take the reciprocal of this and raise it to the positive exponent. So another way you could have thought about this is 5 eighths to the negative 2 power."}, {"video_title": "Negative exponents   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "So this is going to be equal to 1 over 5 eighths times 5 eighths, which is 25 over 64. 1 over 25 over 64 is just going to be 64 over 25. So another way to think about it is you're going to take the reciprocal of this and raise it to the positive exponent. So another way you could have thought about this is 5 eighths to the negative 2 power. Let me just take the reciprocal of this, 8 fifths, and raise it to the positive 2 power. So all of these statements are equivalent. And that would have applied even when you are dealing with non-fractions as your base right over here."}, {"video_title": "Negative exponents   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "So another way you could have thought about this is 5 eighths to the negative 2 power. Let me just take the reciprocal of this, 8 fifths, and raise it to the positive 2 power. So all of these statements are equivalent. And that would have applied even when you are dealing with non-fractions as your base right over here. So 2, you could say, well, this is going to be the same thing. 2 to the negative 4 is going to be the same thing as taking my reciprocal. So this is going to be the same thing as taking the reciprocal of 2, which is 1 over 2, and raising it to the positive 4 power."}, {"video_title": "How to evaluate expressions with two variables   6th grade   Khan Academy.mp3", "Sentence": "And I want to evaluate this expression when a is equal to 7 and b is equal to 2. And I encourage you to pause this and try this on your own. Well, wherever we see the a, we would just replace it with the 7. And wherever we see the b, we'd replace it with the 2. So when a equals 7 and b equals 2, this expression will be 7 plus 2, which, of course, is equal to 9. So this expression would be equal to 9 in this circumstance. Let's do a slightly more complicated one."}, {"video_title": "How to evaluate expressions with two variables   6th grade   Khan Academy.mp3", "Sentence": "And wherever we see the b, we'd replace it with the 2. So when a equals 7 and b equals 2, this expression will be 7 plus 2, which, of course, is equal to 9. So this expression would be equal to 9 in this circumstance. Let's do a slightly more complicated one. Let's say we have the expression x times y minus y plus x. Actually, let's make it plus 3x, or another way of saying it, plus 3 times x. So let's evaluate this when x is equal to 3 and y is equal to 2."}, {"video_title": "How to evaluate expressions with two variables   6th grade   Khan Academy.mp3", "Sentence": "Let's do a slightly more complicated one. Let's say we have the expression x times y minus y plus x. Actually, let's make it plus 3x, or another way of saying it, plus 3 times x. So let's evaluate this when x is equal to 3 and y is equal to 2. And once again, I encourage you to pause this video and try this on your own. Well, everywhere we see an x, let's replace it with a 3. Every place we see a y, let's replace it with a 2."}, {"video_title": "How to evaluate expressions with two variables   6th grade   Khan Academy.mp3", "Sentence": "So let's evaluate this when x is equal to 3 and y is equal to 2. And once again, I encourage you to pause this video and try this on your own. Well, everywhere we see an x, let's replace it with a 3. Every place we see a y, let's replace it with a 2. So this is going to be equal to 3 times y. And y is 2 in this case. 3 times 2 minus 2 plus this 3 times x."}, {"video_title": "How to evaluate expressions with two variables   6th grade   Khan Academy.mp3", "Sentence": "Every place we see a y, let's replace it with a 2. So this is going to be equal to 3 times y. And y is 2 in this case. 3 times 2 minus 2 plus this 3 times x. But x is also now equal to 3. So what is this going to be equal to? Well, this is going to be equal to 3 times 2 is 6."}, {"video_title": "How to evaluate expressions with two variables   6th grade   Khan Academy.mp3", "Sentence": "3 times 2 minus 2 plus this 3 times x. But x is also now equal to 3. So what is this going to be equal to? Well, this is going to be equal to 3 times 2 is 6. This 3 times 3 is 9. So it simplifies to 6 minus 2, which is 4, plus 9, which is equal to 13. So in this case, it is equal to 13."}, {"video_title": "Graphing exponential functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "We'll just try out some values for x and see what we get for y, and then we'll plot those coordinates. So let's try some negative and some positive values, and I'll try to center them around 0. So this will be my x values, this will be my y values. Let's start first with something reasonably negative, but not too negative. So let's say we start with x is equal to negative 2. Then y is equal to 5 to the x power, or 5 to the negative 2 power, which we know is the same thing as 1 over 5 to the positive 2 power, which is just 1 25th. Now let's try another value."}, {"video_title": "Graphing exponential functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "Let's start first with something reasonably negative, but not too negative. So let's say we start with x is equal to negative 2. Then y is equal to 5 to the x power, or 5 to the negative 2 power, which we know is the same thing as 1 over 5 to the positive 2 power, which is just 1 25th. Now let's try another value. What happens when x is equal to negative 1? Then y is 5 to the negative 1 power, which is the same thing as 1 over 5 to the first power, or just 1 5th. Now let's think about what x is equal to 0."}, {"video_title": "Graphing exponential functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "Now let's try another value. What happens when x is equal to negative 1? Then y is 5 to the negative 1 power, which is the same thing as 1 over 5 to the first power, or just 1 5th. Now let's think about what x is equal to 0. Then y is going to be equal to 5 to the 0th power, which we know anything to the 0th power is going to be equal to 1. So this is going to be equal to 1. And then finally we have, well actually let's try a couple more points here."}, {"video_title": "Graphing exponential functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "Now let's think about what x is equal to 0. Then y is going to be equal to 5 to the 0th power, which we know anything to the 0th power is going to be equal to 1. So this is going to be equal to 1. And then finally we have, well actually let's try a couple more points here. Let's try out, let me extend this table a little bit further, let's try out x is equal to 1. Then y is 5 to the first power, which is just equal to 5. Let's do one last value over here."}, {"video_title": "Graphing exponential functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "And then finally we have, well actually let's try a couple more points here. Let's try out, let me extend this table a little bit further, let's try out x is equal to 1. Then y is 5 to the first power, which is just equal to 5. Let's do one last value over here. Let's see what happens when x is equal to 2. Then y is 5 squared, 5 to the second power, which is just equal to 25. And now we can plot it to see how this actually looks."}, {"video_title": "Graphing exponential functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "Let's do one last value over here. Let's see what happens when x is equal to 2. Then y is 5 squared, 5 to the second power, which is just equal to 25. And now we can plot it to see how this actually looks. Let's see how it actually looks. Let me get some graph paper going here. My x's go as low as negative 2, as high as positive 2, and then my y's go all the way from 1 25th all the way to 25."}, {"video_title": "Graphing exponential functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "And now we can plot it to see how this actually looks. Let's see how it actually looks. Let me get some graph paper going here. My x's go as low as negative 2, as high as positive 2, and then my y's go all the way from 1 25th all the way to 25. So I have positive values over here. Let me draw it like this. This could be my x axis."}, {"video_title": "Graphing exponential functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "My x's go as low as negative 2, as high as positive 2, and then my y's go all the way from 1 25th all the way to 25. So I have positive values over here. Let me draw it like this. This could be my x axis. And then let's make this my y axis. Draw it as neatly as I can. Let's make that my y axis."}, {"video_title": "Graphing exponential functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "This could be my x axis. And then let's make this my y axis. Draw it as neatly as I can. Let's make that my y axis. And my x values, this could be negative 2. I actually make my y axis keep going. So that's y."}, {"video_title": "Graphing exponential functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "Let's make that my y axis. And my x values, this could be negative 2. I actually make my y axis keep going. So that's y. This is x. That's negative 2. That's negative 1."}, {"video_title": "Graphing exponential functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "So that's y. This is x. That's negative 2. That's negative 1. That's 0. That is 1. And that is positive 2."}, {"video_title": "Graphing exponential functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "That's negative 1. That's 0. That is 1. And that is positive 2. And let's plot the points. x is negative 2. y is 1 25th. Actually, let me make the scale on the y axis."}, {"video_title": "Graphing exponential functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "And that is positive 2. And let's plot the points. x is negative 2. y is 1 25th. Actually, let me make the scale on the y axis. So let's make this, so we're going to go all the way to 25. So let's say that this is 5. Actually, I have to do it a little bit smaller than that too."}, {"video_title": "Graphing exponential functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "Actually, let me make the scale on the y axis. So let's make this, so we're going to go all the way to 25. So let's say that this is 5. Actually, I have to do it a little bit smaller than that too. So this is going to be 5, 10, 15, 20. And then 25 would be right where I wrote the y, give or take. So now let's plot them."}, {"video_title": "Graphing exponential functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "Actually, I have to do it a little bit smaller than that too. So this is going to be 5, 10, 15, 20. And then 25 would be right where I wrote the y, give or take. So now let's plot them. Negative 2, 1 25th. 1 is going to be like there. So 1 25th is going to be really, really close to the x axis."}, {"video_title": "Graphing exponential functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "So now let's plot them. Negative 2, 1 25th. 1 is going to be like there. So 1 25th is going to be really, really close to the x axis. That's about 1 25th. So that is negative 2, 1 25th. It's not going to be on the x axis."}, {"video_title": "Graphing exponential functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "So 1 25th is going to be really, really close to the x axis. That's about 1 25th. So that is negative 2, 1 25th. It's not going to be on the x axis. 1 25th is obviously greater than 0. So it's going to be really, really, really, really close. Now let's do this point here in orange."}, {"video_title": "Graphing exponential functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "It's not going to be on the x axis. 1 25th is obviously greater than 0. So it's going to be really, really, really, really close. Now let's do this point here in orange. Negative 1, 1 5th. 1 5th on this scale is still pretty close. It's pretty close."}, {"video_title": "Graphing exponential functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "Now let's do this point here in orange. Negative 1, 1 5th. 1 5th on this scale is still pretty close. It's pretty close. So that right over there is negative 1, 1 5th. And now in blue, we have 0, 1. 0, 1 is going to be right about there."}, {"video_title": "Graphing exponential functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "It's pretty close. So that right over there is negative 1, 1 5th. And now in blue, we have 0, 1. 0, 1 is going to be right about there. If this is 2 and 1 half, that looks about right for 1. And then we have 1, 5. 1, 5 puts us right over there."}, {"video_title": "Graphing exponential functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "0, 1 is going to be right about there. If this is 2 and 1 half, that looks about right for 1. And then we have 1, 5. 1, 5 puts us right over there. And then finally we have 2, 25. When x is 2, y is 25. 2, 25 puts us right about there."}, {"video_title": "Graphing exponential functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "1, 5 puts us right over there. And then finally we have 2, 25. When x is 2, y is 25. 2, 25 puts us right about there. And so I think you see what happens with this function, with this graph. It starts off the further negative in the negative x, the further in the negative direction we go, 5 to ever increasing negative powers gets us closer and closer to 0, but never quite. So we're leaving 0, getting slightly further, further, further from 0."}, {"video_title": "Graphing exponential functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "2, 25 puts us right about there. And so I think you see what happens with this function, with this graph. It starts off the further negative in the negative x, the further in the negative direction we go, 5 to ever increasing negative powers gets us closer and closer to 0, but never quite. So we're leaving 0, getting slightly further, further, further from 0. Right at the y-axis, we have y equal 1. And then once the exponent over here, or right at x is equal to 0, we have y is equal to 1. And then once x starts increasing beyond 0, then we start seeing what the exponential is good at, which is just this very rapid increase."}, {"video_title": "Graphing exponential functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "So we're leaving 0, getting slightly further, further, further from 0. Right at the y-axis, we have y equal 1. And then once the exponent over here, or right at x is equal to 0, we have y is equal to 1. And then once x starts increasing beyond 0, then we start seeing what the exponential is good at, which is just this very rapid increase. Some people would call it an exponential increase, which is obviously the case right over here. So then if I just keep this curve going, you see it's just going on kind of this, sometimes called a hockey stick. It just keeps on going up like this at a super fast rate, ever increasing rate."}, {"video_title": "Graphing exponential functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "And then once x starts increasing beyond 0, then we start seeing what the exponential is good at, which is just this very rapid increase. Some people would call it an exponential increase, which is obviously the case right over here. So then if I just keep this curve going, you see it's just going on kind of this, sometimes called a hockey stick. It just keeps on going up like this at a super fast rate, ever increasing rate. So you could keep going forever to the left and get closer and closer and closer to 0 without quite getting to 0. So 5 to the negative billionth power is still not going to get you to 0, but it's going to get you pretty darn close to 0. But obviously if you go to 5 to the positive billionth power, you're going to get to a super huge number that's going to keep skyrocketing up like that."}, {"video_title": "Graphing exponential functions   Exponential and logarithmic functions   Algebra II   Khan Academy.mp3", "Sentence": "It just keeps on going up like this at a super fast rate, ever increasing rate. So you could keep going forever to the left and get closer and closer and closer to 0 without quite getting to 0. So 5 to the negative billionth power is still not going to get you to 0, but it's going to get you pretty darn close to 0. But obviously if you go to 5 to the positive billionth power, you're going to get to a super huge number that's going to keep skyrocketing up like that. So let me just draw the whole curve, just make sure you see it. Over here I'm not actually on 0, although the way I drew it, it might look like that. I'm slightly above 0, I'm increasing above that, increasing above that, and once I get into the positive x's, then I start really, really shooting up."}, {"video_title": "Writing equations in all forms   Algebra I   Khan Academy.mp3", "Sentence": "And the way to think about these, these are just three different ways of writing the same equation. So if you give me one of them, we can manipulate it to get any of the other ones. But just so you know what these are, point-slope form, let's say that x, let's say the point x1, y1, let's say that that is a point on the line. And when someone puts this little subscript here, so if they just write an x, that means that we're talking about a variable that can take on any value. If someone writes x with a subscript 1 and a y with a subscript 1, that's like saying a particular value of x and a particular value of y or a particular coordinate. And you'll see that when we do the example. But point-slope form says that, look, if I know a particular point and if I know the slope of the line, then putting that line in point-slope form would be y minus y1 is equal to m times x minus x1."}, {"video_title": "Writing equations in all forms   Algebra I   Khan Academy.mp3", "Sentence": "And when someone puts this little subscript here, so if they just write an x, that means that we're talking about a variable that can take on any value. If someone writes x with a subscript 1 and a y with a subscript 1, that's like saying a particular value of x and a particular value of y or a particular coordinate. And you'll see that when we do the example. But point-slope form says that, look, if I know a particular point and if I know the slope of the line, then putting that line in point-slope form would be y minus y1 is equal to m times x minus x1. So for example, and we'll see do that in this video, if the point negative 3, 6 is on the line, then we'd say y minus 6 is equal to m times x minus negative 3. So it'll end up becoming x plus 3. So this is a particular x and a particular y."}, {"video_title": "Writing equations in all forms   Algebra I   Khan Academy.mp3", "Sentence": "But point-slope form says that, look, if I know a particular point and if I know the slope of the line, then putting that line in point-slope form would be y minus y1 is equal to m times x minus x1. So for example, and we'll see do that in this video, if the point negative 3, 6 is on the line, then we'd say y minus 6 is equal to m times x minus negative 3. So it'll end up becoming x plus 3. So this is a particular x and a particular y. It could be a negative 3 and 6. So that's point-slope form. Slope-intercept form is y is equal to mx plus b, where once again, m is the slope, b is the y-intercept."}, {"video_title": "Writing equations in all forms   Algebra I   Khan Academy.mp3", "Sentence": "So this is a particular x and a particular y. It could be a negative 3 and 6. So that's point-slope form. Slope-intercept form is y is equal to mx plus b, where once again, m is the slope, b is the y-intercept. Where does the line intersect the y-axis? What value does y take on when x is 0? And then standard form is the form ax plus by is equal to c, where these are just two numbers, essentially."}, {"video_title": "Writing equations in all forms   Algebra I   Khan Academy.mp3", "Sentence": "Slope-intercept form is y is equal to mx plus b, where once again, m is the slope, b is the y-intercept. Where does the line intersect the y-axis? What value does y take on when x is 0? And then standard form is the form ax plus by is equal to c, where these are just two numbers, essentially. They really don't have any interpretation directly on the graph. So let's do this. Let's figure out all of these forms."}, {"video_title": "Writing equations in all forms   Algebra I   Khan Academy.mp3", "Sentence": "And then standard form is the form ax plus by is equal to c, where these are just two numbers, essentially. They really don't have any interpretation directly on the graph. So let's do this. Let's figure out all of these forms. So the first thing we want to do is figure out the slope. Once we figure out the slope, then point-slope form is actually very, very, very straightforward to calculate. So just to remind ourselves, slope, which is equal to m, which is going to be equal to the change in y over the change in x."}, {"video_title": "Writing equations in all forms   Algebra I   Khan Academy.mp3", "Sentence": "Let's figure out all of these forms. So the first thing we want to do is figure out the slope. Once we figure out the slope, then point-slope form is actually very, very, very straightforward to calculate. So just to remind ourselves, slope, which is equal to m, which is going to be equal to the change in y over the change in x. Now, what is the change in y? If we view this as our endpoint, if we imagine that we're going from here to that point, what is the change in y? Well, we have our endpoint, which is 0. y ends up at 0, and y was at 6."}, {"video_title": "Writing equations in all forms   Algebra I   Khan Academy.mp3", "Sentence": "So just to remind ourselves, slope, which is equal to m, which is going to be equal to the change in y over the change in x. Now, what is the change in y? If we view this as our endpoint, if we imagine that we're going from here to that point, what is the change in y? Well, we have our endpoint, which is 0. y ends up at 0, and y was at 6. So our finishing y point is 0. Our starting y point is 6. What was our finishing x point, or x coordinate?"}, {"video_title": "Writing equations in all forms   Algebra I   Khan Academy.mp3", "Sentence": "Well, we have our endpoint, which is 0. y ends up at 0, and y was at 6. So our finishing y point is 0. Our starting y point is 6. What was our finishing x point, or x coordinate? Our finishing x coordinate was 6. Let me make this very clear. I don't want to confuse you."}, {"video_title": "Writing equations in all forms   Algebra I   Khan Academy.mp3", "Sentence": "What was our finishing x point, or x coordinate? Our finishing x coordinate was 6. Let me make this very clear. I don't want to confuse you. So this 0, we have that 0. That is that 0 right there. And then we have this 6, which was our starting y point."}, {"video_title": "Writing equations in all forms   Algebra I   Khan Academy.mp3", "Sentence": "I don't want to confuse you. So this 0, we have that 0. That is that 0 right there. And then we have this 6, which was our starting y point. That is that 6 right there. And then we want our finishing x value. That is that 6 right there."}, {"video_title": "Writing equations in all forms   Algebra I   Khan Academy.mp3", "Sentence": "And then we have this 6, which was our starting y point. That is that 6 right there. And then we want our finishing x value. That is that 6 right there. And we want to subtract from that our starting x value. Well, our starting x value is that right over there. That's that negative 3."}, {"video_title": "Writing equations in all forms   Algebra I   Khan Academy.mp3", "Sentence": "That is that 6 right there. And we want to subtract from that our starting x value. Well, our starting x value is that right over there. That's that negative 3. And just to make sure we know what we're doing, this negative 3 is that negative 3 right there. I'm just saying, if we go from that point to that point, our y went down by 6. We went from 6 to 0."}, {"video_title": "Writing equations in all forms   Algebra I   Khan Academy.mp3", "Sentence": "That's that negative 3. And just to make sure we know what we're doing, this negative 3 is that negative 3 right there. I'm just saying, if we go from that point to that point, our y went down by 6. We went from 6 to 0. Our y went down by 6. So we get 0 minus 6 is negative 6. That makes sense."}, {"video_title": "Writing equations in all forms   Algebra I   Khan Academy.mp3", "Sentence": "We went from 6 to 0. Our y went down by 6. So we get 0 minus 6 is negative 6. That makes sense. y went down by 6. And if we went from that point to that point, what happened to x? We went from negative 3 to 6."}, {"video_title": "Writing equations in all forms   Algebra I   Khan Academy.mp3", "Sentence": "That makes sense. y went down by 6. And if we went from that point to that point, what happened to x? We went from negative 3 to 6. It should go up by 9. And if you calculate this, take your 6 minus negative 3. That's the same thing as 6 plus 3."}, {"video_title": "Writing equations in all forms   Algebra I   Khan Academy.mp3", "Sentence": "We went from negative 3 to 6. It should go up by 9. And if you calculate this, take your 6 minus negative 3. That's the same thing as 6 plus 3. That is 9. And what is negative 6 9? Well, if you simplify it, it is negative 2 3rds."}, {"video_title": "Writing equations in all forms   Algebra I   Khan Academy.mp3", "Sentence": "That's the same thing as 6 plus 3. That is 9. And what is negative 6 9? Well, if you simplify it, it is negative 2 3rds. You divide the numerator and the denominator by 3. So that is our slope, negative 2 3rds. So we're pretty much ready to use point-slope form."}, {"video_title": "Writing equations in all forms   Algebra I   Khan Academy.mp3", "Sentence": "Well, if you simplify it, it is negative 2 3rds. You divide the numerator and the denominator by 3. So that is our slope, negative 2 3rds. So we're pretty much ready to use point-slope form. We have a point. We could pick one of these points. I'll just go with the negative 3 6."}, {"video_title": "Writing equations in all forms   Algebra I   Khan Academy.mp3", "Sentence": "So we're pretty much ready to use point-slope form. We have a point. We could pick one of these points. I'll just go with the negative 3 6. And we have our slope. So let's put it in point-slope form. So point-slope form."}, {"video_title": "Writing equations in all forms   Algebra I   Khan Academy.mp3", "Sentence": "I'll just go with the negative 3 6. And we have our slope. So let's put it in point-slope form. So point-slope form. All we have to do is we say y minus. Now, we could have taken either of these points. I'll take this one."}, {"video_title": "Writing equations in all forms   Algebra I   Khan Academy.mp3", "Sentence": "So point-slope form. All we have to do is we say y minus. Now, we could have taken either of these points. I'll take this one. So y minus the y value over here. So y minus 6 is equal to our slope, which is negative 2 3rds times x minus our x coordinate. Well, our x coordinate."}, {"video_title": "Writing equations in all forms   Algebra I   Khan Academy.mp3", "Sentence": "I'll take this one. So y minus the y value over here. So y minus 6 is equal to our slope, which is negative 2 3rds times x minus our x coordinate. Well, our x coordinate. So x minus our x coordinate is negative 3. And we're done. We can simplify it a little bit."}, {"video_title": "Writing equations in all forms   Algebra I   Khan Academy.mp3", "Sentence": "Well, our x coordinate. So x minus our x coordinate is negative 3. And we're done. We can simplify it a little bit. This becomes y minus 6 is equal to negative 2 3rds times x. x minus negative 3 is the same thing as x plus 3. This is our point-slope form. Now, we can literally just algebraically manipulate this guy right here to put it into our slope-intercept form."}, {"video_title": "Writing equations in all forms   Algebra I   Khan Academy.mp3", "Sentence": "We can simplify it a little bit. This becomes y minus 6 is equal to negative 2 3rds times x. x minus negative 3 is the same thing as x plus 3. This is our point-slope form. Now, we can literally just algebraically manipulate this guy right here to put it into our slope-intercept form. Let's do that. So let's do slope-intercept in orange. So we have slope-intercept."}, {"video_title": "Writing equations in all forms   Algebra I   Khan Academy.mp3", "Sentence": "Now, we can literally just algebraically manipulate this guy right here to put it into our slope-intercept form. Let's do that. So let's do slope-intercept in orange. So we have slope-intercept. So what can we do here to simplify this? Well, we can multiply out the negative 2 3rds. So you get y minus 6 is equal to."}, {"video_title": "Writing equations in all forms   Algebra I   Khan Academy.mp3", "Sentence": "So we have slope-intercept. So what can we do here to simplify this? Well, we can multiply out the negative 2 3rds. So you get y minus 6 is equal to. I'm just distributing the negative 2 3rds. So negative 2 3rds times x is negative 2 3rds x. And then negative 2 3rds times 3 is negative 2."}, {"video_title": "Writing equations in all forms   Algebra I   Khan Academy.mp3", "Sentence": "So you get y minus 6 is equal to. I'm just distributing the negative 2 3rds. So negative 2 3rds times x is negative 2 3rds x. And then negative 2 3rds times 3 is negative 2. And now to get it in slope-intercept form, we just have to add the 6 to both sides so we get rid of it on the left-hand side. So let's add 6 to both sides of this equation. Left-hand side of the equation, we're just left with a y."}, {"video_title": "Writing equations in all forms   Algebra I   Khan Academy.mp3", "Sentence": "And then negative 2 3rds times 3 is negative 2. And now to get it in slope-intercept form, we just have to add the 6 to both sides so we get rid of it on the left-hand side. So let's add 6 to both sides of this equation. Left-hand side of the equation, we're just left with a y. These guys cancel out. You get a y is equal to negative 2 3rds x. Negative 2 plus 6 is plus 4."}, {"video_title": "Writing equations in all forms   Algebra I   Khan Academy.mp3", "Sentence": "Left-hand side of the equation, we're just left with a y. These guys cancel out. You get a y is equal to negative 2 3rds x. Negative 2 plus 6 is plus 4. So there you have it. That is our slope-intercept form, mx plus b. That's our y-intercept."}, {"video_title": "Writing equations in all forms   Algebra I   Khan Academy.mp3", "Sentence": "Negative 2 plus 6 is plus 4. So there you have it. That is our slope-intercept form, mx plus b. That's our y-intercept. Now the last thing we need to do is get it into the standard form. So once again, we just have to algebraically manipulate it so that the x's and the y's are both on this side of the equation. So let's just add 2 3rds x to both sides of this equation."}, {"video_title": "Writing equations in all forms   Algebra I   Khan Academy.mp3", "Sentence": "That's our y-intercept. Now the last thing we need to do is get it into the standard form. So once again, we just have to algebraically manipulate it so that the x's and the y's are both on this side of the equation. So let's just add 2 3rds x to both sides of this equation. So let me start here. So we have y is equal to negative 2 3rds x plus 4. That's slope-intercept form."}, {"video_title": "Writing equations in all forms   Algebra I   Khan Academy.mp3", "Sentence": "So let's just add 2 3rds x to both sides of this equation. So let me start here. So we have y is equal to negative 2 3rds x plus 4. That's slope-intercept form. Let's add 2 3rds x. So plus 2 3rds x to both sides of this equation. I'm doing that so I don't have this 2 3rds x on the right-hand side, this negative 2 3rds x."}, {"video_title": "Writing equations in all forms   Algebra I   Khan Academy.mp3", "Sentence": "That's slope-intercept form. Let's add 2 3rds x. So plus 2 3rds x to both sides of this equation. I'm doing that so I don't have this 2 3rds x on the right-hand side, this negative 2 3rds x. So the left-hand side of the equation, I squenched it up a little bit, maybe more than I should have. The left-hand side of this equation is what? It is 2 3rds x, because 2 over 3x plus this y, that's my left-hand side, is equal to, these guys cancel out, is equal to 4."}, {"video_title": "Writing equations in all forms   Algebra I   Khan Academy.mp3", "Sentence": "I'm doing that so I don't have this 2 3rds x on the right-hand side, this negative 2 3rds x. So the left-hand side of the equation, I squenched it up a little bit, maybe more than I should have. The left-hand side of this equation is what? It is 2 3rds x, because 2 over 3x plus this y, that's my left-hand side, is equal to, these guys cancel out, is equal to 4. So this by itself, we are in standard form. This is the standard form of the equation. If we wanted to make it look extra clean, have no fractions here, we could multiply both sides of this equation by 3."}, {"video_title": "Writing equations in all forms   Algebra I   Khan Academy.mp3", "Sentence": "It is 2 3rds x, because 2 over 3x plus this y, that's my left-hand side, is equal to, these guys cancel out, is equal to 4. So this by itself, we are in standard form. This is the standard form of the equation. If we wanted to make it look extra clean, have no fractions here, we could multiply both sides of this equation by 3. If we do that, what do we get? 2 3rds x times 3 is just 2x. y times 3 is 3y."}, {"video_title": "Writing equations in all forms   Algebra I   Khan Academy.mp3", "Sentence": "If we wanted to make it look extra clean, have no fractions here, we could multiply both sides of this equation by 3. If we do that, what do we get? 2 3rds x times 3 is just 2x. y times 3 is 3y. And then 4 times 3 is 12. These are the same equations, I just multiplied every term by 3. If you do it to the left-hand side, you can do it to the right-hand side, or you have to do it to the right-hand side."}, {"video_title": "Converting from slope-intercept to standard form   Algebra I   Khan Academy.mp3", "Sentence": "The equation must be simplified, which means all numbers must be integers that do not share a common factor other than one. All right, we'll worry about this second part in a little bit. Let's see if we can rewrite this. So it's y equals 2 3rds x plus 4 7ths. Let me get my scratch pad out. So it's y is equal to 2 3rds x plus 4 7ths. So the way that it's written right now, this is slope-intercept form."}, {"video_title": "Converting from slope-intercept to standard form   Algebra I   Khan Academy.mp3", "Sentence": "So it's y equals 2 3rds x plus 4 7ths. Let me get my scratch pad out. So it's y is equal to 2 3rds x plus 4 7ths. So the way that it's written right now, this is slope-intercept form. It's written in the form y is equal to mx plus b, where m in this case is 2 3rds and b is 4 7ths. It's very easy to figure out what the slope and what the y-intercept is from this equation. But we want to write this in standard form, which would be the form ax plus by is equal to c. And that extra text they were saying, the equation must be simplified, which means all numbers must be integers that do not share a common factor other than one."}, {"video_title": "Converting from slope-intercept to standard form   Algebra I   Khan Academy.mp3", "Sentence": "So the way that it's written right now, this is slope-intercept form. It's written in the form y is equal to mx plus b, where m in this case is 2 3rds and b is 4 7ths. It's very easy to figure out what the slope and what the y-intercept is from this equation. But we want to write this in standard form, which would be the form ax plus by is equal to c. And that extra text they were saying, the equation must be simplified, which means all numbers must be integers that do not share a common factor other than one. That just means that a, b, and c, in the standard form they want, need to be integers, and they want them to not have any common factors. So if we got to the point of, say, 4x plus 2y is equal to 10, well, this number, this number, and this number are all divisible by 2. They all have the common factor of 2."}, {"video_title": "Converting from slope-intercept to standard form   Algebra I   Khan Academy.mp3", "Sentence": "But we want to write this in standard form, which would be the form ax plus by is equal to c. And that extra text they were saying, the equation must be simplified, which means all numbers must be integers that do not share a common factor other than one. That just means that a, b, and c, in the standard form they want, need to be integers, and they want them to not have any common factors. So if we got to the point of, say, 4x plus 2y is equal to 10, well, this number, this number, and this number are all divisible by 2. They all have the common factor of 2. So we would want to simplify it more, divide them all by 2, and then you would get to, so you divide this by 2, you get 2x. Divide this by 2, you get plus y is equal to 5. So this is the form that they're asking for, and probably because it's just easier for the site to know that this is the right answer, because there's obviously a bunch of forms in this way."}, {"video_title": "Converting from slope-intercept to standard form   Algebra I   Khan Academy.mp3", "Sentence": "They all have the common factor of 2. So we would want to simplify it more, divide them all by 2, and then you would get to, so you divide this by 2, you get 2x. Divide this by 2, you get plus y is equal to 5. So this is the form that they're asking for, and probably because it's just easier for the site to know that this is the right answer, because there's obviously a bunch of forms in this way. So let's see if we can do that. Let's see if we can write it in standard form. So the first thing I would want to do is, well, there's a bunch of ways that you could approach it."}, {"video_title": "Converting from slope-intercept to standard form   Algebra I   Khan Academy.mp3", "Sentence": "So this is the form that they're asking for, and probably because it's just easier for the site to know that this is the right answer, because there's obviously a bunch of forms in this way. So let's see if we can do that. Let's see if we can write it in standard form. So the first thing I would want to do is, well, there's a bunch of ways that you could approach it. The first thing you could try to do is, well, let's get rid of all of these, let's get rid of all of these fractions, and the best way to get rid of the fractions is to multiply by 3 and to multiply by 7. If you multiply by 3, you get rid of this fraction. You multiply by 7, you get rid of this fraction."}, {"video_title": "Converting from slope-intercept to standard form   Algebra I   Khan Academy.mp3", "Sentence": "So the first thing I would want to do is, well, there's a bunch of ways that you could approach it. The first thing you could try to do is, well, let's get rid of all of these, let's get rid of all of these fractions, and the best way to get rid of the fractions is to multiply by 3 and to multiply by 7. If you multiply by 3, you get rid of this fraction. You multiply by 7, you get rid of this fraction. So if you multiply by 3 and you multiply by 7, actually, let me just rewrite it over here. Actually, I'll show a couple of ways that we could do this. So if you multiply, so one way to do it, so we start with y is equal to 2 thirds x plus 4 sevenths."}, {"video_title": "Converting from slope-intercept to standard form   Algebra I   Khan Academy.mp3", "Sentence": "You multiply by 7, you get rid of this fraction. So if you multiply by 3 and you multiply by 7, actually, let me just rewrite it over here. Actually, I'll show a couple of ways that we could do this. So if you multiply, so one way to do it, so we start with y is equal to 2 thirds x plus 4 sevenths. So if I multiply this side by 3 and I multiply by 7, I have to do that to this side as well. So this is going to be multiplied by 3 and multiplied by 7. So the left-hand side becomes 21y."}, {"video_title": "Converting from slope-intercept to standard form   Algebra I   Khan Academy.mp3", "Sentence": "So if you multiply, so one way to do it, so we start with y is equal to 2 thirds x plus 4 sevenths. So if I multiply this side by 3 and I multiply by 7, I have to do that to this side as well. So this is going to be multiplied by 3 and multiplied by 7. So the left-hand side becomes 21y. 3 times 7, of course, is 21. We just figured that out. So we would distribute the 21."}, {"video_title": "Converting from slope-intercept to standard form   Algebra I   Khan Academy.mp3", "Sentence": "So the left-hand side becomes 21y. 3 times 7, of course, is 21. We just figured that out. So we would distribute the 21. 21 times 2 thirds. Well, let's see, 21 divided by 3 is 7, times 2 is 14. So it's going to be 14x."}, {"video_title": "Converting from slope-intercept to standard form   Algebra I   Khan Academy.mp3", "Sentence": "So we would distribute the 21. 21 times 2 thirds. Well, let's see, 21 divided by 3 is 7, times 2 is 14. So it's going to be 14x. And then 21 divided by 7 is 3, times 4 is 12. So just like that, I was able to get rid of the fractions. And now let's try to get, I want to get all of the, I want to get all the x's and y's on one side."}, {"video_title": "Converting from slope-intercept to standard form   Algebra I   Khan Academy.mp3", "Sentence": "So it's going to be 14x. And then 21 divided by 7 is 3, times 4 is 12. So just like that, I was able to get rid of the fractions. And now let's try to get, I want to get all of the, I want to get all the x's and y's on one side. So I want to get this 14x onto the left side. So let's see if I can do that. So I'm going to do that by, to get rid of this, I would want to subtract 14x."}, {"video_title": "Converting from slope-intercept to standard form   Algebra I   Khan Academy.mp3", "Sentence": "And now let's try to get, I want to get all of the, I want to get all the x's and y's on one side. So I want to get this 14x onto the left side. So let's see if I can do that. So I'm going to do that by, to get rid of this, I would want to subtract 14x. I can't just do it on the right-hand side. I have to do it on the left-hand side as well. So I want to subtract 14x."}, {"video_title": "Converting from slope-intercept to standard form   Algebra I   Khan Academy.mp3", "Sentence": "So I'm going to do that by, to get rid of this, I would want to subtract 14x. I can't just do it on the right-hand side. I have to do it on the left-hand side as well. So I want to subtract 14x. And then what am I left with? Let me give myself a little bit more space. So on the left-hand side, I have negative 14x, negative 14x plus 21y, plus 21y is equal to, let's see, I subtracted 14x to get rid of this."}, {"video_title": "Converting from slope-intercept to standard form   Algebra I   Khan Academy.mp3", "Sentence": "So I want to subtract 14x. And then what am I left with? Let me give myself a little bit more space. So on the left-hand side, I have negative 14x, negative 14x plus 21y, plus 21y is equal to, let's see, I subtracted 14x to get rid of this. And then I have just is equal to 12. Now let's see, do these, am I done? Do these share any, do 14, 21, and 12 share any common factors?"}, {"video_title": "Converting from slope-intercept to standard form   Algebra I   Khan Academy.mp3", "Sentence": "So on the left-hand side, I have negative 14x, negative 14x plus 21y, plus 21y is equal to, let's see, I subtracted 14x to get rid of this. And then I have just is equal to 12. Now let's see, do these, am I done? Do these share any, do 14, 21, and 12 share any common factors? Let's see, 14 is divisible by 2 and 7. 21 is divisible by 3 and 7. 12 is divisible by 2, 6, 3, 4."}, {"video_title": "Converting from slope-intercept to standard form   Algebra I   Khan Academy.mp3", "Sentence": "Do these share any, do 14, 21, and 12 share any common factors? Let's see, 14 is divisible by 2 and 7. 21 is divisible by 3 and 7. 12 is divisible by 2, 6, 3, 4. But all of these aren't divisible by the same number. 14, let's see, 14 is divisible by 2, so is 12, but 21 isn't. 14 is divisible by 7, so is 21, but 12 isn't."}, {"video_title": "Converting from slope-intercept to standard form   Algebra I   Khan Academy.mp3", "Sentence": "12 is divisible by 2, 6, 3, 4. But all of these aren't divisible by the same number. 14, let's see, 14 is divisible by 2, so is 12, but 21 isn't. 14 is divisible by 7, so is 21, but 12 isn't. And yeah, so, and 21 and 12 are divisible by 3, but 14 isn't. So I think this is about as simplified as I could get. If there was a common factor for all three of these numbers, then I would divide all of them the way I did in that previous example."}, {"video_title": "Converting from slope-intercept to standard form   Algebra I   Khan Academy.mp3", "Sentence": "14 is divisible by 7, so is 21, but 12 isn't. And yeah, so, and 21 and 12 are divisible by 3, but 14 isn't. So I think this is about as simplified as I could get. If there was a common factor for all three of these numbers, then I would divide all of them the way I did in that previous example. But that's not the case right over here. So it's negative 14x plus 21y is equal to 12. So let me see if I can remember that and type that in."}, {"video_title": "Converting from slope-intercept to standard form   Algebra I   Khan Academy.mp3", "Sentence": "If there was a common factor for all three of these numbers, then I would divide all of them the way I did in that previous example. But that's not the case right over here. So it's negative 14x plus 21y is equal to 12. So let me see if I can remember that and type that in. So it is, it is negative 14x plus 21y is equal to, is equal to 12. Let's see if I got it right. It worked out."}, {"video_title": "Visual introduction to parabolas.mp3", "Sentence": "Now in terms of why it is called the parabola, I've seen multiple explanations for it. It comes from Greek, para, that root word, similar to parable, you could view of something beside, alongside, something in parallel. Bola, same root as when we're talking about ballistics, throwing something. So you could interpret it as beside, alongside, something that is being thrown. Now how does that relate to curves like this? Well my brain immediately imagines, well this is the trajectory, this is the path that is a pretty good approximation for the path of things that are actually thrown. When you study physics, you will see the path, you will approximate the paths of objects being thrown as parabolas, so maybe that's where it comes from."}, {"video_title": "Visual introduction to parabolas.mp3", "Sentence": "So you could interpret it as beside, alongside, something that is being thrown. Now how does that relate to curves like this? Well my brain immediately imagines, well this is the trajectory, this is the path that is a pretty good approximation for the path of things that are actually thrown. When you study physics, you will see the path, you will approximate the paths of objects being thrown as parabolas, so maybe that's where it comes from. But there are other potential explanations for why it is actually called the parabola. It has been lost to history. But what exactly is a parabola?"}, {"video_title": "Visual introduction to parabolas.mp3", "Sentence": "When you study physics, you will see the path, you will approximate the paths of objects being thrown as parabolas, so maybe that's where it comes from. But there are other potential explanations for why it is actually called the parabola. It has been lost to history. But what exactly is a parabola? In future videos, we're gonna describe it a little bit more algebraically. In this one, we just wanna get a sense for what parabolas look like, and introduce ourselves to some terminology around a parabola. So these three curves, they are all hand-drawn versions of a parabola."}, {"video_title": "Visual introduction to parabolas.mp3", "Sentence": "But what exactly is a parabola? In future videos, we're gonna describe it a little bit more algebraically. In this one, we just wanna get a sense for what parabolas look like, and introduce ourselves to some terminology around a parabola. So these three curves, they are all hand-drawn versions of a parabola. And so you immediately notice some interesting things about them. Some of them are open upwards, like this yellow one and this pink one, and some of them are open downwards. And you will hear people say things like open, open down, open downwards, or open down, or open upwards."}, {"video_title": "Visual introduction to parabolas.mp3", "Sentence": "So these three curves, they are all hand-drawn versions of a parabola. And so you immediately notice some interesting things about them. Some of them are open upwards, like this yellow one and this pink one, and some of them are open downwards. And you will hear people say things like open, open down, open downwards, or open down, or open upwards. So it's good to know what they're talking about. And it's hopefully fairly self-explanatory. Open upwards, the parabola is open towards the top of our graph paper."}, {"video_title": "Visual introduction to parabolas.mp3", "Sentence": "And you will hear people say things like open, open down, open downwards, or open down, or open upwards. So it's good to know what they're talking about. And it's hopefully fairly self-explanatory. Open upwards, the parabola is open towards the top of our graph paper. Here it's open towards the bottom of our graph paper. This looks like a right-side up U. This looks like an upside down U right over here."}, {"video_title": "Visual introduction to parabolas.mp3", "Sentence": "Open upwards, the parabola is open towards the top of our graph paper. Here it's open towards the bottom of our graph paper. This looks like a right-side up U. This looks like an upside down U right over here. This pink one would be open upwards. Now another term that you'll see associated with a parabola, and once again, in the future we'll learn how to calculate these things and find them precisely, is the vertex. And the vertex you should view as the maximum or minimum point on a parabola."}, {"video_title": "Visual introduction to parabolas.mp3", "Sentence": "This looks like an upside down U right over here. This pink one would be open upwards. Now another term that you'll see associated with a parabola, and once again, in the future we'll learn how to calculate these things and find them precisely, is the vertex. And the vertex you should view as the maximum or minimum point on a parabola. So if a parabola opens upwards, like these two on the right, the vertex is the minimum point. The vertex is the minimum point right over there. And so someone said, what is the vertex of this yellow parabola?"}, {"video_title": "Visual introduction to parabolas.mp3", "Sentence": "And the vertex you should view as the maximum or minimum point on a parabola. So if a parabola opens upwards, like these two on the right, the vertex is the minimum point. The vertex is the minimum point right over there. And so someone said, what is the vertex of this yellow parabola? Well it looks like the X coordinate is 3 1\u20442. So it is 3 1\u20442. It looks like the Y coordinate, it looks like it is about negative 3 1\u20442."}, {"video_title": "Visual introduction to parabolas.mp3", "Sentence": "And so someone said, what is the vertex of this yellow parabola? Well it looks like the X coordinate is 3 1\u20442. So it is 3 1\u20442. It looks like the Y coordinate, it looks like it is about negative 3 1\u20442. And once again, once we start representing these things with equations, we'll have techniques for calculating them more precisely. But the vertex of this other upward-opening parabola, it is the minimum point, it is the low point. There is no maximum point on an upward-opening parabola."}, {"video_title": "Visual introduction to parabolas.mp3", "Sentence": "It looks like the Y coordinate, it looks like it is about negative 3 1\u20442. And once again, once we start representing these things with equations, we'll have techniques for calculating them more precisely. But the vertex of this other upward-opening parabola, it is the minimum point, it is the low point. There is no maximum point on an upward-opening parabola. It just keeps increasing as X gets larger in the positive or the negative direction. Now if your parabola opens downward, then your vertex is going to be your maximum point. Now related to the idea of a vertex is the idea of an axis of symmetry."}, {"video_title": "Visual introduction to parabolas.mp3", "Sentence": "There is no maximum point on an upward-opening parabola. It just keeps increasing as X gets larger in the positive or the negative direction. Now if your parabola opens downward, then your vertex is going to be your maximum point. Now related to the idea of a vertex is the idea of an axis of symmetry. And in general, when we're talking about, well not just two dimensions, but even three dimensions, but especially in two dimensions, you can imagine a line over which you can flip the graph, and so it folds onto itself. And so the axis of symmetry for this yellow graph right over here, for this yellow parabola, it would be this line. I could draw it a little bit better."}, {"video_title": "Visual introduction to parabolas.mp3", "Sentence": "Now related to the idea of a vertex is the idea of an axis of symmetry. And in general, when we're talking about, well not just two dimensions, but even three dimensions, but especially in two dimensions, you can imagine a line over which you can flip the graph, and so it folds onto itself. And so the axis of symmetry for this yellow graph right over here, for this yellow parabola, it would be this line. I could draw it a little bit better. It would be, it would be that line right over there. You could fold the parabola over that line and it would meet itself. And that line, I didn't draw it as neat as I should, that should go directly through the vertex."}, {"video_title": "Visual introduction to parabolas.mp3", "Sentence": "I could draw it a little bit better. It would be, it would be that line right over there. You could fold the parabola over that line and it would meet itself. And that line, I didn't draw it as neat as I should, that should go directly through the vertex. So to describe that line, you would say that line is x is equal to 3.5. Similarly, the axis of symmetry for this pink parabola, it should go through the line x equals negative one. So let me do that."}, {"video_title": "Visual introduction to parabolas.mp3", "Sentence": "And that line, I didn't draw it as neat as I should, that should go directly through the vertex. So to describe that line, you would say that line is x is equal to 3.5. Similarly, the axis of symmetry for this pink parabola, it should go through the line x equals negative one. So let me do that. That's the axis of symmetry. It goes through the vertex. And if you were to fold the parabola over it, it would meet itself."}, {"video_title": "Visual introduction to parabolas.mp3", "Sentence": "So let me do that. That's the axis of symmetry. It goes through the vertex. And if you were to fold the parabola over it, it would meet itself. The axis of symmetry for this green one, it should once again go through the vertex. It looks like it is x is equal to negative six. This is, let me write that down."}, {"video_title": "Visual introduction to parabolas.mp3", "Sentence": "And if you were to fold the parabola over it, it would meet itself. The axis of symmetry for this green one, it should once again go through the vertex. It looks like it is x is equal to negative six. This is, let me write that down. That is the axis of symmetry. Now another concept that isn't unique to parabolas but we'll talk a lot about it in the context of parabolas are intercepts. So when people say y-intercept, and you saw this when you first graphed lines, they're saying where does the graph, where does the curve intercept or intersect the y-axis?"}, {"video_title": "Visual introduction to parabolas.mp3", "Sentence": "This is, let me write that down. That is the axis of symmetry. Now another concept that isn't unique to parabolas but we'll talk a lot about it in the context of parabolas are intercepts. So when people say y-intercept, and you saw this when you first graphed lines, they're saying where does the graph, where does the curve intercept or intersect the y-axis? So the y-intercept of this yellow line would be right there. It looks like it's the point zero comma three. Zero comma three."}, {"video_title": "Visual introduction to parabolas.mp3", "Sentence": "So when people say y-intercept, and you saw this when you first graphed lines, they're saying where does the graph, where does the curve intercept or intersect the y-axis? So the y-intercept of this yellow line would be right there. It looks like it's the point zero comma three. Zero comma three. The y-intercept for the pink one is right over there. We at least on this graph paper, we don't see the y-intercept, but it eventually will intersect the y-axis. It just will be way off of this screen."}, {"video_title": "Visual introduction to parabolas.mp3", "Sentence": "Zero comma three. The y-intercept for the pink one is right over there. We at least on this graph paper, we don't see the y-intercept, but it eventually will intersect the y-axis. It just will be way off of this screen. Now you might also be familiar with the term x-intercept. And that's especially interesting with parabolas as we'll see in the future. X-intercept is where you intercept or intersect the x-axis."}, {"video_title": "Visual introduction to parabolas.mp3", "Sentence": "It just will be way off of this screen. Now you might also be familiar with the term x-intercept. And that's especially interesting with parabolas as we'll see in the future. X-intercept is where you intercept or intersect the x-axis. And here, this yellow one, you see it does it two places. And this is where it gets interesting. Lines will only intersect the x-axis once at most."}, {"video_title": "Visual introduction to parabolas.mp3", "Sentence": "X-intercept is where you intercept or intersect the x-axis. And here, this yellow one, you see it does it two places. And this is where it gets interesting. Lines will only intersect the x-axis once at most. But here, we see that a parabola can intersect the x-axis twice because it curves back around to intersect it again. And so for here, the x-intercepts are going to be the point one comma zero and six comma zero. Now you might already notice something interesting."}, {"video_title": "Visual introduction to parabolas.mp3", "Sentence": "Lines will only intersect the x-axis once at most. But here, we see that a parabola can intersect the x-axis twice because it curves back around to intersect it again. And so for here, the x-intercepts are going to be the point one comma zero and six comma zero. Now you might already notice something interesting. The x-intercepts are symmetric around the axis of symmetry. So they should be equal distant from that axis of symmetry. And you can see they indeed are."}, {"video_title": "Visual introduction to parabolas.mp3", "Sentence": "Now you might already notice something interesting. The x-intercepts are symmetric around the axis of symmetry. So they should be equal distant from that axis of symmetry. And you can see they indeed are. They're both exactly two and a half away from that axis of symmetry. And so if you know where the intercepts are, you just take, you could say the midpoint of the x-coordinates, and then you're going to have the axis of symmetry, the x-coordinate of the axis of symmetry, and the x-coordinate of the actual vertex. Similarly, the x-intercepts here, looks like it's negative, the points are negative seven comma zero and negative five comma zero."}, {"video_title": "Visual introduction to parabolas.mp3", "Sentence": "And you can see they indeed are. They're both exactly two and a half away from that axis of symmetry. And so if you know where the intercepts are, you just take, you could say the midpoint of the x-coordinates, and then you're going to have the axis of symmetry, the x-coordinate of the axis of symmetry, and the x-coordinate of the actual vertex. Similarly, the x-intercepts here, looks like it's negative, the points are negative seven comma zero and negative five comma zero. And the x-coordinate of the vertex or the line of symmetry is right in between those two points. Now it's worth noting, not every parabola is going to intersect the x-axis. Notice this pink upward opening parabola, its low point is above the x-axis."}, {"video_title": "Visual introduction to parabolas.mp3", "Sentence": "Similarly, the x-intercepts here, looks like it's negative, the points are negative seven comma zero and negative five comma zero. And the x-coordinate of the vertex or the line of symmetry is right in between those two points. Now it's worth noting, not every parabola is going to intersect the x-axis. Notice this pink upward opening parabola, its low point is above the x-axis. So it's never going to intersect the actual x-axis. So this is actually not going to have any x-intercepts. So I'll leave you there."}, {"video_title": "Visual introduction to parabolas.mp3", "Sentence": "Notice this pink upward opening parabola, its low point is above the x-axis. So it's never going to intersect the actual x-axis. So this is actually not going to have any x-intercepts. So I'll leave you there. Those are actually the core ideas or the core visual themes around parabolas. And we're going to discuss them in a lot more detail when we represent them with equations. And as you'll see, these equations are going to involve second degree terms."}, {"video_title": "Visual introduction to parabolas.mp3", "Sentence": "So I'll leave you there. Those are actually the core ideas or the core visual themes around parabolas. And we're going to discuss them in a lot more detail when we represent them with equations. And as you'll see, these equations are going to involve second degree terms. So the most simple parabola is going to be y is equal to x squared. But then you can complicate it a little bit. You could have things like y is equal to two x squared minus five x plus seven."}, {"video_title": "Solving systems of equations graphically   Algebra II   Khan Academy.mp3", "Sentence": "And this is to solve systems of equations visually. So they say right over here, graph this system of equations and solve. And they give us two equations, this first one in blue. Y is equal to 7 5ths x minus five, and then this one in green. Y is equal to 3 5ths x minus one. So let's graph each of these, and we'll do it in the corresponding color. So first let's graph this first equation."}, {"video_title": "Solving systems of equations graphically   Algebra II   Khan Academy.mp3", "Sentence": "Y is equal to 7 5ths x minus five, and then this one in green. Y is equal to 3 5ths x minus one. So let's graph each of these, and we'll do it in the corresponding color. So first let's graph this first equation. So the first thing I see is that it's y-intercept is negative five. Or another way to think about it, when x is equal to zero, y is going to be negative five. So let's try this out."}, {"video_title": "Solving systems of equations graphically   Algebra II   Khan Academy.mp3", "Sentence": "So first let's graph this first equation. So the first thing I see is that it's y-intercept is negative five. Or another way to think about it, when x is equal to zero, y is going to be negative five. So let's try this out. So when x is equal to zero, y is going to be equal to negative five. So that makes sense. And then we see its slope is 7 5ths."}, {"video_title": "Solving systems of equations graphically   Algebra II   Khan Academy.mp3", "Sentence": "So let's try this out. So when x is equal to zero, y is going to be equal to negative five. So that makes sense. And then we see its slope is 7 5ths. This was conveniently placed in slope intercept form for us. So it's rise over run. So for every time it moves five to the right, it's going to move seven up."}, {"video_title": "Solving systems of equations graphically   Algebra II   Khan Academy.mp3", "Sentence": "And then we see its slope is 7 5ths. This was conveniently placed in slope intercept form for us. So it's rise over run. So for every time it moves five to the right, it's going to move seven up. So if it moves one, two, three, four, five to the right, it's going to move seven up. One, two, three, four, five, six, seven. So it'll get right over there."}, {"video_title": "Solving systems of equations graphically   Algebra II   Khan Academy.mp3", "Sentence": "So for every time it moves five to the right, it's going to move seven up. So if it moves one, two, three, four, five to the right, it's going to move seven up. One, two, three, four, five, six, seven. So it'll get right over there. Another way you could have done it is you could have just tested out some values. You could have said, oh, when x is equal to zero, y is equal to negative five. When x is equal to five, 7 5ths times five is seven, minus five is two."}, {"video_title": "Solving systems of equations graphically   Algebra II   Khan Academy.mp3", "Sentence": "So it'll get right over there. Another way you could have done it is you could have just tested out some values. You could have said, oh, when x is equal to zero, y is equal to negative five. When x is equal to five, 7 5ths times five is seven, minus five is two. So I think we've properly graphed this top one. Let's try this bottom one right over here. So we have negative, when x is equal to zero, y is equal to negative one."}, {"video_title": "Solving systems of equations graphically   Algebra II   Khan Academy.mp3", "Sentence": "When x is equal to five, 7 5ths times five is seven, minus five is two. So I think we've properly graphed this top one. Let's try this bottom one right over here. So we have negative, when x is equal to zero, y is equal to negative one. So when x is equal to zero, y is equal to negative one. And the slope is 3 5ths. So if we move over five to the right, if we move over five to the right, we will move up three."}, {"video_title": "Solving systems of equations graphically   Algebra II   Khan Academy.mp3", "Sentence": "So we have negative, when x is equal to zero, y is equal to negative one. So when x is equal to zero, y is equal to negative one. And the slope is 3 5ths. So if we move over five to the right, if we move over five to the right, we will move up three. So we will go right over there. And it looks like they intersect right at that point, right at the point x is equal to five, y is equal to two. So we'll type in x is equal to five, y is equal to two."}, {"video_title": "Solving systems of equations graphically   Algebra II   Khan Academy.mp3", "Sentence": "So if we move over five to the right, if we move over five to the right, we will move up three. So we will go right over there. And it looks like they intersect right at that point, right at the point x is equal to five, y is equal to two. So we'll type in x is equal to five, y is equal to two. And you could even verify by substituting those values into both equations to show that it does satisfy both constraints. So let's check our answer. And it worked."}, {"video_title": "Finding intercepts from an equation   Algebra I   Khan Academy.mp3", "Sentence": "So we have to find the intercepts. And then use the intercepts to graph this line on the coordinate plane. So then graph the line. So whenever someone talks about intercepts, they're talking about where you're intersecting the x and the y axes. So let me label my axes here. So this is the x-axis, and that is the y-axis there. And when I intersect the x-axis, what's going on?"}, {"video_title": "Finding intercepts from an equation   Algebra I   Khan Academy.mp3", "Sentence": "So whenever someone talks about intercepts, they're talking about where you're intersecting the x and the y axes. So let me label my axes here. So this is the x-axis, and that is the y-axis there. And when I intersect the x-axis, what's going on? What is my y value when I'm at the x-axis? Well, my y value is 0. I'm not above or below the x-axis."}, {"video_title": "Finding intercepts from an equation   Algebra I   Khan Academy.mp3", "Sentence": "And when I intersect the x-axis, what's going on? What is my y value when I'm at the x-axis? Well, my y value is 0. I'm not above or below the x-axis. So the x-intercept is when y is equal to 0. And then by that same argument, what's the y-intercept? Well, if I'm somewhere along the y-axis, what's my x value?"}, {"video_title": "Finding intercepts from an equation   Algebra I   Khan Academy.mp3", "Sentence": "I'm not above or below the x-axis. So the x-intercept is when y is equal to 0. And then by that same argument, what's the y-intercept? Well, if I'm somewhere along the y-axis, what's my x value? Well, I'm not to the right or the left, so my x value has to be 0. So the y-intercept occurs when x is equal to 0. So to figure out the intercepts, let's set y equal to 0 in this equation and solve for x."}, {"video_title": "Finding intercepts from an equation   Algebra I   Khan Academy.mp3", "Sentence": "Well, if I'm somewhere along the y-axis, what's my x value? Well, I'm not to the right or the left, so my x value has to be 0. So the y-intercept occurs when x is equal to 0. So to figure out the intercepts, let's set y equal to 0 in this equation and solve for x. And then let's set x is equal to 0 and then solve for y. So when y is equal to 0, what does this equation become? I'll do it in orange."}, {"video_title": "Finding intercepts from an equation   Algebra I   Khan Academy.mp3", "Sentence": "So to figure out the intercepts, let's set y equal to 0 in this equation and solve for x. And then let's set x is equal to 0 and then solve for y. So when y is equal to 0, what does this equation become? I'll do it in orange. You get negative 5x plus 4y. Well, we're saying y is 0, so 4 times 0 is equal to 20. 4 times 0 is just 0, so we can just not write that."}, {"video_title": "Finding intercepts from an equation   Algebra I   Khan Academy.mp3", "Sentence": "I'll do it in orange. You get negative 5x plus 4y. Well, we're saying y is 0, so 4 times 0 is equal to 20. 4 times 0 is just 0, so we can just not write that. So let me just rewrite it. So we have negative 5x is equal to 20. We can divide both sides of this equation by negative 5."}, {"video_title": "Finding intercepts from an equation   Algebra I   Khan Academy.mp3", "Sentence": "4 times 0 is just 0, so we can just not write that. So let me just rewrite it. So we have negative 5x is equal to 20. We can divide both sides of this equation by negative 5. The negative 5's cancel out. That was the whole point behind dividing by negative 5. And we get x is equal to 20 divided by negative 5 is negative 4."}, {"video_title": "Finding intercepts from an equation   Algebra I   Khan Academy.mp3", "Sentence": "We can divide both sides of this equation by negative 5. The negative 5's cancel out. That was the whole point behind dividing by negative 5. And we get x is equal to 20 divided by negative 5 is negative 4. So when y is equal to 0, we saw that right there, x is equal to negative 4. Or if we wanted to plot that point, we always put the x-coordinate for it first. So that would be the point negative 4 comma 0."}, {"video_title": "Finding intercepts from an equation   Algebra I   Khan Academy.mp3", "Sentence": "And we get x is equal to 20 divided by negative 5 is negative 4. So when y is equal to 0, we saw that right there, x is equal to negative 4. Or if we wanted to plot that point, we always put the x-coordinate for it first. So that would be the point negative 4 comma 0. So let me graph that. So if we go 1, 2, 3, 4, that's negative 4. And then the y value is just 0, so that point is right over there."}, {"video_title": "Finding intercepts from an equation   Algebra I   Khan Academy.mp3", "Sentence": "So that would be the point negative 4 comma 0. So let me graph that. So if we go 1, 2, 3, 4, that's negative 4. And then the y value is just 0, so that point is right over there. That is the x-intercept. y is 0, x is negative 4. Notice we're intersecting the x-axis."}, {"video_title": "Finding intercepts from an equation   Algebra I   Khan Academy.mp3", "Sentence": "And then the y value is just 0, so that point is right over there. That is the x-intercept. y is 0, x is negative 4. Notice we're intersecting the x-axis. Now let's do the exact same thing for the y-intercept. Let's set x equal to 0. So if we set x is equal to 0, we have negative 5 times 0 plus 4y is equal to 20."}, {"video_title": "Finding intercepts from an equation   Algebra I   Khan Academy.mp3", "Sentence": "Notice we're intersecting the x-axis. Now let's do the exact same thing for the y-intercept. Let's set x equal to 0. So if we set x is equal to 0, we have negative 5 times 0 plus 4y is equal to 20. Well, anything times 0 is 0, so we can just put that out of the way. And remember, this was setting x is equal to 0. We're doing the y-intercept now."}, {"video_title": "Finding intercepts from an equation   Algebra I   Khan Academy.mp3", "Sentence": "So if we set x is equal to 0, we have negative 5 times 0 plus 4y is equal to 20. Well, anything times 0 is 0, so we can just put that out of the way. And remember, this was setting x is equal to 0. We're doing the y-intercept now. So this just simplifies to 4y is equal to 20. We can divide both sides of this equation by 4 to get rid of this 4 right there. And you get y is equal to 20 over 4, which is 5."}, {"video_title": "Finding intercepts from an equation   Algebra I   Khan Academy.mp3", "Sentence": "We're doing the y-intercept now. So this just simplifies to 4y is equal to 20. We can divide both sides of this equation by 4 to get rid of this 4 right there. And you get y is equal to 20 over 4, which is 5. So when x is equal to 0, y is equal to 5. So the point 0, 5 is on the graph for this line. So 0, 5, 0, x is 0, and y is 1, 2, 3, 4, 5 right over there."}, {"video_title": "Finding intercepts from an equation   Algebra I   Khan Academy.mp3", "Sentence": "And you get y is equal to 20 over 4, which is 5. So when x is equal to 0, y is equal to 5. So the point 0, 5 is on the graph for this line. So 0, 5, 0, x is 0, and y is 1, 2, 3, 4, 5 right over there. And notice, when x is 0, we're right on the y-axis. This is our y-intercept right over there. And if we graph the line, all you need is two points to graph any line."}, {"video_title": "Finding intercepts from an equation   Algebra I   Khan Academy.mp3", "Sentence": "So 0, 5, 0, x is 0, and y is 1, 2, 3, 4, 5 right over there. And notice, when x is 0, we're right on the y-axis. This is our y-intercept right over there. And if we graph the line, all you need is two points to graph any line. So we just have to connect the dots. And that is our line. So let me connect the dots, try my best to draw as straight of a line as I can."}, {"video_title": "Finding intercepts from an equation   Algebra I   Khan Academy.mp3", "Sentence": "And if we graph the line, all you need is two points to graph any line. So we just have to connect the dots. And that is our line. So let me connect the dots, try my best to draw as straight of a line as I can. Well, I could do a better job than that. To draw as straight of a line as I can. And that's the graph of this equation using the x-intercept and the y-intercept."}, {"video_title": "Finding a percentage   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "And I encourage you to pause this video and to try it out yourself. So when you're saying what percent of 16 is 4, percent is another way of saying what fraction of 16 is 4. And we just need to write it as a percent, as per 100. So if you said what fraction of 16 is 4, you would say, well, look, this is the same thing as 4 16ths, which is the same thing as 1 4th. But this is saying what fraction 4 is of 16. You'd say, well, 4 is 1 4th of 16. But it still doesn't answer our question."}, {"video_title": "Finding a percentage   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "So if you said what fraction of 16 is 4, you would say, well, look, this is the same thing as 4 16ths, which is the same thing as 1 4th. But this is saying what fraction 4 is of 16. You'd say, well, 4 is 1 4th of 16. But it still doesn't answer our question. What percent? So in order to write this as a percent, we literally have to write it as something over 100. Percent literally means per cent."}, {"video_title": "Finding a percentage   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "But it still doesn't answer our question. What percent? So in order to write this as a percent, we literally have to write it as something over 100. Percent literally means per cent. The word cent you know from cents and century. It relates to the number 100. So it's per 100."}, {"video_title": "Finding a percentage   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "Percent literally means per cent. The word cent you know from cents and century. It relates to the number 100. So it's per 100. So you could say, well, this is going to be equal to question mark over 100, the part of 100. And there's a bunch of ways that you could think about this. You could say, well, look, if in the denominator to go from 4 to 100, I have to multiply by 25."}, {"video_title": "Finding a percentage   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "So it's per 100. So you could say, well, this is going to be equal to question mark over 100, the part of 100. And there's a bunch of ways that you could think about this. You could say, well, look, if in the denominator to go from 4 to 100, I have to multiply by 25. In the numerator, I need to also multiply by 25 in order to have an equivalent fraction. So I'm also going to multiply by 25. So 1 4th is the same thing as 25 over 100."}, {"video_title": "Finding a percentage   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "You could say, well, look, if in the denominator to go from 4 to 100, I have to multiply by 25. In the numerator, I need to also multiply by 25 in order to have an equivalent fraction. So I'm also going to multiply by 25. So 1 4th is the same thing as 25 over 100. And another way of saying 25 over 100 is this is 25 per 100, or 25 per cent. So this is equal to 25 per cent. Now, there's a couple of other ways you could have thought about it."}, {"video_title": "Finding a percentage   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "So 1 4th is the same thing as 25 over 100. And another way of saying 25 over 100 is this is 25 per 100, or 25 per cent. So this is equal to 25 per cent. Now, there's a couple of other ways you could have thought about it. You could have said, well, 4 over 16, this is literally 4 divided by 16. Well, let me just do the division and convert to a decimal, which is very easy to convert to a percentage. So let's try to actually do this division right over here."}, {"video_title": "Finding a percentage   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "Now, there's a couple of other ways you could have thought about it. You could have said, well, 4 over 16, this is literally 4 divided by 16. Well, let me just do the division and convert to a decimal, which is very easy to convert to a percentage. So let's try to actually do this division right over here. So we're going to literally divide 4 by 16. Now, 16 goes into 4 0 times. 0 times 16 is 0."}, {"video_title": "Finding a percentage   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "So let's try to actually do this division right over here. So we're going to literally divide 4 by 16. Now, 16 goes into 4 0 times. 0 times 16 is 0. You subtract, and you get a 4. And we're not satisfied just having this remainder. We want to keep adding 0's to get a decimal answer right over here."}, {"video_title": "Finding a percentage   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "0 times 16 is 0. You subtract, and you get a 4. And we're not satisfied just having this remainder. We want to keep adding 0's to get a decimal answer right over here. So let's put a decimal right over here. We're going into the tenths place. And let's throw some 0's right over here."}, {"video_title": "Finding a percentage   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "We want to keep adding 0's to get a decimal answer right over here. So let's put a decimal right over here. We're going into the tenths place. And let's throw some 0's right over here. The decimal will make sure we keep track of the fact that we are now in the tenths, and then the hundredths, and then the thousandths place if we have to go that far. But let's bring another 0 down. 16 goes into 40 2 times."}, {"video_title": "Finding a percentage   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "And let's throw some 0's right over here. The decimal will make sure we keep track of the fact that we are now in the tenths, and then the hundredths, and then the thousandths place if we have to go that far. But let's bring another 0 down. 16 goes into 40 2 times. 2 times 16 is 32. You subtract, you get 8. And you could bring down another 0."}, {"video_title": "Finding a percentage   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "16 goes into 40 2 times. 2 times 16 is 32. You subtract, you get 8. And you could bring down another 0. And we have 16 goes into 80 5 times. 5 times 16 is 80. You subtract, you have no remainder, and you're done."}, {"video_title": "Finding a percentage   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "And you could bring down another 0. And we have 16 goes into 80 5 times. 5 times 16 is 80. You subtract, you have no remainder, and you're done. 4 16ths is the same thing as 0.25. Now, 0.25 is the same thing as 25 hundredths. Or this is the same thing as 25 over 100, which is the same thing as 25%."}, {"video_title": "Finding the x-intercept of a line   Algebra I   Khan Academy.mp3", "Sentence": "Or it's the x value where our graph actually intersects the x-axis. Notice right over here our y value is exactly zero. We're sitting on the x-axis. So let's think about what this x value must be. Well just looking at it from, just trying to eyeball a little bit, it looks like it's a little over two. It's between two and three. It looks like it's less than two and a half."}, {"video_title": "Finding the x-intercept of a line   Algebra I   Khan Academy.mp3", "Sentence": "So let's think about what this x value must be. Well just looking at it from, just trying to eyeball a little bit, it looks like it's a little over two. It's between two and three. It looks like it's less than two and a half. But we don't know the exact value. So let's go turn to the equation to figure out the exact value. So we essentially have to figure out what x value when y is equal to zero will have this equation be true."}, {"video_title": "Finding the x-intercept of a line   Algebra I   Khan Academy.mp3", "Sentence": "It looks like it's less than two and a half. But we don't know the exact value. So let's go turn to the equation to figure out the exact value. So we essentially have to figure out what x value when y is equal to zero will have this equation be true. So we could just say two times zero plus 3x is equal to seven. Well two times zero is just gonna be zero. So we have 3x is equal to seven."}, {"video_title": "Finding the x-intercept of a line   Algebra I   Khan Academy.mp3", "Sentence": "So we essentially have to figure out what x value when y is equal to zero will have this equation be true. So we could just say two times zero plus 3x is equal to seven. Well two times zero is just gonna be zero. So we have 3x is equal to seven. And then we can divide both sides by three to solve for x. And we get x is equal to seven over three. Now does that look like 7 3rds?"}, {"video_title": "Finding the x-intercept of a line   Algebra I   Khan Academy.mp3", "Sentence": "So we have 3x is equal to seven. And then we can divide both sides by three to solve for x. And we get x is equal to seven over three. Now does that look like 7 3rds? Well we just have to remind ourselves that seven over three is the same thing as six over three plus one over three. And six over three is two. So this is the same thing as two and 1 3rd."}, {"video_title": "Finding the x-intercept of a line   Algebra I   Khan Academy.mp3", "Sentence": "Now does that look like 7 3rds? Well we just have to remind ourselves that seven over three is the same thing as six over three plus one over three. And six over three is two. So this is the same thing as two and 1 3rd. Another way you could think about it is three goes into seven two times. And then you have a remainder of one. So you still gotta divide that one by three."}, {"video_title": "Finding the x-intercept of a line   Algebra I   Khan Academy.mp3", "Sentence": "So this is the same thing as two and 1 3rd. Another way you could think about it is three goes into seven two times. And then you have a remainder of one. So you still gotta divide that one by three. So it's two full times and then a 1 3rd. So this looks like two and 1 3rd. And so that's its x intercept."}, {"video_title": "Finding the x-intercept of a line   Algebra I   Khan Academy.mp3", "Sentence": "So you still gotta divide that one by three. So it's two full times and then a 1 3rd. So this looks like two and 1 3rd. And so that's its x intercept. Seven 7 3rds. If I was doing this on the exercise on Khan Academy, it's always a little easier to type in the improper fraction. So I would put in 7 3rds."}, {"video_title": "Multiplying decimals example   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "And then you count the number of spaces behind the decimal you have in your two numbers you're multiplying and you're going to have that many spaces in your product. And let me show you what I'm talking about. So let's just multiply these two characters. So we have 32.12 times 0.5. And when you write them out, you can just push both of them all the way to the right. You can almost ignore the decimal. Right now, you should write the decimal where they belong."}, {"video_title": "Multiplying decimals example   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "So we have 32.12 times 0.5. And when you write them out, you can just push both of them all the way to the right. You can almost ignore the decimal. Right now, you should write the decimal where they belong. But you can almost pretend that this is 3,212 times 5. And then we'll worry about the decimals in a second. So let's get started."}, {"video_title": "Multiplying decimals example   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "Right now, you should write the decimal where they belong. But you can almost pretend that this is 3,212 times 5. And then we'll worry about the decimals in a second. So let's get started. So if we were just multiplying 5 times 3,212, we would say, well, 5 times 2 is 10. Regroup the 1. 5 times 1 is 5."}, {"video_title": "Multiplying decimals example   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "So let's get started. So if we were just multiplying 5 times 3,212, we would say, well, 5 times 2 is 10. Regroup the 1. 5 times 1 is 5. Plus 1 is 6. 5 times 2 is 10. Regroup the 1."}, {"video_title": "Multiplying decimals example   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "5 times 1 is 5. Plus 1 is 6. 5 times 2 is 10. Regroup the 1. And then finally, you have 5 times 3 is 15. Plus 1 is 16. And then we don't have any other places."}, {"video_title": "Multiplying decimals example   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "Regroup the 1. And then finally, you have 5 times 3 is 15. Plus 1 is 16. And then we don't have any other places. This 0 really isn't. If we were just doing this as 0,5, we wouldn't multiply 0 times this whole thing. We'd just get 0 anyway."}, {"video_title": "Multiplying decimals example   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "And then we don't have any other places. This 0 really isn't. If we were just doing this as 0,5, we wouldn't multiply 0 times this whole thing. We'd just get 0 anyway. So just 5 times 3,212 gives us this number. But now we want to care about the decimals. We just have to count the total number of spaces or places we have behind the decimal points in the two numbers we're multiplying."}, {"video_title": "Multiplying decimals example   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "We'd just get 0 anyway. So just 5 times 3,212 gives us this number. But now we want to care about the decimals. We just have to count the total number of spaces or places we have behind the decimal points in the two numbers we're multiplying. So we have 1, 2, 3 spaces or 3 numbers to the right of the decimals in the two numbers that we're multiplying. So we need that many numbers to the right of the decimal in our answer. So we go 1, 2, 3."}, {"video_title": "Multiplying decimals example   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "We just have to count the total number of spaces or places we have behind the decimal points in the two numbers we're multiplying. So we have 1, 2, 3 spaces or 3 numbers to the right of the decimals in the two numbers that we're multiplying. So we need that many numbers to the right of the decimal in our answer. So we go 1, 2, 3. Put the decimal right over there. So 32.12 times 0.5 is 16.060. And this trailing 0 right here, we can ignore because it's really not adding any information there."}, {"video_title": "Multiplying decimals example   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "So we go 1, 2, 3. Put the decimal right over there. So 32.12 times 0.5 is 16.060. And this trailing 0 right here, we can ignore because it's really not adding any information there. So we can just write this as 16.06. Now the last thing you want to do is just kind of make sure that this makes sense. You have a number that's almost 32 and we're multiplying it by 0.5."}, {"video_title": "Multiplying decimals example   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "And this trailing 0 right here, we can ignore because it's really not adding any information there. So we can just write this as 16.06. Now the last thing you want to do is just kind of make sure that this makes sense. You have a number that's almost 32 and we're multiplying it by 0.5. Remember, 0.5 is the same thing as 5 over 10, which is the same thing as 1 half. So we're really multiplying 32.12 times 1 half. We're trying to figure out what 1 half of 32.12 is."}, {"video_title": "Extending arithmetic sequences   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "We're told the first four terms of an arithmetic sequence are given. So it goes from negative eight to negative 14 to negative 20 to negative 26. What is the fifth term in the sequence? We just need to figure out the next term. Well, an arithmetic sequence, each successive term is separated by the same amount. So when we go from negative eight to negative 14, we went down by six. And then we go down by six again to go to negative 20."}, {"video_title": "Extending arithmetic sequences   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "We just need to figure out the next term. Well, an arithmetic sequence, each successive term is separated by the same amount. So when we go from negative eight to negative 14, we went down by six. And then we go down by six again to go to negative 20. And then we go down by six again to go to negative 26. And so we're gonna go down by six again to get to negative 32. Negative 32."}, {"video_title": "Extending arithmetic sequences   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "And then we go down by six again to go to negative 20. And then we go down by six again to go to negative 26. And so we're gonna go down by six again to get to negative 32. Negative 32. Let's do a couple more of these. The first five terms of the sequence are given. What is the sixth term?"}, {"video_title": "Extending arithmetic sequences   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "Negative 32. Let's do a couple more of these. The first five terms of the sequence are given. What is the sixth term? Well, let's see, to go from two to negative one, you subtracted three. Two minus three is negative one. Negative one minus three is negative four."}, {"video_title": "Extending arithmetic sequences   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "What is the sixth term? Well, let's see, to go from two to negative one, you subtracted three. Two minus three is negative one. Negative one minus three is negative four. Negative four minus three is negative seven. Negative seven minus three is negative 10. And so negative 10 minus three is negative 13."}, {"video_title": "How to evaluate an expression with variables   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "The individual cost of participating in the raffle is given by the following expression. 5t plus 3, or 5 times t plus 3, where t represents the number of tickets someone purchases. Evaluate the expression when t is equal to 1, t is equal to 8, and t is equal to 10. So let's first take the situation where t is equal to 1. Then this expression right over here becomes, and I'll use that same color, becomes 5 times 1 plus 3. And we know from order of operations, you do the multiplication before you do the addition. So this will be 5 times 1 is 5 plus 3."}, {"video_title": "How to evaluate an expression with variables   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "So let's first take the situation where t is equal to 1. Then this expression right over here becomes, and I'll use that same color, becomes 5 times 1 plus 3. And we know from order of operations, you do the multiplication before you do the addition. So this will be 5 times 1 is 5 plus 3. And then this is clearly equal to 8. Now let's do it when t is equal to 8. So when t is equal to 8, this expression becomes, and I'll do the same colors again, 5 times 8 plus 3."}, {"video_title": "How to evaluate an expression with variables   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "So this will be 5 times 1 is 5 plus 3. And then this is clearly equal to 8. Now let's do it when t is equal to 8. So when t is equal to 8, this expression becomes, and I'll do the same colors again, 5 times 8 plus 3. And once again, 5 times 8 is 40. And then we have the plus 3 there. So this is equal to 43."}, {"video_title": "How to evaluate an expression with variables   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "So when t is equal to 8, this expression becomes, and I'll do the same colors again, 5 times 8 plus 3. And once again, 5 times 8 is 40. And then we have the plus 3 there. So this is equal to 43. And so we have the last situation. With t is equal to 10, I'll do that in blue. So we have 5 times 10."}, {"video_title": "How to evaluate an expression with variables   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "So this is equal to 43. And so we have the last situation. With t is equal to 10, I'll do that in blue. So we have 5 times 10. So 5t is 5 times 10. Instead of a t, we put a 10 there. 5 times 10 plus 3."}, {"video_title": "How to evaluate an expression with variables   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "So we have 5 times 10. So 5t is 5 times 10. Instead of a t, we put a 10 there. 5 times 10 plus 3. That's a slightly differentiated green, but I think you get the idea. 5 times 10 is 50. And then we're going to have to add 3 to that."}, {"video_title": "Solving two-variable inequalities word problem   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "Ezra enjoys gardening. Every sunflower plant he waters requires 0.7 liters of water, and every lily plant he waters requires 0.5 liters of water. Ezra has a total of 11 liters of water for these plants. In the following inequality, S represents the number of sunflower plants, and L represents the number of lily plants Ezra can water. See if this makes sense. So it looks like the number of sunflower plants is S, and he would have to use 7 tenths of a liter per plant. So this first term right over here, this is how much water used to water the sunflower plants, and then the second term is how much water used to water the lily plants, because he needs half a liter for each of the L lily plants."}, {"video_title": "Solving two-variable inequalities word problem   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "In the following inequality, S represents the number of sunflower plants, and L represents the number of lily plants Ezra can water. See if this makes sense. So it looks like the number of sunflower plants is S, and he would have to use 7 tenths of a liter per plant. So this first term right over here, this is how much water used to water the sunflower plants, and then the second term is how much water used to water the lily plants, because he needs half a liter for each of the L lily plants. And he has a total of 11 liters, so this sum right over here has to be less than or equal to 11. All right. Ezra waters 10 lily plants, so that's L is equal to 10."}, {"video_title": "Solving two-variable inequalities word problem   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "So this first term right over here, this is how much water used to water the sunflower plants, and then the second term is how much water used to water the lily plants, because he needs half a liter for each of the L lily plants. And he has a total of 11 liters, so this sum right over here has to be less than or equal to 11. All right. Ezra waters 10 lily plants, so that's L is equal to 10. How many sunflower plants can he water at most with the remaining amount of water? Well, let's think about this a little bit. We have 0.7 times the number of sunflower plants he waters, plus, so we're going to assume L is equal to 10."}, {"video_title": "Solving two-variable inequalities word problem   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "Ezra waters 10 lily plants, so that's L is equal to 10. How many sunflower plants can he water at most with the remaining amount of water? Well, let's think about this a little bit. We have 0.7 times the number of sunflower plants he waters, plus, so we're going to assume L is equal to 10. He waters 10 lily plants, and each of them he gives half a liter to. So 0.5 times 10 is going to be 5. So he's going to be 5 liters watering the 10 lily plants, and this has to be less than or equal to 11."}, {"video_title": "Solving two-variable inequalities word problem   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "We have 0.7 times the number of sunflower plants he waters, plus, so we're going to assume L is equal to 10. He waters 10 lily plants, and each of them he gives half a liter to. So 0.5 times 10 is going to be 5. So he's going to be 5 liters watering the 10 lily plants, and this has to be less than or equal to 11. So we can try to isolate the S on the left-hand side. So let's do that, the number of sunflowers. So we could first subtract 5 from both sides, and we are going to be left with, on the left-hand side, we're going to just have 0.7 times S. So this expression, this is the total amount of water he's going to spend on the sunflower plants, has to be less than or equal to 11 minus 5 is equal to 6."}, {"video_title": "Solving two-variable inequalities word problem   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "So he's going to be 5 liters watering the 10 lily plants, and this has to be less than or equal to 11. So we can try to isolate the S on the left-hand side. So let's do that, the number of sunflowers. So we could first subtract 5 from both sides, and we are going to be left with, on the left-hand side, we're going to just have 0.7 times S. So this expression, this is the total amount of water he's going to spend on the sunflower plants, has to be less than or equal to 11 minus 5 is equal to 6. And so now we can divide both sides by 0.7. And we don't have to change the inequality because we are dividing by a positive number, 0.7. So we're going to get S is less than or equal to 6 divided by 0.7."}, {"video_title": "Solving two-variable inequalities word problem   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "So we could first subtract 5 from both sides, and we are going to be left with, on the left-hand side, we're going to just have 0.7 times S. So this expression, this is the total amount of water he's going to spend on the sunflower plants, has to be less than or equal to 11 minus 5 is equal to 6. And so now we can divide both sides by 0.7. And we don't have to change the inequality because we are dividing by a positive number, 0.7. So we're going to get S is less than or equal to 6 divided by 0.7. Or we could even write this as, well, this is 6 divided by 0.7. Let me write everything as a fraction. This is the same thing as 6 divided by 7 tenths, which is equal to 6 times 10 over 7, which is equal to 60 sevenths, which is equal to, what is this, 8 and 4 sevenths."}, {"video_title": "Solving two-variable inequalities word problem   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "So we're going to get S is less than or equal to 6 divided by 0.7. Or we could even write this as, well, this is 6 divided by 0.7. Let me write everything as a fraction. This is the same thing as 6 divided by 7 tenths, which is equal to 6 times 10 over 7, which is equal to 60 sevenths, which is equal to, what is this, 8 and 4 sevenths. So S has to be less than 8 and 4 sevenths. Now we're assuming that he's only, we're only thinking in terms of whole plant, integer number of plants, or actually whole numbers of plants. And so if S has to be less than 8 and 4 sevenths, how many sunflower plants can he water at most with the remaining amount of water?"}, {"video_title": "Solving two-variable inequalities word problem   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "This is the same thing as 6 divided by 7 tenths, which is equal to 6 times 10 over 7, which is equal to 60 sevenths, which is equal to, what is this, 8 and 4 sevenths. So S has to be less than 8 and 4 sevenths. Now we're assuming that he's only, we're only thinking in terms of whole plant, integer number of plants, or actually whole numbers of plants. And so if S has to be less than 8 and 4 sevenths, how many sunflower plants can he water at most with the remaining amount of water? Well, we'll say that he, at most, he can water 8 plants. We're going to assume that he can't water just a fraction of the plant. He either waters something or he doesn't."}, {"video_title": "Expression for compound or exponential growth.mp3", "Sentence": "The bank will provide 1.8% interest on the money in the account every year. Another way of saying that, the money in the savings account will grow by 1.8% per year. Write an expression that describes how much money will be in the account in 15 years. So let's just think about this a little bit. Let's just think about the starting amount. So in the start, we're just gonna put $3,800. We could view that as year zero."}, {"video_title": "Expression for compound or exponential growth.mp3", "Sentence": "So let's just think about this a little bit. Let's just think about the starting amount. So in the start, we're just gonna put $3,800. We could view that as year zero. Year, actually let me write it that way. So the start is the same thing as year zero and we're gonna start with $3,800. Now let's think about year one."}, {"video_title": "Expression for compound or exponential growth.mp3", "Sentence": "We could view that as year zero. Year, actually let me write it that way. So the start is the same thing as year zero and we're gonna start with $3,800. Now let's think about year one. How much money will we have after one year? Well, we would have the original amount that we put, $3,800 and then we're going to get the amount that we get in interest. And they say that the bank will provide 1.8% interest on the money in the account."}, {"video_title": "Expression for compound or exponential growth.mp3", "Sentence": "Now let's think about year one. How much money will we have after one year? Well, we would have the original amount that we put, $3,800 and then we're going to get the amount that we get in interest. And they say that the bank will provide 1.8% interest on the money in the account. So it'll be plus 1.8% times $3,800. And we could also write this as a decimal. This is equal to 3,800 plus, and I'll just write, I'll switch the order of multiplication here, plus 3,800 times 0.018, 1.8% is the same thing as 18,000ths or 1.8 hundredths, depending on how you want to pronounce it."}, {"video_title": "Expression for compound or exponential growth.mp3", "Sentence": "And they say that the bank will provide 1.8% interest on the money in the account. So it'll be plus 1.8% times $3,800. And we could also write this as a decimal. This is equal to 3,800 plus, and I'll just write, I'll switch the order of multiplication here, plus 3,800 times 0.018, 1.8% is the same thing as 18,000ths or 1.8 hundredths, depending on how you want to pronounce it. And so here you might say, well, there's kind of an interesting potential simplification mathematically here. I could factor 3,800 out of each of these terms. I have a 3,800 here, I have a 3,800 here."}, {"video_title": "Expression for compound or exponential growth.mp3", "Sentence": "This is equal to 3,800 plus, and I'll just write, I'll switch the order of multiplication here, plus 3,800 times 0.018, 1.8% is the same thing as 18,000ths or 1.8 hundredths, depending on how you want to pronounce it. And so here you might say, well, there's kind of an interesting potential simplification mathematically here. I could factor 3,800 out of each of these terms. I have a 3,800 here, I have a 3,800 here. So why don't I factor it out, essentially undistribute it. So this is going to be 3,800 times, when you factor it out here, you get a one, plus, when you factor it out here, you get 0.018. And so I could just rewrite this as 3,800 times 1.018."}, {"video_title": "Expression for compound or exponential growth.mp3", "Sentence": "I have a 3,800 here, I have a 3,800 here. So why don't I factor it out, essentially undistribute it. So this is going to be 3,800 times, when you factor it out here, you get a one, plus, when you factor it out here, you get 0.018. And so I could just rewrite this as 3,800 times 1.018. So this is an interesting time to pause. We're not at the full answer yet. How much do we have in 15 years?"}, {"video_title": "Expression for compound or exponential growth.mp3", "Sentence": "And so I could just rewrite this as 3,800 times 1.018. So this is an interesting time to pause. We're not at the full answer yet. How much do we have in 15 years? But we have an interesting expression for how much we have after one year. Notice that if the money is growing by 1.8%, or another way, it was growing by 0.018, that's equivalent to multiplying the amount that we started the year with, by one plus the amount that it's growing by, or 1.018. And once again, why does this make intuitive sense?"}, {"video_title": "Expression for compound or exponential growth.mp3", "Sentence": "How much do we have in 15 years? But we have an interesting expression for how much we have after one year. Notice that if the money is growing by 1.8%, or another way, it was growing by 0.018, that's equivalent to multiplying the amount that we started the year with, by one plus the amount that it's growing by, or 1.018. And once again, why does this make intuitive sense? Because at the end of the year, you're going to have the original amount that you put, that's what that one really represents, and then plus, you're gonna have the amount that you grew by. So you multiply both the sum here times the original amount you put, and that's how much you'll have at the end of year one. What about year two?"}, {"video_title": "Expression for compound or exponential growth.mp3", "Sentence": "And once again, why does this make intuitive sense? Because at the end of the year, you're going to have the original amount that you put, that's what that one really represents, and then plus, you're gonna have the amount that you grew by. So you multiply both the sum here times the original amount you put, and that's how much you'll have at the end of year one. What about year two? So year two. Well, we know what we're going to start with in year two. We're gonna start with whatever we finished year one with."}, {"video_title": "Expression for compound or exponential growth.mp3", "Sentence": "What about year two? So year two. Well, we know what we're going to start with in year two. We're gonna start with whatever we finished year one with. So we're gonna start with 3,800 times 1.018, but then it's gonna grow by 1.8%, or grow by 0.018. And we already said, if you're gonna grow by that amount, that's equivalent to multiplying it by 1.018. Well, this is the same thing as 3,800 times 1.018 to the second power."}, {"video_title": "Expression for compound or exponential growth.mp3", "Sentence": "We're gonna start with whatever we finished year one with. So we're gonna start with 3,800 times 1.018, but then it's gonna grow by 1.8%, or grow by 0.018. And we already said, if you're gonna grow by that amount, that's equivalent to multiplying it by 1.018. Well, this is the same thing as 3,800 times 1.018 to the second power. I think you see where this is going. Every time we grow by 1.8%, we're gonna multiply by 1.018. And if we're thinking about 15 years in the future, we're gonna do that 15 times."}, {"video_title": "Expression for compound or exponential growth.mp3", "Sentence": "Well, this is the same thing as 3,800 times 1.018 to the second power. I think you see where this is going. Every time we grow by 1.8%, we're gonna multiply by 1.018. And if we're thinking about 15 years in the future, we're gonna do that 15 times. So one year in the future, your exponent here is essentially one. Two years, your exponent is two. So year 15, I can just cut to the chase here."}, {"video_title": "Expression for compound or exponential growth.mp3", "Sentence": "And if we're thinking about 15 years in the future, we're gonna do that 15 times. So one year in the future, your exponent here is essentially one. Two years, your exponent is two. So year 15, I can just cut to the chase here. So year 15, well, that's just going to be, we're going to have the original amount that we invested, and we are going to grow 1.018 15 times. So we're gonna multiply by this amount 15 times to get the final amount. And one of the fun things, this is actually called compound growth, where every year you grow on top of the amount that you had before."}, {"video_title": "Expression for compound or exponential growth.mp3", "Sentence": "So year 15, I can just cut to the chase here. So year 15, well, that's just going to be, we're going to have the original amount that we invested, and we are going to grow 1.018 15 times. So we're gonna multiply by this amount 15 times to get the final amount. And one of the fun things, this is actually called compound growth, where every year you grow on top of the amount that you had before. You'll see, if you type this into a calculator, even though 1.8% per year does not seem like a lot, over 15 years, it actually would amount to a reasonable amount. But this is the expression. They're not asking us to calculate it."}, {"video_title": "Checking solutions of two-variable linear inequalities example   Algebra I   Khan Academy.mp3", "Sentence": "So we can literally just try each of these ordered pairs. We could try what happens when x is equal to 3 and y is equal to 5 in this inequality and see if it satisfies it. And then we could try it for 1 and negative 7. So let's do that first. Let's do it first for 3 and 5. So when x is 3, y is 5. Let's see if this actually gets satisfied."}, {"video_title": "Checking solutions of two-variable linear inequalities example   Algebra I   Khan Academy.mp3", "Sentence": "So let's do that first. Let's do it first for 3 and 5. So when x is 3, y is 5. Let's see if this actually gets satisfied. So we get 5 times 3 minus 3 times 5. Let's see if this is greater than or equal to 25. So 5 times 3 is 15."}, {"video_title": "Checking solutions of two-variable linear inequalities example   Algebra I   Khan Academy.mp3", "Sentence": "Let's see if this actually gets satisfied. So we get 5 times 3 minus 3 times 5. Let's see if this is greater than or equal to 25. So 5 times 3 is 15. And then from that, we're going to subtract 15. And let's see if that is greater than or equal to 25. Put that question mark there because we don't know."}, {"video_title": "Checking solutions of two-variable linear inequalities example   Algebra I   Khan Academy.mp3", "Sentence": "So 5 times 3 is 15. And then from that, we're going to subtract 15. And let's see if that is greater than or equal to 25. Put that question mark there because we don't know. And 15 minus 15, that is 0. So we get the expression 0 is greater than or equal to 25. This is not true."}, {"video_title": "Checking solutions of two-variable linear inequalities example   Algebra I   Khan Academy.mp3", "Sentence": "Put that question mark there because we don't know. And 15 minus 15, that is 0. So we get the expression 0 is greater than or equal to 25. This is not true. 0 is less than 25. So this is not true. So this ordered pair is not a solution to the inequality."}, {"video_title": "Checking solutions of two-variable linear inequalities example   Algebra I   Khan Academy.mp3", "Sentence": "This is not true. 0 is less than 25. So this is not true. So this ordered pair is not a solution to the inequality. So this is not a solution. You put in x is 3, y is 5. You get 0 is greater than or equal to 25, which is absolutely not true."}, {"video_title": "Checking solutions of two-variable linear inequalities example   Algebra I   Khan Academy.mp3", "Sentence": "So this ordered pair is not a solution to the inequality. So this is not a solution. You put in x is 3, y is 5. You get 0 is greater than or equal to 25, which is absolutely not true. Now let's try it with 1 and negative 7. So we have 5 times 1 minus 3 times negative 7 needs to be greater than or equal to 25. 5 times 1 is 5."}, {"video_title": "Checking solutions of two-variable linear inequalities example   Algebra I   Khan Academy.mp3", "Sentence": "You get 0 is greater than or equal to 25, which is absolutely not true. Now let's try it with 1 and negative 7. So we have 5 times 1 minus 3 times negative 7 needs to be greater than or equal to 25. 5 times 1 is 5. And then minus 3 times negative 7 is negative 21. So it becomes minus negative 21 needs to be greater than or equal to 25. This is the same thing as 5 plus 21."}, {"video_title": "Checking solutions of two-variable linear inequalities example   Algebra I   Khan Academy.mp3", "Sentence": "5 times 1 is 5. And then minus 3 times negative 7 is negative 21. So it becomes minus negative 21 needs to be greater than or equal to 25. This is the same thing as 5 plus 21. Subtracting a negative, same thing as adding the positive, is greater than or equal to 25. And 5 plus 21 is 26 is indeed greater than or equal to 25. So this works out."}, {"video_title": "Checking solutions of two-variable linear inequalities example   Algebra I   Khan Academy.mp3", "Sentence": "This is the same thing as 5 plus 21. Subtracting a negative, same thing as adding the positive, is greater than or equal to 25. And 5 plus 21 is 26 is indeed greater than or equal to 25. So this works out. So this is a solution. And just to see if we can visualize this a little bit better, I'm going to graph this inequality. I'm not going to show you exactly how I do it this time."}, {"video_title": "Checking solutions of two-variable linear inequalities example   Algebra I   Khan Academy.mp3", "Sentence": "So this works out. So this is a solution. And just to see if we can visualize this a little bit better, I'm going to graph this inequality. I'm not going to show you exactly how I do it this time. But I'm going to show you where these points lie relative to this solution. So first let's put this in point. So we have 5."}, {"video_title": "Checking solutions of two-variable linear inequalities example   Algebra I   Khan Academy.mp3", "Sentence": "I'm not going to show you exactly how I do it this time. But I'm going to show you where these points lie relative to this solution. So first let's put this in point. So we have 5. We just need a new color. So we have 5x. That's not a new color."}, {"video_title": "Checking solutions of two-variable linear inequalities example   Algebra I   Khan Academy.mp3", "Sentence": "So we have 5. We just need a new color. So we have 5x. That's not a new color. Having trouble switching colors today. We have 5x minus 3y is greater than or equal to 25. Let me write this inequality in kind of our slope intercept form."}, {"video_title": "Checking solutions of two-variable linear inequalities example   Algebra I   Khan Academy.mp3", "Sentence": "That's not a new color. Having trouble switching colors today. We have 5x minus 3y is greater than or equal to 25. Let me write this inequality in kind of our slope intercept form. So this would be the same thing. If we subtract 5x from both sides, we get negative 3y is greater than or equal to negative 5x plus 25. I just subtracted 5x from both sides."}, {"video_title": "Checking solutions of two-variable linear inequalities example   Algebra I   Khan Academy.mp3", "Sentence": "Let me write this inequality in kind of our slope intercept form. So this would be the same thing. If we subtract 5x from both sides, we get negative 3y is greater than or equal to negative 5x plus 25. I just subtracted 5x from both sides. So that gets eliminated. And you have a negative 5x over here. Now let's divide both sides of this equation, or I should say this inequality, by negative 3."}, {"video_title": "Checking solutions of two-variable linear inequalities example   Algebra I   Khan Academy.mp3", "Sentence": "I just subtracted 5x from both sides. So that gets eliminated. And you have a negative 5x over here. Now let's divide both sides of this equation, or I should say this inequality, by negative 3. And when you divide both sides of an inequality by a negative number, multiply or divide by a negative number, it swaps the inequality. So if you divide both sides by negative 3, you get y is less than or equal to negative 5 divided by negative 3 is 5 over 3x. And then 25 divided by negative 3 is minus 25 over 3."}, {"video_title": "Checking solutions of two-variable linear inequalities example   Algebra I   Khan Academy.mp3", "Sentence": "Now let's divide both sides of this equation, or I should say this inequality, by negative 3. And when you divide both sides of an inequality by a negative number, multiply or divide by a negative number, it swaps the inequality. So if you divide both sides by negative 3, you get y is less than or equal to negative 5 divided by negative 3 is 5 over 3x. And then 25 divided by negative 3 is minus 25 over 3. So if I were to graph this, so this is now the expression, or the inequality, y is less than or equal to 5 thirds x minus 25 over 3. So if I wanted to graph this, I'll try to draw a relatively rough graph here, but really just so that we can visualize this. So our y-intercept is negative 25 over 3."}, {"video_title": "Checking solutions of two-variable linear inequalities example   Algebra I   Khan Academy.mp3", "Sentence": "And then 25 divided by negative 3 is minus 25 over 3. So if I were to graph this, so this is now the expression, or the inequality, y is less than or equal to 5 thirds x minus 25 over 3. So if I wanted to graph this, I'll try to draw a relatively rough graph here, but really just so that we can visualize this. So our y-intercept is negative 25 over 3. That's the same thing as negative 8 and 1 third. So that's 1, 2, 3, 4, 5, 6, 7, 8, and a little bit more than 8. So our y-intercept is negative 8 and 1 third, like that, and has a slope of 5 thirds."}, {"video_title": "Checking solutions of two-variable linear inequalities example   Algebra I   Khan Academy.mp3", "Sentence": "So our y-intercept is negative 25 over 3. That's the same thing as negative 8 and 1 third. So that's 1, 2, 3, 4, 5, 6, 7, 8, and a little bit more than 8. So our y-intercept is negative 8 and 1 third, like that, and has a slope of 5 thirds. So that means for every 3 it goes to the right, it rises 5. So it goes 1, 2, 3, it rises 5. So the line's going to look something like this."}, {"video_title": "Checking solutions of two-variable linear inequalities example   Algebra I   Khan Academy.mp3", "Sentence": "So our y-intercept is negative 8 and 1 third, like that, and has a slope of 5 thirds. So that means for every 3 it goes to the right, it rises 5. So it goes 1, 2, 3, it rises 5. So the line's going to look something like this. I'm drawing a very rough version of it. So the line will look something like that. This line over here."}, {"video_title": "Checking solutions of two-variable linear inequalities example   Algebra I   Khan Academy.mp3", "Sentence": "So the line's going to look something like this. I'm drawing a very rough version of it. So the line will look something like that. This line over here. That's if it was y is equal to 5 thirds x minus 25 over 3. But here we have an inequality. It's y is less than or equal to."}, {"video_title": "Checking solutions of two-variable linear inequalities example   Algebra I   Khan Academy.mp3", "Sentence": "This line over here. That's if it was y is equal to 5 thirds x minus 25 over 3. But here we have an inequality. It's y is less than or equal to. So for any x, the y's that satisfy it are the y that equals 5 thirds x minus 25 over 3. That would be on the line. So it would be that point there."}, {"video_title": "Checking solutions of two-variable linear inequalities example   Algebra I   Khan Academy.mp3", "Sentence": "It's y is less than or equal to. So for any x, the y's that satisfy it are the y that equals 5 thirds x minus 25 over 3. That would be on the line. So it would be that point there. And all the y's less than it. So the solution is this whole area right over here. Since it's less than or equal to, we can include the line."}, {"video_title": "Checking solutions of two-variable linear inequalities example   Algebra I   Khan Academy.mp3", "Sentence": "So it would be that point there. And all the y's less than it. So the solution is this whole area right over here. Since it's less than or equal to, we can include the line. The equal to allows us to include the line. And less than tells us we're going to go below the line. We can verify that by looking at these two points over here."}, {"video_title": "Checking solutions of two-variable linear inequalities example   Algebra I   Khan Academy.mp3", "Sentence": "Since it's less than or equal to, we can include the line. The equal to allows us to include the line. And less than tells us we're going to go below the line. We can verify that by looking at these two points over here. We saw that 3 comma 5 is not part of the solution. So 3 comma 5 is 1, 2, it's right about there. And then up 5."}, {"video_title": "Checking solutions of two-variable linear inequalities example   Algebra I   Khan Academy.mp3", "Sentence": "We can verify that by looking at these two points over here. We saw that 3 comma 5 is not part of the solution. So 3 comma 5 is 1, 2, it's right about there. And then up 5. So 3 comma 5 is right above here. It's in this region above the line. And notice, not part of the solution."}, {"video_title": "Checking solutions of two-variable linear inequalities example   Algebra I   Khan Academy.mp3", "Sentence": "And then up 5. So 3 comma 5 is right above here. It's in this region above the line. And notice, not part of the solution. And then 1 comma negative 7 is going to be right over here. It's almost on the line. So 1 comma negative 7 is going to be right over there."}, {"video_title": "Checking solutions of two-variable linear inequalities example   Algebra I   Khan Academy.mp3", "Sentence": "And notice, not part of the solution. And then 1 comma negative 7 is going to be right over here. It's almost on the line. So 1 comma negative 7 is going to be right over there. But it is at least within this solution area. So hopefully that gives you a little bit more sense of how to visualize these things. And we'll cover this in more detail in future videos."}, {"video_title": "Constructing and solving a one-step inequality   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "And it's less than $1,000, not less than or equal to $1,000. The size of each tile is 1 square foot. Write an inequality that represents the number of tiles he can purchase with a $1,000 limit, and then figure out how large the stone patio can be. So let x be equal to the number of tiles purchased. And so the cost of purchasing x tiles, they're going to be $3 each, so it's going to be 3x. So 3x is going to be the total cost of purchasing the tiles, and he wants to spend less than $1,000. 3x is how much he spends if he buys x tiles."}, {"video_title": "Constructing and solving a one-step inequality   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So let x be equal to the number of tiles purchased. And so the cost of purchasing x tiles, they're going to be $3 each, so it's going to be 3x. So 3x is going to be the total cost of purchasing the tiles, and he wants to spend less than $1,000. 3x is how much he spends if he buys x tiles. It has to be less than $1,000. We say it right there. If it was less than or equal to, we'd have a little equal sign right there."}, {"video_title": "Constructing and solving a one-step inequality   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "3x is how much he spends if he buys x tiles. It has to be less than $1,000. We say it right there. If it was less than or equal to, we'd have a little equal sign right there. So if we want to solve for x, how many tiles can he buy? We can divide both sides of this inequality by 3. And because we're dividing or multiplying, you can imagine we're multiplying by 1 third or dividing by 3, because this is a positive number, we do not have to swap the inequality sign."}, {"video_title": "Constructing and solving a one-step inequality   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "If it was less than or equal to, we'd have a little equal sign right there. So if we want to solve for x, how many tiles can he buy? We can divide both sides of this inequality by 3. And because we're dividing or multiplying, you can imagine we're multiplying by 1 third or dividing by 3, because this is a positive number, we do not have to swap the inequality sign. So we are left with x is less than 1,000 over 3, which is 333 and 1 third. So he has to buy less than 333 and 1 third tiles. That's how many tiles."}, {"video_title": "Constructing and solving a one-step inequality   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "And because we're dividing or multiplying, you can imagine we're multiplying by 1 third or dividing by 3, because this is a positive number, we do not have to swap the inequality sign. So we are left with x is less than 1,000 over 3, which is 333 and 1 third. So he has to buy less than 333 and 1 third tiles. That's how many tiles. And each tile is 1 square foot. So if he can buy less than 333 and 1 third tiles, then the patio has to be less than 333 and 1 third square feet. And we're done."}, {"video_title": "Area of a circle   Perimeter, area, and volume   Geometry   Khan Academy.mp3", "Sentence": "What is the area of each candy? So the candy, they say it's the shape of circular disks. And they tell us that the diameter of each wafer is 16 millimeters. If I draw a line across the circle that goes through the center, the length of that line all the way across the circle through the center is 16 millimeters. So let me write that. So diameter here is 16 millimeters. And they want us to figure out the area of the surface of this candy, or essentially the area of this circle."}, {"video_title": "Area of a circle   Perimeter, area, and volume   Geometry   Khan Academy.mp3", "Sentence": "If I draw a line across the circle that goes through the center, the length of that line all the way across the circle through the center is 16 millimeters. So let me write that. So diameter here is 16 millimeters. And they want us to figure out the area of the surface of this candy, or essentially the area of this circle. And so when we think about area, we know that the area of a circle is equal to pi times the radius of the circle squared. And you say, well, they gave us a diameter. What is the radius?"}, {"video_title": "Area of a circle   Perimeter, area, and volume   Geometry   Khan Academy.mp3", "Sentence": "And they want us to figure out the area of the surface of this candy, or essentially the area of this circle. And so when we think about area, we know that the area of a circle is equal to pi times the radius of the circle squared. And you say, well, they gave us a diameter. What is the radius? Well, you might remember the radius is half of diameter, so the distance from the center of the circle to the outside, to the boundary of the circle. So it would be this distance right over here, which is exactly half of the diameter. So it would be 8 millimeters."}, {"video_title": "Area of a circle   Perimeter, area, and volume   Geometry   Khan Academy.mp3", "Sentence": "What is the radius? Well, you might remember the radius is half of diameter, so the distance from the center of the circle to the outside, to the boundary of the circle. So it would be this distance right over here, which is exactly half of the diameter. So it would be 8 millimeters. So where we see the radius, we could put 8 millimeters. So the area is going to be equal to pi times 8 millimeters squared, which would be 64 square millimeters. And typically, this is written with pi after the 64."}, {"video_title": "Area of a circle   Perimeter, area, and volume   Geometry   Khan Academy.mp3", "Sentence": "So it would be 8 millimeters. So where we see the radius, we could put 8 millimeters. So the area is going to be equal to pi times 8 millimeters squared, which would be 64 square millimeters. And typically, this is written with pi after the 64. So you might often see it as this is equal to 64 pi millimeters squared. Now, this is the answer, 64 pi millimeters squared. But sometimes it's not so satisfying to just leave it as pi."}, {"video_title": "Area of a circle   Perimeter, area, and volume   Geometry   Khan Academy.mp3", "Sentence": "And typically, this is written with pi after the 64. So you might often see it as this is equal to 64 pi millimeters squared. Now, this is the answer, 64 pi millimeters squared. But sometimes it's not so satisfying to just leave it as pi. You might say, well, I want to get an estimate of what number this is close to. I want a decimal representation of this. And so we could start to use approximate values of pi."}, {"video_title": "Area of a circle   Perimeter, area, and volume   Geometry   Khan Academy.mp3", "Sentence": "But sometimes it's not so satisfying to just leave it as pi. You might say, well, I want to get an estimate of what number this is close to. I want a decimal representation of this. And so we could start to use approximate values of pi. So the most rough approximate value that tends to be used is saying that pi, a very rough approximation, is equal to 3.14. So in that case, we could say that this is going to be equal to 64 times 3.14 millimeters squared. And we can get our calculator to figure out what this will be in decimal form."}, {"video_title": "Area of a circle   Perimeter, area, and volume   Geometry   Khan Academy.mp3", "Sentence": "And so we could start to use approximate values of pi. So the most rough approximate value that tends to be used is saying that pi, a very rough approximation, is equal to 3.14. So in that case, we could say that this is going to be equal to 64 times 3.14 millimeters squared. And we can get our calculator to figure out what this will be in decimal form. So we have 64 times 3.14 gives us 200.96. So we could say that the area is approximately equal to 200.96 square millimeters. Now, if we want to get a more accurate representation of this, pi actually just keeps going on and on and on forever, we could use the calculator's internal representation of pi."}, {"video_title": "Area of a circle   Perimeter, area, and volume   Geometry   Khan Academy.mp3", "Sentence": "And we can get our calculator to figure out what this will be in decimal form. So we have 64 times 3.14 gives us 200.96. So we could say that the area is approximately equal to 200.96 square millimeters. Now, if we want to get a more accurate representation of this, pi actually just keeps going on and on and on forever, we could use the calculator's internal representation of pi. In which case, we'll say 64 times, and then we have to look for the pi in the calculator. It's up here in this yellow, so I'll do this little second function, get the pi there. Every calculator will be a little different."}, {"video_title": "Area of a circle   Perimeter, area, and volume   Geometry   Khan Academy.mp3", "Sentence": "Now, if we want to get a more accurate representation of this, pi actually just keeps going on and on and on forever, we could use the calculator's internal representation of pi. In which case, we'll say 64 times, and then we have to look for the pi in the calculator. It's up here in this yellow, so I'll do this little second function, get the pi there. Every calculator will be a little different. But 64 times pi. Now we're going to use the calculator's internal approximation of pi, which is going to be more precise than what I had in the last one. And you get 201."}, {"video_title": "Area of a circle   Perimeter, area, and volume   Geometry   Khan Academy.mp3", "Sentence": "Every calculator will be a little different. But 64 times pi. Now we're going to use the calculator's internal approximation of pi, which is going to be more precise than what I had in the last one. And you get 201. So let me put it over here so I can write it down. So a more precise is 201. And I'll round to the nearest hundredth."}, {"video_title": "Area of a circle   Perimeter, area, and volume   Geometry   Khan Academy.mp3", "Sentence": "And you get 201. So let me put it over here so I can write it down. So a more precise is 201. And I'll round to the nearest hundredth. So you get 201.06. So a more precise is 201.06 square millimeters. So this is closer to the actual answer, because the calculator's representation is more precise than this very rough approximation of what pi is."}, {"video_title": "Converting decimals to fractions example 3   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "The way I like to do it is to say, well, 0.36, this is the same thing as 36 hundredths. Or one way to think about it is, this is in the hundredths place. Hundredths. Hundredths place. This is in the tens place. Or you could view this as 30 hundredths. You could view this as 3 tenths or 30 hundredths."}, {"video_title": "Converting decimals to fractions example 3   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "Hundredths place. This is in the tens place. Or you could view this as 30 hundredths. You could view this as 3 tenths or 30 hundredths. So we could say that this is the same thing as 36 hundredths, or this is equal to 36 over 100. And we've already expressed it as a fraction, but now we can actually simplify it because both 36 and 100 have some common factors. They are both divisible by, well, it looks like they're both divisible by four."}, {"video_title": "Converting decimals to fractions example 3   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "You could view this as 3 tenths or 30 hundredths. So we could say that this is the same thing as 36 hundredths, or this is equal to 36 over 100. And we've already expressed it as a fraction, but now we can actually simplify it because both 36 and 100 have some common factors. They are both divisible by, well, it looks like they're both divisible by four. So if we can divide the numerator by four and the denominator by four, we get we're doing the same thing to both, so we're not changing the value of the fraction. 36 divided by four is nine. And then 100 divided by four is 25."}, {"video_title": "Circles radius, diameter, circumference and Pi   Geometry   Khan Academy.mp3", "Sentence": "The circle is arguably the most fundamental shape in our universe, whether you look at the shapes of orbits of planets, whether you look at wheels, whether you look at things on kind of a molecular level, the circle just keeps showing up over and over and over again. So it's probably worthwhile for us to understand some of the properties of the circle. So the first thing when people kind of discovered the circle, and you just have to look at the moon to see a circle, but the first time they said, well, what are the properties of any circle? So the first one they might want to say is, well, a circle is all of the points that are equal distance from the center of the circle, right? All of these points along the edge are equal distance from that center right there. So one of the first things someone might want to ask is, what is that distance, that equal distance that everything is from the center right there? We call that the radius."}, {"video_title": "Circles radius, diameter, circumference and Pi   Geometry   Khan Academy.mp3", "Sentence": "So the first one they might want to say is, well, a circle is all of the points that are equal distance from the center of the circle, right? All of these points along the edge are equal distance from that center right there. So one of the first things someone might want to ask is, what is that distance, that equal distance that everything is from the center right there? We call that the radius. The radius of the circle. It's just the distance from the center out to the edge. If that radius is 3 centimeters, then this radius is going to be 3 centimeters."}, {"video_title": "Circles radius, diameter, circumference and Pi   Geometry   Khan Academy.mp3", "Sentence": "We call that the radius. The radius of the circle. It's just the distance from the center out to the edge. If that radius is 3 centimeters, then this radius is going to be 3 centimeters. And this radius is going to be 3 centimeters. It's never going to change. By definition, a circle is all of the points that are equidistant from the center point."}, {"video_title": "Circles radius, diameter, circumference and Pi   Geometry   Khan Academy.mp3", "Sentence": "If that radius is 3 centimeters, then this radius is going to be 3 centimeters. And this radius is going to be 3 centimeters. It's never going to change. By definition, a circle is all of the points that are equidistant from the center point. And that distance is the radius. Now, the next most interesting thing about that, people might say, well, how fat is the circle? How wide is it along its widest point?"}, {"video_title": "Circles radius, diameter, circumference and Pi   Geometry   Khan Academy.mp3", "Sentence": "By definition, a circle is all of the points that are equidistant from the center point. And that distance is the radius. Now, the next most interesting thing about that, people might say, well, how fat is the circle? How wide is it along its widest point? Or if you just want to cut it along its widest point, what is that distance right there? And it doesn't have to be just right there. I could have just as easily cut it along its widest point right there."}, {"video_title": "Circles radius, diameter, circumference and Pi   Geometry   Khan Academy.mp3", "Sentence": "How wide is it along its widest point? Or if you just want to cut it along its widest point, what is that distance right there? And it doesn't have to be just right there. I could have just as easily cut it along its widest point right there. I just wouldn't be cutting it someplace like that, because that wouldn't be along its widest point. There's multiple places where I could cut it along its widest point. Well, we just saw the radius."}, {"video_title": "Circles radius, diameter, circumference and Pi   Geometry   Khan Academy.mp3", "Sentence": "I could have just as easily cut it along its widest point right there. I just wouldn't be cutting it someplace like that, because that wouldn't be along its widest point. There's multiple places where I could cut it along its widest point. Well, we just saw the radius. And we see that widest point goes through the center and just keeps going. So it's essentially 2 radii. You get one radius there, and then you have another radius over there."}, {"video_title": "Circles radius, diameter, circumference and Pi   Geometry   Khan Academy.mp3", "Sentence": "Well, we just saw the radius. And we see that widest point goes through the center and just keeps going. So it's essentially 2 radii. You get one radius there, and then you have another radius over there. And we call this distance along the widest point of the circle the diameter. So that is the diameter of the circle. It has a very easy relationship with the radius."}, {"video_title": "Circles radius, diameter, circumference and Pi   Geometry   Khan Academy.mp3", "Sentence": "You get one radius there, and then you have another radius over there. And we call this distance along the widest point of the circle the diameter. So that is the diameter of the circle. It has a very easy relationship with the radius. The diameter is equal to 2 times the radius. Now, the next most interesting thing that you might be wondering about a circle is how far is it around the circle? So if you were to get your tape measure out, and you were to measure around the circle like that, what's that distance?"}, {"video_title": "Circles radius, diameter, circumference and Pi   Geometry   Khan Academy.mp3", "Sentence": "It has a very easy relationship with the radius. The diameter is equal to 2 times the radius. Now, the next most interesting thing that you might be wondering about a circle is how far is it around the circle? So if you were to get your tape measure out, and you were to measure around the circle like that, what's that distance? We call that word the circumference of the circle. Now, we know how the diameter and the radius relates, but how does the circumference relate to, say, the diameter? If we know how it relates to the diameter, it's very easy to figure out how it relates to the radius."}, {"video_title": "Circles radius, diameter, circumference and Pi   Geometry   Khan Academy.mp3", "Sentence": "So if you were to get your tape measure out, and you were to measure around the circle like that, what's that distance? We call that word the circumference of the circle. Now, we know how the diameter and the radius relates, but how does the circumference relate to, say, the diameter? If we know how it relates to the diameter, it's very easy to figure out how it relates to the radius. Well, many thousands of years ago, people took their tape measures out, and they keep measuring circumferences and radiuses. And let's say when their tape measures weren't so good, let's say they measured the circumference of the circle, and they would get, well, it looks like it's about 3. And then they measure the radius of this circle right here, or the diameter of that circle, and they'd say, well, the diameter looks like it's about 1."}, {"video_title": "Circles radius, diameter, circumference and Pi   Geometry   Khan Academy.mp3", "Sentence": "If we know how it relates to the diameter, it's very easy to figure out how it relates to the radius. Well, many thousands of years ago, people took their tape measures out, and they keep measuring circumferences and radiuses. And let's say when their tape measures weren't so good, let's say they measured the circumference of the circle, and they would get, well, it looks like it's about 3. And then they measure the radius of this circle right here, or the diameter of that circle, and they'd say, well, the diameter looks like it's about 1. So they would say, let me write this down. So we're worried about the ratio of the circumference to the diameter. So let's say that somebody had some circle over here."}, {"video_title": "Circles radius, diameter, circumference and Pi   Geometry   Khan Academy.mp3", "Sentence": "And then they measure the radius of this circle right here, or the diameter of that circle, and they'd say, well, the diameter looks like it's about 1. So they would say, let me write this down. So we're worried about the ratio of the circumference to the diameter. So let's say that somebody had some circle over here. Let's say they had this circle. And the first time, not that good of a tape measure, they measured around the circle. And they said, hey, it's roughly equal to 3 meters when I go around it."}, {"video_title": "Circles radius, diameter, circumference and Pi   Geometry   Khan Academy.mp3", "Sentence": "So let's say that somebody had some circle over here. Let's say they had this circle. And the first time, not that good of a tape measure, they measured around the circle. And they said, hey, it's roughly equal to 3 meters when I go around it. And when I measure the diameter of the circle, it's roughly equal to 1. OK, that's interesting. It may be the ratio of the circumference to the diameter is 3, so maybe the circumference is always 3 times the diameter."}, {"video_title": "Circles radius, diameter, circumference and Pi   Geometry   Khan Academy.mp3", "Sentence": "And they said, hey, it's roughly equal to 3 meters when I go around it. And when I measure the diameter of the circle, it's roughly equal to 1. OK, that's interesting. It may be the ratio of the circumference to the diameter is 3, so maybe the circumference is always 3 times the diameter. Well, that was just for this circle, but let's say they measured some other circle here. Let's say it's like this. I drew it smaller."}, {"video_title": "Circles radius, diameter, circumference and Pi   Geometry   Khan Academy.mp3", "Sentence": "It may be the ratio of the circumference to the diameter is 3, so maybe the circumference is always 3 times the diameter. Well, that was just for this circle, but let's say they measured some other circle here. Let's say it's like this. I drew it smaller. But let's say that on this circle, they measured around it, and they found out that the circumference is 6 centimeters, roughly. They have a bad tape measure right then. And then they find out that the diameter is roughly 2 centimeters."}, {"video_title": "Circles radius, diameter, circumference and Pi   Geometry   Khan Academy.mp3", "Sentence": "I drew it smaller. But let's say that on this circle, they measured around it, and they found out that the circumference is 6 centimeters, roughly. They have a bad tape measure right then. And then they find out that the diameter is roughly 2 centimeters. And once again, the ratio of the circumference to the diameter was roughly 3. So OK, this is a neat property of circles. Maybe the ratio of the circumference to the diameter is always fixed for any circle."}, {"video_title": "Circles radius, diameter, circumference and Pi   Geometry   Khan Academy.mp3", "Sentence": "And then they find out that the diameter is roughly 2 centimeters. And once again, the ratio of the circumference to the diameter was roughly 3. So OK, this is a neat property of circles. Maybe the ratio of the circumference to the diameter is always fixed for any circle. So they said, let me study this further. So they got better tape measures. When they got better tape measures, they measured, hey, you know, my diameter is definitely 1."}, {"video_title": "Circles radius, diameter, circumference and Pi   Geometry   Khan Academy.mp3", "Sentence": "Maybe the ratio of the circumference to the diameter is always fixed for any circle. So they said, let me study this further. So they got better tape measures. When they got better tape measures, they measured, hey, you know, my diameter is definitely 1. But when I measure my circumference a little bit, I realize it's closer to 3.1. And the same thing with this over here. They noticed that this ratio is closer to 3.1."}, {"video_title": "Circles radius, diameter, circumference and Pi   Geometry   Khan Academy.mp3", "Sentence": "When they got better tape measures, they measured, hey, you know, my diameter is definitely 1. But when I measure my circumference a little bit, I realize it's closer to 3.1. And the same thing with this over here. They noticed that this ratio is closer to 3.1. And then they kept measuring it better and better and better, and then they realized that they were getting this number. They just kept measuring it better and better and better, and they were getting this number 3.14159. And they just kept adding digits, and it would never repeat, and it was this strange, fascinating, metaphysical number that kept showing up."}, {"video_title": "Circles radius, diameter, circumference and Pi   Geometry   Khan Academy.mp3", "Sentence": "They noticed that this ratio is closer to 3.1. And then they kept measuring it better and better and better, and then they realized that they were getting this number. They just kept measuring it better and better and better, and they were getting this number 3.14159. And they just kept adding digits, and it would never repeat, and it was this strange, fascinating, metaphysical number that kept showing up. And so since this number was so fundamental to our universe, because a circle is so fundamental to our universe, and it just showed up for every circle, the ratio of the circumference to the diameter was this kind of magical number, they gave it a name. They called it pi. Or you could just give it the Latin or the Greek letter pi, just like that."}, {"video_title": "Circles radius, diameter, circumference and Pi   Geometry   Khan Academy.mp3", "Sentence": "And they just kept adding digits, and it would never repeat, and it was this strange, fascinating, metaphysical number that kept showing up. And so since this number was so fundamental to our universe, because a circle is so fundamental to our universe, and it just showed up for every circle, the ratio of the circumference to the diameter was this kind of magical number, they gave it a name. They called it pi. Or you could just give it the Latin or the Greek letter pi, just like that. That represents this number, which is arguably the most fascinating number in our universe. It first shows up as the ratio of the circumference to the diameter, but you're going to learn as you go through your mathematical journey that it shows up everywhere. It's one of these fundamental things about the universe that just makes you think that there's some order to it."}, {"video_title": "Circles radius, diameter, circumference and Pi   Geometry   Khan Academy.mp3", "Sentence": "Or you could just give it the Latin or the Greek letter pi, just like that. That represents this number, which is arguably the most fascinating number in our universe. It first shows up as the ratio of the circumference to the diameter, but you're going to learn as you go through your mathematical journey that it shows up everywhere. It's one of these fundamental things about the universe that just makes you think that there's some order to it. But anyway, how can we use this in our basic mathematics? So we know, or I'm telling you, that the ratio of the circumference to the diameter, when I say the ratio, literally I'm just saying if you divide the circumference by the diameter, you're going to get pi. Pi is just this number."}, {"video_title": "Circles radius, diameter, circumference and Pi   Geometry   Khan Academy.mp3", "Sentence": "It's one of these fundamental things about the universe that just makes you think that there's some order to it. But anyway, how can we use this in our basic mathematics? So we know, or I'm telling you, that the ratio of the circumference to the diameter, when I say the ratio, literally I'm just saying if you divide the circumference by the diameter, you're going to get pi. Pi is just this number. I could write 3.14159 and just keep going on and on and on, but that would be a waste of space, and it would just be hard to deal with. So people just write this Greek letter pi there. So how can we relate this?"}, {"video_title": "Circles radius, diameter, circumference and Pi   Geometry   Khan Academy.mp3", "Sentence": "Pi is just this number. I could write 3.14159 and just keep going on and on and on, but that would be a waste of space, and it would just be hard to deal with. So people just write this Greek letter pi there. So how can we relate this? We can multiply both sides of this by the diameter, and we could say that the circumference is equal to pi times the diameter. Or, since the diameter is equal to 2 times the radius, we could say that the circumference is equal to pi times 2 times the radius. Or the form that you're most likely to see it, it's equal to 2 pi r. So let's see if we can apply that to some problems."}, {"video_title": "Circles radius, diameter, circumference and Pi   Geometry   Khan Academy.mp3", "Sentence": "So how can we relate this? We can multiply both sides of this by the diameter, and we could say that the circumference is equal to pi times the diameter. Or, since the diameter is equal to 2 times the radius, we could say that the circumference is equal to pi times 2 times the radius. Or the form that you're most likely to see it, it's equal to 2 pi r. So let's see if we can apply that to some problems. So let's say I have a circle. I have a circle just like that. And I were to tell you it has a radius."}, {"video_title": "Circles radius, diameter, circumference and Pi   Geometry   Khan Academy.mp3", "Sentence": "Or the form that you're most likely to see it, it's equal to 2 pi r. So let's see if we can apply that to some problems. So let's say I have a circle. I have a circle just like that. And I were to tell you it has a radius. Its radius right there is 3. So 3, let me write this down. So the radius is equal to 3."}, {"video_title": "Circles radius, diameter, circumference and Pi   Geometry   Khan Academy.mp3", "Sentence": "And I were to tell you it has a radius. Its radius right there is 3. So 3, let me write this down. So the radius is equal to 3. Maybe it's 3 meters. Put some units in there. What is the circumference of this circle?"}, {"video_title": "Circles radius, diameter, circumference and Pi   Geometry   Khan Academy.mp3", "Sentence": "So the radius is equal to 3. Maybe it's 3 meters. Put some units in there. What is the circumference of this circle? The circumference is equal to 2 times pi times the radius. So it's going to be equal to 2 times pi times the radius, times 3 meters, which is equal to 6 meters times pi, or 6 pi meters. Now I could multiply this out."}, {"video_title": "Circles radius, diameter, circumference and Pi   Geometry   Khan Academy.mp3", "Sentence": "What is the circumference of this circle? The circumference is equal to 2 times pi times the radius. So it's going to be equal to 2 times pi times the radius, times 3 meters, which is equal to 6 meters times pi, or 6 pi meters. Now I could multiply this out. Remember, pi is just a number. Pi is 3.14159, and going on and on and on. So if I multiply 6 times that, maybe I'll get 18 point something, something, something."}, {"video_title": "Circles radius, diameter, circumference and Pi   Geometry   Khan Academy.mp3", "Sentence": "Now I could multiply this out. Remember, pi is just a number. Pi is 3.14159, and going on and on and on. So if I multiply 6 times that, maybe I'll get 18 point something, something, something. If you have your calculator, you might want to do it. But for simplicity, people just tend to leave our numbers in terms of pi. I don't know what this is."}, {"video_title": "Circles radius, diameter, circumference and Pi   Geometry   Khan Academy.mp3", "Sentence": "So if I multiply 6 times that, maybe I'll get 18 point something, something, something. If you have your calculator, you might want to do it. But for simplicity, people just tend to leave our numbers in terms of pi. I don't know what this is. If you multiply 6 times 3.14159, I don't know if you get something close to 19 or 18. Maybe it's approximately 18 point something, something, something. I don't have my calculator in front of me."}, {"video_title": "Circles radius, diameter, circumference and Pi   Geometry   Khan Academy.mp3", "Sentence": "I don't know what this is. If you multiply 6 times 3.14159, I don't know if you get something close to 19 or 18. Maybe it's approximately 18 point something, something, something. I don't have my calculator in front of me. But instead of writing that number, you just write 6 pi there. Actually, I think it wouldn't quite cross the threshold to 19 yet. Now, what if I were to go there?"}, {"video_title": "Circles radius, diameter, circumference and Pi   Geometry   Khan Academy.mp3", "Sentence": "I don't have my calculator in front of me. But instead of writing that number, you just write 6 pi there. Actually, I think it wouldn't quite cross the threshold to 19 yet. Now, what if I were to go there? Well, let's ask another question. What is the diameter of this circle? What is the diameter?"}, {"video_title": "Circles radius, diameter, circumference and Pi   Geometry   Khan Academy.mp3", "Sentence": "Now, what if I were to go there? Well, let's ask another question. What is the diameter of this circle? What is the diameter? Well, if this radius is 3, the diameter is just twice that. So it's just going to be 3 times 2, or 3 plus 3, which is equal to 6 meters. So the circumference is 6 pi meters."}, {"video_title": "Circles radius, diameter, circumference and Pi   Geometry   Khan Academy.mp3", "Sentence": "What is the diameter? Well, if this radius is 3, the diameter is just twice that. So it's just going to be 3 times 2, or 3 plus 3, which is equal to 6 meters. So the circumference is 6 pi meters. The diameter is 6 meters. The radius is 3 meters. Now let's go the other way."}, {"video_title": "Circles radius, diameter, circumference and Pi   Geometry   Khan Academy.mp3", "Sentence": "So the circumference is 6 pi meters. The diameter is 6 meters. The radius is 3 meters. Now let's go the other way. Let's say I have another circle. Let's say I have another circle here. And I were to tell you that its circumference is equal to 10 meters."}, {"video_title": "Circles radius, diameter, circumference and Pi   Geometry   Khan Academy.mp3", "Sentence": "Now let's go the other way. Let's say I have another circle. Let's say I have another circle here. And I were to tell you that its circumference is equal to 10 meters. That's the circumference of the circle, if you were to put a tape measure to go around it. And someone were to ask you, what is the diameter of the circle? Well, we know that the diameter times pi, we know that pi times the diameter is equal to the circumference, is equal to 10 meters."}, {"video_title": "Circles radius, diameter, circumference and Pi   Geometry   Khan Academy.mp3", "Sentence": "And I were to tell you that its circumference is equal to 10 meters. That's the circumference of the circle, if you were to put a tape measure to go around it. And someone were to ask you, what is the diameter of the circle? Well, we know that the diameter times pi, we know that pi times the diameter is equal to the circumference, is equal to 10 meters. So to solve for this, we would just divide both sides of this equation by pi. So the diameter would equal 10 meters over pi, or 10 over pi meters, and that is just a number. If you have your calculator, you could actually divide 10 divided by 3.14159."}, {"video_title": "Circles radius, diameter, circumference and Pi   Geometry   Khan Academy.mp3", "Sentence": "Well, we know that the diameter times pi, we know that pi times the diameter is equal to the circumference, is equal to 10 meters. So to solve for this, we would just divide both sides of this equation by pi. So the diameter would equal 10 meters over pi, or 10 over pi meters, and that is just a number. If you have your calculator, you could actually divide 10 divided by 3.14159. You're going to get 3 point something, something, something meters. I can't do it in my head. But this is just a number."}, {"video_title": "Circles radius, diameter, circumference and Pi   Geometry   Khan Academy.mp3", "Sentence": "If you have your calculator, you could actually divide 10 divided by 3.14159. You're going to get 3 point something, something, something meters. I can't do it in my head. But this is just a number. But for simplicity, we often just leave it that way. Now, what is the radius? Well, the radius is equal to 1 half the diameter."}, {"video_title": "Circles radius, diameter, circumference and Pi   Geometry   Khan Academy.mp3", "Sentence": "But this is just a number. But for simplicity, we often just leave it that way. Now, what is the radius? Well, the radius is equal to 1 half the diameter. So this whole distance right here is 10 over pi meters. And if we just want half of that, if we just want the radius, we just multiply it times half. So you have 1 half times 10 over pi, which is equal to 1 half times 10, or you could just divide the numerator and the denominator by 2."}, {"video_title": "Circles radius, diameter, circumference and Pi   Geometry   Khan Academy.mp3", "Sentence": "Well, the radius is equal to 1 half the diameter. So this whole distance right here is 10 over pi meters. And if we just want half of that, if we just want the radius, we just multiply it times half. So you have 1 half times 10 over pi, which is equal to 1 half times 10, or you could just divide the numerator and the denominator by 2. You get a 5 there, so you get 5 over pi. So the radius over here is 5 over pi. Nothing super fancy about this."}, {"video_title": "Circles radius, diameter, circumference and Pi   Geometry   Khan Academy.mp3", "Sentence": "So you have 1 half times 10 over pi, which is equal to 1 half times 10, or you could just divide the numerator and the denominator by 2. You get a 5 there, so you get 5 over pi. So the radius over here is 5 over pi. Nothing super fancy about this. I think the thing that confuses people the most is to just realize that pi is a number. Pi is just 3.14159. It just keeps going on and on and on."}, {"video_title": "Circles radius, diameter, circumference and Pi   Geometry   Khan Academy.mp3", "Sentence": "Nothing super fancy about this. I think the thing that confuses people the most is to just realize that pi is a number. Pi is just 3.14159. It just keeps going on and on and on. You could write, there's actually thousands of books written about pi. So it's not like, I don't know if there's thousands, but maybe I'm exaggerating. But you could write books about this number."}, {"video_title": "Circles radius, diameter, circumference and Pi   Geometry   Khan Academy.mp3", "Sentence": "It just keeps going on and on and on. You could write, there's actually thousands of books written about pi. So it's not like, I don't know if there's thousands, but maybe I'm exaggerating. But you could write books about this number. But it's just a number. It's a very special number. And if you wanted to write it in a way that you're used to writing numbers, you could literally just multiply this out."}, {"video_title": "Circles radius, diameter, circumference and Pi   Geometry   Khan Academy.mp3", "Sentence": "But you could write books about this number. But it's just a number. It's a very special number. And if you wanted to write it in a way that you're used to writing numbers, you could literally just multiply this out. But most of the time, people just realize they like leaving things in terms of pi. Anyway, I'll leave you there. And in the next video, we'll figure out the area of a circle."}, {"video_title": "Trolls, tolls, and systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "You can't swim across it. It's a very rough river. So you have to cross this bridge. And so as you approach the bridge, this troll shows up. That's the troll. And he says, well, I'm a reasonable troll. You just have to pay $5."}, {"video_title": "Trolls, tolls, and systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "And so as you approach the bridge, this troll shows up. That's the troll. And he says, well, I'm a reasonable troll. You just have to pay $5. And when you look a little bit more carefully, you see that there actually was a sign there that says $5 toll to cross the bridge. Now, unfortunately for you, you do not have any money in your pocket. And the troll says, well, you can't cross."}, {"video_title": "Trolls, tolls, and systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "You just have to pay $5. And when you look a little bit more carefully, you see that there actually was a sign there that says $5 toll to cross the bridge. Now, unfortunately for you, you do not have any money in your pocket. And the troll says, well, you can't cross. But you say I need to really, really get to that castle. And so the troll says, well, I'll take some pity on you. Instead of paying the $5, I will give you a riddle."}, {"video_title": "Trolls, tolls, and systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "And the troll says, well, you can't cross. But you say I need to really, really get to that castle. And so the troll says, well, I'll take some pity on you. Instead of paying the $5, I will give you a riddle. And the riddle is this. And now I'm speaking as the troll. I am a rich troll because I get to charge $5 from everyone who crosses the bridge."}, {"video_title": "Trolls, tolls, and systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "Instead of paying the $5, I will give you a riddle. And the riddle is this. And now I'm speaking as the troll. I am a rich troll because I get to charge $5 from everyone who crosses the bridge. And actually, I only accept $5 or $10 bills. It's a bit of a riddle why they accept American currency in this fantasy land. But let's just take that as a given for now."}, {"video_title": "Trolls, tolls, and systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "I am a rich troll because I get to charge $5 from everyone who crosses the bridge. And actually, I only accept $5 or $10 bills. It's a bit of a riddle why they accept American currency in this fantasy land. But let's just take that as a given for now. So I only take $5 or $10 bills. I'm being the troll. Obviously, if you give me a 10, I'll give you 5 back."}, {"video_title": "Trolls, tolls, and systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "But let's just take that as a given for now. So I only take $5 or $10 bills. I'm being the troll. Obviously, if you give me a 10, I'll give you 5 back. And I know, because I count my money on a daily basis, I like to save my money as the troll, I know that I have a total of 900 bills. Let me write that down. I have a total of 900 bills."}, {"video_title": "Trolls, tolls, and systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "Obviously, if you give me a 10, I'll give you 5 back. And I know, because I count my money on a daily basis, I like to save my money as the troll, I know that I have a total of 900 bills. Let me write that down. I have a total of 900 bills. Total of 900 $5 and $10 bills. And he says, I will give you, because I'm very sympathetic, I'll give you another piece of information. He says, if you were to add up the value of all of my money, which is all in $5 and $10 bills, value of all the $5 and $10 bills that I have, I, speaking as a troll, bought dollar bills, is $5,500."}, {"video_title": "Trolls, tolls, and systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "I have a total of 900 bills. Total of 900 $5 and $10 bills. And he says, I will give you, because I'm very sympathetic, I'll give you another piece of information. He says, if you were to add up the value of all of my money, which is all in $5 and $10 bills, value of all the $5 and $10 bills that I have, I, speaking as a troll, bought dollar bills, is $5,500. This is a rich troll. Is $5,500. And so the riddle is, exactly, and if you give the wrong answer, and if you're not able to solve it in 10 minutes, he's just going to push you into the river and do something horrible to you."}, {"video_title": "Trolls, tolls, and systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "He says, if you were to add up the value of all of my money, which is all in $5 and $10 bills, value of all the $5 and $10 bills that I have, I, speaking as a troll, bought dollar bills, is $5,500. This is a rich troll. Is $5,500. And so the riddle is, exactly, and if you give the wrong answer, and if you're not able to solve it in 10 minutes, he's just going to push you into the river and do something horrible to you. He says, exactly, how many $5s and $10s do I, the troll, have? So the first thing I'm going to have you think about is, is this even a solvable problem? Because it's not a solvable problem."}, {"video_title": "Trolls, tolls, and systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "And so the riddle is, exactly, and if you give the wrong answer, and if you're not able to solve it in 10 minutes, he's just going to push you into the river and do something horrible to you. He says, exactly, how many $5s and $10s do I, the troll, have? So the first thing I'm going to have you think about is, is this even a solvable problem? Because it's not a solvable problem. You should probably run as fast as you can in the other direction. So now I will tell you, yes, it is a solvable problem. And let's start thinking about it a little bit algebraically."}, {"video_title": "Trolls, tolls, and systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "Because it's not a solvable problem. You should probably run as fast as you can in the other direction. So now I will tell you, yes, it is a solvable problem. And let's start thinking about it a little bit algebraically. And to do that, let's just set some variables. And I will set the variables to be what we're really trying to solve for. We're trying to solve for the number of $5 bills we have and the number of $10 bills that we have."}, {"video_title": "Trolls, tolls, and systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "And let's start thinking about it a little bit algebraically. And to do that, let's just set some variables. And I will set the variables to be what we're really trying to solve for. We're trying to solve for the number of $5 bills we have and the number of $10 bills that we have. So let's just define some variables. I'll say f for 5. Let's let f equal the number of $5 bills that we have."}, {"video_title": "Trolls, tolls, and systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "We're trying to solve for the number of $5 bills we have and the number of $10 bills that we have. So let's just define some variables. I'll say f for 5. Let's let f equal the number of $5 bills that we have. And I'll use the same idea. Let's let t is equal to the number of $10 bills that we have. Now given this information, and now I'm not sure if I'm speaking as a troll."}, {"video_title": "Trolls, tolls, and systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "Let's let f equal the number of $5 bills that we have. And I'll use the same idea. Let's let t is equal to the number of $10 bills that we have. Now given this information, and now I'm not sure if I'm speaking as a troll. Let's say I'm still speaking as a troll. I'm a very sympathetic troll. And I'm going to give you hints."}, {"video_title": "Trolls, tolls, and systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "Now given this information, and now I'm not sure if I'm speaking as a troll. Let's say I'm still speaking as a troll. I'm a very sympathetic troll. And I'm going to give you hints. Given this information, setting these variables in this way, can I represent the clues in the riddle mathematically? So let's focus on the first clue. Can I represent this clue, that the total of 900 $5 and $10 bills, or can I represent that mathematically, that I have a total of 900 $5 and $10 bills?"}, {"video_title": "Trolls, tolls, and systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "And I'm going to give you hints. Given this information, setting these variables in this way, can I represent the clues in the riddle mathematically? So let's focus on the first clue. Can I represent this clue, that the total of 900 $5 and $10 bills, or can I represent that mathematically, that I have a total of 900 $5 and $10 bills? Well, what's going to be our total of bills? It's going to be the number of 5's that we have, which is f. The number of 5's that we have is f. And then the number of 10's that we have is t. The total number of 5's plus the total number of 10's, that's our total number of bills. So that's going to be equal to 900."}, {"video_title": "Trolls, tolls, and systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "Can I represent this clue, that the total of 900 $5 and $10 bills, or can I represent that mathematically, that I have a total of 900 $5 and $10 bills? Well, what's going to be our total of bills? It's going to be the number of 5's that we have, which is f. The number of 5's that we have is f. And then the number of 10's that we have is t. The total number of 5's plus the total number of 10's, that's our total number of bills. So that's going to be equal to 900. So this statement, this first clue in our riddle, can be written mathematically like this if we defined the variables like that. And I just said f for 5, because f for 5 and t for 10. Now let's look at the second clue."}, {"video_title": "Trolls, tolls, and systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "So that's going to be equal to 900. So this statement, this first clue in our riddle, can be written mathematically like this if we defined the variables like that. And I just said f for 5, because f for 5 and t for 10. Now let's look at the second clue. Can we represent this one mathematically, given these variable definitions that we created? Well, let's think separately about the value of the $5 bills and the value of the $10 bills. What's the value of all of the $5 bills?"}, {"video_title": "Trolls, tolls, and systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "Now let's look at the second clue. Can we represent this one mathematically, given these variable definitions that we created? Well, let's think separately about the value of the $5 bills and the value of the $10 bills. What's the value of all of the $5 bills? Well, each $5 bill is worth $5. So it's going to be 5 times the number of $5 bills that we have. So if I have one $5 bill, it will be $5."}, {"video_title": "Trolls, tolls, and systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "What's the value of all of the $5 bills? Well, each $5 bill is worth $5. So it's going to be 5 times the number of $5 bills that we have. So if I have one $5 bill, it will be $5. If I have 100 $5 bills, that's going to be $500. However many $5 bills, I just multiply it by 5. That's the value of the $5 bills."}, {"video_title": "Trolls, tolls, and systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "So if I have one $5 bill, it will be $5. If I have 100 $5 bills, that's going to be $500. However many $5 bills, I just multiply it by 5. That's the value of the $5 bills. Let me write that down. Value of the $5 bills. Now, same logic."}, {"video_title": "Trolls, tolls, and systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "That's the value of the $5 bills. Let me write that down. Value of the $5 bills. Now, same logic. What's the value of the $10 bills? Well, the value of the $10 bills is just going to be 10 times however many bills I have. Value of the $10 bills."}, {"video_title": "Trolls, tolls, and systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "Now, same logic. What's the value of the $10 bills? Well, the value of the $10 bills is just going to be 10 times however many bills I have. Value of the $10 bills. So what's going to be the total value of my bills? What's going to be the value of the $5 bills plus the value of the $10 bills? And he tells me what that total value is."}, {"video_title": "Trolls, tolls, and systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "Value of the $10 bills. So what's going to be the total value of my bills? What's going to be the value of the $5 bills plus the value of the $10 bills? And he tells me what that total value is. It's $5,500. So if I add these two things, they're going to add up to be $5,500. So the second statement we can represent mathematically with this second equation right over here."}, {"video_title": "Trolls, tolls, and systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "And he tells me what that total value is. It's $5,500. So if I add these two things, they're going to add up to be $5,500. So the second statement we can represent mathematically with this second equation right over here. And what we essentially have right over here, we have two equations. Each of them have two unknowns. And just using one of these equations, we can't really figure out what f and t are."}, {"video_title": "Trolls, tolls, and systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "So the second statement we can represent mathematically with this second equation right over here. And what we essentially have right over here, we have two equations. Each of them have two unknowns. And just using one of these equations, we can't really figure out what f and t are. You can pick a bunch of different combinations that add up to 900 here. You could pick a bunch of different combinations where if you work out all the math, you get 5,500 here. So independently, these equations, you don't know what f and t are."}, {"video_title": "Trolls, tolls, and systems of equations   Algebra II   Khan Academy.mp3", "Sentence": "And just using one of these equations, we can't really figure out what f and t are. You can pick a bunch of different combinations that add up to 900 here. You could pick a bunch of different combinations where if you work out all the math, you get 5,500 here. So independently, these equations, you don't know what f and t are. But what we will see over the next several videos is that if you use both of this information, if you say that there's an f and a t that has to satisfy both of these equations, then you can find a solution. And this is called a system of equations. This is a system of equations."}, {"video_title": "Consistent and inconsistent systems   Algebra II   Khan Academy.mp3", "Sentence": "Is the system of linear equations below consistent or inconsistent? And they give us x plus 2y is equal to 13, and 3x minus y is equal to negative 11. So to answer this question, we need to know what it means to be consistent or inconsistent. So a consistent system of equations has at least one solution. And an inconsistent system of equations, as you can imagine, has no solutions. So if we think about it graphically, what would the graph of a consistent system look like? Well either, let me just draw a really rough graph, so that's my x axis, and that is my y axis."}, {"video_title": "Consistent and inconsistent systems   Algebra II   Khan Academy.mp3", "Sentence": "So a consistent system of equations has at least one solution. And an inconsistent system of equations, as you can imagine, has no solutions. So if we think about it graphically, what would the graph of a consistent system look like? Well either, let me just draw a really rough graph, so that's my x axis, and that is my y axis. So if I have just two different lines that intersect, that would be consistent. So that's one line, and then that's another line. They clearly have that one solution where they both intersect, so that would be a consistent system."}, {"video_title": "Consistent and inconsistent systems   Algebra II   Khan Academy.mp3", "Sentence": "Well either, let me just draw a really rough graph, so that's my x axis, and that is my y axis. So if I have just two different lines that intersect, that would be consistent. So that's one line, and then that's another line. They clearly have that one solution where they both intersect, so that would be a consistent system. Another consistent system would be if they're the same line, because then they would intersect at a ton of points. Actually an infinite number of points. So let's say one of the lines looks like that, and then the other line is actually the exact same line."}, {"video_title": "Consistent and inconsistent systems   Algebra II   Khan Academy.mp3", "Sentence": "They clearly have that one solution where they both intersect, so that would be a consistent system. Another consistent system would be if they're the same line, because then they would intersect at a ton of points. Actually an infinite number of points. So let's say one of the lines looks like that, and then the other line is actually the exact same line. So it's exactly right on top of it. So those two intersect at every point along those lines, so that also would be consistent. An inconsistent system would have no solutions."}, {"video_title": "Consistent and inconsistent systems   Algebra II   Khan Academy.mp3", "Sentence": "So let's say one of the lines looks like that, and then the other line is actually the exact same line. So it's exactly right on top of it. So those two intersect at every point along those lines, so that also would be consistent. An inconsistent system would have no solutions. So let me again draw my axes. It will have no solutions. The only way that you're going to have two lines in two dimensions have no solutions is if they don't intersect, or if they are parallel."}, {"video_title": "Consistent and inconsistent systems   Algebra II   Khan Academy.mp3", "Sentence": "An inconsistent system would have no solutions. So let me again draw my axes. It will have no solutions. The only way that you're going to have two lines in two dimensions have no solutions is if they don't intersect, or if they are parallel. So one line could look like this, and then the other line would have the same slope, but it would be shifted over. It would have a different y intercept. So it would look like this."}, {"video_title": "Consistent and inconsistent systems   Algebra II   Khan Academy.mp3", "Sentence": "The only way that you're going to have two lines in two dimensions have no solutions is if they don't intersect, or if they are parallel. So one line could look like this, and then the other line would have the same slope, but it would be shifted over. It would have a different y intercept. So it would look like this. So that's what an inconsistent system would look like. You have parallel lines. This right here is inconsistent."}, {"video_title": "Consistent and inconsistent systems   Algebra II   Khan Academy.mp3", "Sentence": "So it would look like this. So that's what an inconsistent system would look like. You have parallel lines. This right here is inconsistent. So what we could do is just do a rough graph of both of these lines and see if they intersect. Another way to do it is you could look at the slope, and if they have the same slope and different y intercepts, then you'd also have an inconsistent system. But let's just graph them."}, {"video_title": "Consistent and inconsistent systems   Algebra II   Khan Academy.mp3", "Sentence": "This right here is inconsistent. So what we could do is just do a rough graph of both of these lines and see if they intersect. Another way to do it is you could look at the slope, and if they have the same slope and different y intercepts, then you'd also have an inconsistent system. But let's just graph them. So let me draw my x-axis, and let me draw my y-axis. So this is x, and then this is y. And then there's a couple of ways we could do it."}, {"video_title": "Consistent and inconsistent systems   Algebra II   Khan Academy.mp3", "Sentence": "But let's just graph them. So let me draw my x-axis, and let me draw my y-axis. So this is x, and then this is y. And then there's a couple of ways we could do it. The easiest way is really just find two points that satisfy each of these equations, and that's enough to define a line. So for this first one, let's just make a little table of x's and y's. When x is 0, you have 2y is equal to 13."}, {"video_title": "Consistent and inconsistent systems   Algebra II   Khan Academy.mp3", "Sentence": "And then there's a couple of ways we could do it. The easiest way is really just find two points that satisfy each of these equations, and that's enough to define a line. So for this first one, let's just make a little table of x's and y's. When x is 0, you have 2y is equal to 13. So when x is 0, you have 2y is equal to 13, or y is equal to 13 over 2, which is the same thing as 6 1\u20442. So when x is 0, y is 6 1\u20442. I'll just put it right over here."}, {"video_title": "Consistent and inconsistent systems   Algebra II   Khan Academy.mp3", "Sentence": "When x is 0, you have 2y is equal to 13. So when x is 0, you have 2y is equal to 13, or y is equal to 13 over 2, which is the same thing as 6 1\u20442. So when x is 0, y is 6 1\u20442. I'll just put it right over here. So this is 0, 13 over 2. And then let's just see what happens when y is 0. When y is 0, then 2 times y is 0."}, {"video_title": "Consistent and inconsistent systems   Algebra II   Khan Academy.mp3", "Sentence": "I'll just put it right over here. So this is 0, 13 over 2. And then let's just see what happens when y is 0. When y is 0, then 2 times y is 0. You have x equaling 13. So we have the point 13, 0. So this is 0, 6 1\u20442."}, {"video_title": "Consistent and inconsistent systems   Algebra II   Khan Academy.mp3", "Sentence": "When y is 0, then 2 times y is 0. You have x equaling 13. So we have the point 13, 0. So this is 0, 6 1\u20442. So 13, 0 would be right about there. I'm just trying to approximate. 13, 0."}, {"video_title": "Consistent and inconsistent systems   Algebra II   Khan Academy.mp3", "Sentence": "So this is 0, 6 1\u20442. So 13, 0 would be right about there. I'm just trying to approximate. 13, 0. And so this line right up here, this equation, can be represented by this line. Let me try my best to draw it. It would look something like that."}, {"video_title": "Consistent and inconsistent systems   Algebra II   Khan Academy.mp3", "Sentence": "13, 0. And so this line right up here, this equation, can be represented by this line. Let me try my best to draw it. It would look something like that. Now let's worry about this one. Let's worry about that one. So once again, let's make a little table, x's and y's."}, {"video_title": "Consistent and inconsistent systems   Algebra II   Khan Academy.mp3", "Sentence": "It would look something like that. Now let's worry about this one. Let's worry about that one. So once again, let's make a little table, x's and y's. I'm really just looking for two points on this graph. So when x is equal to 0, you get this 3 times 0 is just 0. So you get negative y is equal to negative 11, or you get y is equal to 11."}, {"video_title": "Consistent and inconsistent systems   Algebra II   Khan Academy.mp3", "Sentence": "So once again, let's make a little table, x's and y's. I'm really just looking for two points on this graph. So when x is equal to 0, you get this 3 times 0 is just 0. So you get negative y is equal to negative 11, or you get y is equal to 11. So you have the point 0, 11. So that's maybe right over there. 0, 11 is on that line."}, {"video_title": "Consistent and inconsistent systems   Algebra II   Khan Academy.mp3", "Sentence": "So you get negative y is equal to negative 11, or you get y is equal to 11. So you have the point 0, 11. So that's maybe right over there. 0, 11 is on that line. And then when y is 0, you have 3x minus 0 is equal to negative 11. Or 3x is equal to negative 11. Or if you divide both sides by 3, you get x is equal to negative 11 over 3."}, {"video_title": "Consistent and inconsistent systems   Algebra II   Khan Academy.mp3", "Sentence": "0, 11 is on that line. And then when y is 0, you have 3x minus 0 is equal to negative 11. Or 3x is equal to negative 11. Or if you divide both sides by 3, you get x is equal to negative 11 over 3. And this is the exact same thing as negative 3 and 2 thirds. So when y is 0, you have x being negative 3 and 2 thirds. So maybe this is about 6, so negative 3 and 2 thirds would be right about here."}, {"video_title": "Consistent and inconsistent systems   Algebra II   Khan Academy.mp3", "Sentence": "Or if you divide both sides by 3, you get x is equal to negative 11 over 3. And this is the exact same thing as negative 3 and 2 thirds. So when y is 0, you have x being negative 3 and 2 thirds. So maybe this is about 6, so negative 3 and 2 thirds would be right about here. So this is the point negative 11 thirds, comma, 0. And so the second equation will look something like this. It will look something like that."}, {"video_title": "Consistent and inconsistent systems   Algebra II   Khan Academy.mp3", "Sentence": "So maybe this is about 6, so negative 3 and 2 thirds would be right about here. So this is the point negative 11 thirds, comma, 0. And so the second equation will look something like this. It will look something like that. Now clearly, and I might have not been completely precise when I did this hand-drawn graph, clearly these two guys intersect. They intersect right over here. And to answer their question, you don't even have to find the point that they intersect at."}, {"video_title": "Consistent and inconsistent systems   Algebra II   Khan Academy.mp3", "Sentence": "It will look something like that. Now clearly, and I might have not been completely precise when I did this hand-drawn graph, clearly these two guys intersect. They intersect right over here. And to answer their question, you don't even have to find the point that they intersect at. We just have to see very clearly that these two lines intersect. So this is a consistent system of equations. It has one solution."}, {"video_title": "Solving proportions 2 exercise examples   Algebra Basics   Khan Academy.mp3", "Sentence": "We have the proportion x minus 9 over 12 is equal to 2 over 3. And we want to solve for the x that satisfies this proportion. Now there's a bunch of ways that you could do it. A lot of people, as soon as they see a proportion like this, they want to cross multiply. They want to say, hey, 3 times x minus 9 is going to be equal to 2 times 12. And that's completely legitimate. You would get, let me write that down."}, {"video_title": "Solving proportions 2 exercise examples   Algebra Basics   Khan Academy.mp3", "Sentence": "A lot of people, as soon as they see a proportion like this, they want to cross multiply. They want to say, hey, 3 times x minus 9 is going to be equal to 2 times 12. And that's completely legitimate. You would get, let me write that down. 3 times x minus 9 is equal to 2 times 12. So it would be equal to 2 times 12. And then you can distribute the 3."}, {"video_title": "Solving proportions 2 exercise examples   Algebra Basics   Khan Academy.mp3", "Sentence": "You would get, let me write that down. 3 times x minus 9 is equal to 2 times 12. So it would be equal to 2 times 12. And then you can distribute the 3. You'd get 3x minus 27 is equal to 24. And then you could add 27 to both sides. And you would get, let me actually do that."}, {"video_title": "Solving proportions 2 exercise examples   Algebra Basics   Khan Academy.mp3", "Sentence": "And then you can distribute the 3. You'd get 3x minus 27 is equal to 24. And then you could add 27 to both sides. And you would get, let me actually do that. So let me add 27 to both sides. And we are left with 3x is equal to 51. And then x would be equal to 17."}, {"video_title": "Solving proportions 2 exercise examples   Algebra Basics   Khan Academy.mp3", "Sentence": "And you would get, let me actually do that. So let me add 27 to both sides. And we are left with 3x is equal to 51. And then x would be equal to 17. And you can verify that this works. 17 minus 9 is 8. 8 twelfths is the same thing as 2 thirds."}, {"video_title": "Solving proportions 2 exercise examples   Algebra Basics   Khan Academy.mp3", "Sentence": "And then x would be equal to 17. And you can verify that this works. 17 minus 9 is 8. 8 twelfths is the same thing as 2 thirds. So this checks out. Another way you could do that, instead of just straight up doing the cross multiplication, you could say, look, I want to get rid of this 12 in the denominator right over here. Multiply both sides by 12."}, {"video_title": "Solving proportions 2 exercise examples   Algebra Basics   Khan Academy.mp3", "Sentence": "8 twelfths is the same thing as 2 thirds. So this checks out. Another way you could do that, instead of just straight up doing the cross multiplication, you could say, look, I want to get rid of this 12 in the denominator right over here. Multiply both sides by 12. So if you multiply both sides by 12, on your left-hand side, you are just left with x minus 9. And on your right-hand side, 2 thirds times 12, well, 2 thirds of 12 is just 8. And you could do the actual multiplication, 2 thirds times 12 over 1."}, {"video_title": "Solving proportions 2 exercise examples   Algebra Basics   Khan Academy.mp3", "Sentence": "Multiply both sides by 12. So if you multiply both sides by 12, on your left-hand side, you are just left with x minus 9. And on your right-hand side, 2 thirds times 12, well, 2 thirds of 12 is just 8. And you could do the actual multiplication, 2 thirds times 12 over 1. 12 and 3. So 12 divided by 3 is 4. 3 divided by 3 is 1."}, {"video_title": "Solving proportions 2 exercise examples   Algebra Basics   Khan Academy.mp3", "Sentence": "And you could do the actual multiplication, 2 thirds times 12 over 1. 12 and 3. So 12 divided by 3 is 4. 3 divided by 3 is 1. So it becomes 2 times 4 over 1, which is just 8. And then you add 9 to both sides. So the fun of algebra is that as long as you do something that's logically consistent, you will get the right answer."}, {"video_title": "Solving proportions 2 exercise examples   Algebra Basics   Khan Academy.mp3", "Sentence": "3 divided by 3 is 1. So it becomes 2 times 4 over 1, which is just 8. And then you add 9 to both sides. So the fun of algebra is that as long as you do something that's logically consistent, you will get the right answer. There's no one way of doing it. So here you get x is equal to 17 again. And you could also multiply both sides by 12 and both sides by 3."}, {"video_title": "Solving proportions 2 exercise examples   Algebra Basics   Khan Academy.mp3", "Sentence": "So the fun of algebra is that as long as you do something that's logically consistent, you will get the right answer. There's no one way of doing it. So here you get x is equal to 17 again. And you could also multiply both sides by 12 and both sides by 3. And then that would be functionally equivalent to cross multiplying. Let's do one more. So here, another proportion."}, {"video_title": "Solving proportions 2 exercise examples   Algebra Basics   Khan Academy.mp3", "Sentence": "And you could also multiply both sides by 12 and both sides by 3. And then that would be functionally equivalent to cross multiplying. Let's do one more. So here, another proportion. And this time, the x is in the denominator. But just like before, if we want, we can cross multiply. And just to see where cross multiplying comes from, that it's not some voodoo, that you still are doing logical algebra, that you're doing the same thing to both sides of the equation, you just need to appreciate that we're just multiplying both sides by both denominators."}, {"video_title": "Solving proportions 2 exercise examples   Algebra Basics   Khan Academy.mp3", "Sentence": "So here, another proportion. And this time, the x is in the denominator. But just like before, if we want, we can cross multiply. And just to see where cross multiplying comes from, that it's not some voodoo, that you still are doing logical algebra, that you're doing the same thing to both sides of the equation, you just need to appreciate that we're just multiplying both sides by both denominators. So we have this 8 right over here on the left-hand side. If we want to get rid of this 8 on the left-hand side in the denominator, we can multiply the left-hand side by 8. But in order for the equality to hold true, I can't do something to just one side."}, {"video_title": "Solving proportions 2 exercise examples   Algebra Basics   Khan Academy.mp3", "Sentence": "And just to see where cross multiplying comes from, that it's not some voodoo, that you still are doing logical algebra, that you're doing the same thing to both sides of the equation, you just need to appreciate that we're just multiplying both sides by both denominators. So we have this 8 right over here on the left-hand side. If we want to get rid of this 8 on the left-hand side in the denominator, we can multiply the left-hand side by 8. But in order for the equality to hold true, I can't do something to just one side. I have to do it to both sides. Similarly, if I want to get this x plus 1 out of the denominator, I could multiply by x plus 1 right over here. But I have to do that on both sides if I want my equality to hold true."}, {"video_title": "Solving proportions 2 exercise examples   Algebra Basics   Khan Academy.mp3", "Sentence": "But in order for the equality to hold true, I can't do something to just one side. I have to do it to both sides. Similarly, if I want to get this x plus 1 out of the denominator, I could multiply by x plus 1 right over here. But I have to do that on both sides if I want my equality to hold true. And notice, when you do what we just did, this is going to be equivalent to cross multiplying. Because these 8's cancel out, and this x plus 1 cancels with that x plus 1 right over there. And you are left with x plus 1 times 7."}, {"video_title": "Solving proportions 2 exercise examples   Algebra Basics   Khan Academy.mp3", "Sentence": "But I have to do that on both sides if I want my equality to hold true. And notice, when you do what we just did, this is going to be equivalent to cross multiplying. Because these 8's cancel out, and this x plus 1 cancels with that x plus 1 right over there. And you are left with x plus 1 times 7. And I could write it as 7 times x plus 1 is equal to 5 times 8. Notice, this is exactly what you would have done if you were to cross multiply. Cross multiplication is just a shortcut of multiplying both sides by both denominators."}, {"video_title": "Solving proportions 2 exercise examples   Algebra Basics   Khan Academy.mp3", "Sentence": "And you are left with x plus 1 times 7. And I could write it as 7 times x plus 1 is equal to 5 times 8. Notice, this is exactly what you would have done if you were to cross multiply. Cross multiplication is just a shortcut of multiplying both sides by both denominators. We have 7 times x plus 1 is equal to 5 times 8. And now we can go and solve the algebra. So distributing the 7, we get 7x plus 7 is equal to 40."}, {"video_title": "Solving proportions 2 exercise examples   Algebra Basics   Khan Academy.mp3", "Sentence": "Cross multiplication is just a shortcut of multiplying both sides by both denominators. We have 7 times x plus 1 is equal to 5 times 8. And now we can go and solve the algebra. So distributing the 7, we get 7x plus 7 is equal to 40. And then subtracting 7 from both sides, we are left with 7x is equal to 33. Dividing both sides by 7, we are left with x is equal to 33 over 7. And if we want to write that as a mixed number, this is the same thing as 4 and 5 7's."}, {"video_title": "Exponents of decimals.mp3", "Sentence": "So let's say that I have 0.2 to the third power. Pause this video, see if you can figure out what that is going to be. Well, this would just mean if I take something to the third power, that means I take three of that number and I multiply them together. So it's 0.2 times 0.2 times 0.2. Well, what is this going to be equal to? Well, if I take 0.2 times 0.2, that is going to be 0.04. One way to think about it, two times two is four, and then I have one number behind the decimal to the right of the decimal here."}, {"video_title": "Exponents of decimals.mp3", "Sentence": "So it's 0.2 times 0.2 times 0.2. Well, what is this going to be equal to? Well, if I take 0.2 times 0.2, that is going to be 0.04. One way to think about it, two times two is four, and then I have one number behind the decimal to the right of the decimal here. I have another digit to the right of the decimal right over here, so my product is gonna have two digits to the right of the decimal, so it'd be 0.04. And then if I were to multiply that times 0.2, so if I were to multiply that together, what is that going to be equal to? Well, four times two is equal to eight, and now I have one, two, three numbers to the right of the decimal point, so my product is gonna have one, two, three numbers to the right of the decimal point."}, {"video_title": "Exponents of decimals.mp3", "Sentence": "One way to think about it, two times two is four, and then I have one number behind the decimal to the right of the decimal here. I have another digit to the right of the decimal right over here, so my product is gonna have two digits to the right of the decimal, so it'd be 0.04. And then if I were to multiply that times 0.2, so if I were to multiply that together, what is that going to be equal to? Well, four times two is equal to eight, and now I have one, two, three numbers to the right of the decimal point, so my product is gonna have one, two, three numbers to the right of the decimal point. So now that we've had a little bit of practice with that, let's do another example. So let's say that I were to ask you, what is 0.9 squared? Pause this video and see if you can figure that out."}, {"video_title": "Exponents of decimals.mp3", "Sentence": "Well, four times two is equal to eight, and now I have one, two, three numbers to the right of the decimal point, so my product is gonna have one, two, three numbers to the right of the decimal point. So now that we've had a little bit of practice with that, let's do another example. So let's say that I were to ask you, what is 0.9 squared? Pause this video and see if you can figure that out. All right, well, this is just going to be 0.9 times 0.9. And what's that going to be equal to? Well, you could just say nine times nine is going to be equal to 81."}, {"video_title": "Exponents of decimals.mp3", "Sentence": "Pause this video and see if you can figure that out. All right, well, this is just going to be 0.9 times 0.9. And what's that going to be equal to? Well, you could just say nine times nine is going to be equal to 81. And so let's see, in the two numbers that I'm multiplying, I have a total of one, two numbers, or two digits, to the right of the decimal point, so my answer's going to have one, two digits to the right of the decimal point, so I'd put the decimal right over there, and I'll put the zero, so 0.81. Another way to think about it is 9 tenths of 9 tenths is 81 hundredths, but there you go. Using exponents or taking exponents of decimals, it's the same as when we're taking it of integers, it's just in this case, you just have to do a little bit of a decimal multiplication."}, {"video_title": "Order of operations PEMDAS   Arithmetic properties   Pre-Algebra   Khan Academy (2).mp3", "Sentence": "And remember, we're going to do parentheses first, parentheses, p for parentheses, then exponents. Don't worry if you don't know what exponents are, because this has no exponents in them. Then you're going to do multiplication and division. They're at the same level. Then you do addition and subtraction. So some people remember PEMDAS. But if you remember PEMDAS, remember multiplication, division, same level, addition and subtraction, also at the same level."}, {"video_title": "Order of operations PEMDAS   Arithmetic properties   Pre-Algebra   Khan Academy (2).mp3", "Sentence": "They're at the same level. Then you do addition and subtraction. So some people remember PEMDAS. But if you remember PEMDAS, remember multiplication, division, same level, addition and subtraction, also at the same level. So let's figure what order of operations say that this should evaluate to. So the first thing we're going to do is our parentheses, and we have a lot of parentheses here. We have this expression in parentheses right there, and then even within that, we have these parentheses."}, {"video_title": "Order of operations PEMDAS   Arithmetic properties   Pre-Algebra   Khan Academy (2).mp3", "Sentence": "But if you remember PEMDAS, remember multiplication, division, same level, addition and subtraction, also at the same level. So let's figure what order of operations say that this should evaluate to. So the first thing we're going to do is our parentheses, and we have a lot of parentheses here. We have this expression in parentheses right there, and then even within that, we have these parentheses. So our order of operations say, look, do your parentheses first, but in order to evaluate this outer parentheses, this orange thing, we're going to have to evaluate this thing in yellow right there. So let's evaluate this whole thing. So how can we simplify it?"}, {"video_title": "Order of operations PEMDAS   Arithmetic properties   Pre-Algebra   Khan Academy (2).mp3", "Sentence": "We have this expression in parentheses right there, and then even within that, we have these parentheses. So our order of operations say, look, do your parentheses first, but in order to evaluate this outer parentheses, this orange thing, we're going to have to evaluate this thing in yellow right there. So let's evaluate this whole thing. So how can we simplify it? Well, if we look at just inside of it, the first thing we want to do is simplify the parentheses inside the parentheses. So you see this 5 minus 2 right there. We're going to do that first no matter what, and that's easy to evaluate."}, {"video_title": "Order of operations PEMDAS   Arithmetic properties   Pre-Algebra   Khan Academy (2).mp3", "Sentence": "So how can we simplify it? Well, if we look at just inside of it, the first thing we want to do is simplify the parentheses inside the parentheses. So you see this 5 minus 2 right there. We're going to do that first no matter what, and that's easy to evaluate. 5 minus 2 is 3. And so this simplifies to, I'll do it step by step. Once you get the hang of it, you can do multiple steps at once."}, {"video_title": "Order of operations PEMDAS   Arithmetic properties   Pre-Algebra   Khan Academy (2).mp3", "Sentence": "We're going to do that first no matter what, and that's easy to evaluate. 5 minus 2 is 3. And so this simplifies to, I'll do it step by step. Once you get the hang of it, you can do multiple steps at once. So this is going to be 7 plus 3 times the 5 minus 2, which is 3. And then you have, and all of those have parentheses around it. And of course, you have all this stuff on either side, the divide for, no, whoops, that's not what I want."}, {"video_title": "Order of operations PEMDAS   Arithmetic properties   Pre-Algebra   Khan Academy (2).mp3", "Sentence": "Once you get the hang of it, you can do multiple steps at once. So this is going to be 7 plus 3 times the 5 minus 2, which is 3. And then you have, and all of those have parentheses around it. And of course, you have all this stuff on either side, the divide for, no, whoops, that's not what I want. I wanted to copy and paste that right there. So copy, and then, no, that's giving me the wrong thing. It would have been easier, let me just rewrite it."}, {"video_title": "Order of operations PEMDAS   Arithmetic properties   Pre-Algebra   Khan Academy (2).mp3", "Sentence": "And of course, you have all this stuff on either side, the divide for, no, whoops, that's not what I want. I wanted to copy and paste that right there. So copy, and then, no, that's giving me the wrong thing. It would have been easier, let me just rewrite it. That's the easiest thing. I'm having technical difficulties. So divided by 4 times 2, and on this side, you had that 7 times 2 plus this thing in orange parentheses there."}, {"video_title": "Order of operations PEMDAS   Arithmetic properties   Pre-Algebra   Khan Academy (2).mp3", "Sentence": "It would have been easier, let me just rewrite it. That's the easiest thing. I'm having technical difficulties. So divided by 4 times 2, and on this side, you had that 7 times 2 plus this thing in orange parentheses there. Now, at any step, you just look again. We always want to do parentheses first. We keep wanting to do this until there's really no parentheses left."}, {"video_title": "Order of operations PEMDAS   Arithmetic properties   Pre-Algebra   Khan Academy (2).mp3", "Sentence": "So divided by 4 times 2, and on this side, you had that 7 times 2 plus this thing in orange parentheses there. Now, at any step, you just look again. We always want to do parentheses first. We keep wanting to do this until there's really no parentheses left. So we have to evaluate this parentheses in orange here. So we have to evaluate this thing first. But in order to evaluate this thing, we have to look inside of it."}, {"video_title": "Order of operations PEMDAS   Arithmetic properties   Pre-Algebra   Khan Academy (2).mp3", "Sentence": "We keep wanting to do this until there's really no parentheses left. So we have to evaluate this parentheses in orange here. So we have to evaluate this thing first. But in order to evaluate this thing, we have to look inside of it. And when you look inside of it, you have 7 plus 3 times 3. So if you just had 7 plus 3 times 3, how would you evaluate it? Well, look back to your order of operations."}, {"video_title": "Order of operations PEMDAS   Arithmetic properties   Pre-Algebra   Khan Academy (2).mp3", "Sentence": "But in order to evaluate this thing, we have to look inside of it. And when you look inside of it, you have 7 plus 3 times 3. So if you just had 7 plus 3 times 3, how would you evaluate it? Well, look back to your order of operations. We're inside the parentheses here, so inside of it, there are no longer any parentheses. So the next thing we should do is, there are no exponents, there is multiplication. So we do that before we do any addition or subtraction."}, {"video_title": "Order of operations PEMDAS   Arithmetic properties   Pre-Algebra   Khan Academy (2).mp3", "Sentence": "Well, look back to your order of operations. We're inside the parentheses here, so inside of it, there are no longer any parentheses. So the next thing we should do is, there are no exponents, there is multiplication. So we do that before we do any addition or subtraction. So we want to do the 3 times 3 before we add the 7. So this is going to be 7 plus, and the 3 times 3 we want to do first. We want to do the multiplication first."}, {"video_title": "Order of operations PEMDAS   Arithmetic properties   Pre-Algebra   Khan Academy (2).mp3", "Sentence": "So we do that before we do any addition or subtraction. So we want to do the 3 times 3 before we add the 7. So this is going to be 7 plus, and the 3 times 3 we want to do first. We want to do the multiplication first. 7 plus 9. That's going to be in the orange parentheses. And then you have the 7 times 2 plus that on the left-hand side."}, {"video_title": "Order of operations PEMDAS   Arithmetic properties   Pre-Algebra   Khan Academy (2).mp3", "Sentence": "We want to do the multiplication first. 7 plus 9. That's going to be in the orange parentheses. And then you have the 7 times 2 plus that on the left-hand side. You have the divided by 4 times 2 on the right-hand side. And now this, the thing in parentheses, because we still want to do the parentheses first, pretty easy to evaluate. What's 7 plus 9?"}, {"video_title": "Order of operations PEMDAS   Arithmetic properties   Pre-Algebra   Khan Academy (2).mp3", "Sentence": "And then you have the 7 times 2 plus that on the left-hand side. You have the divided by 4 times 2 on the right-hand side. And now this, the thing in parentheses, because we still want to do the parentheses first, pretty easy to evaluate. What's 7 plus 9? 7 plus 9 is 16. And so everything we have simplifies to 7 times 2 plus 16 divided by 4 times 2. Now, we don't have any parentheses left, so we don't have to worry about the p in PEMDAS."}, {"video_title": "Order of operations PEMDAS   Arithmetic properties   Pre-Algebra   Khan Academy (2).mp3", "Sentence": "What's 7 plus 9? 7 plus 9 is 16. And so everything we have simplifies to 7 times 2 plus 16 divided by 4 times 2. Now, we don't have any parentheses left, so we don't have to worry about the p in PEMDAS. We have no e, no exponents in this, so then we go straight to multiplication and division. We have a multiplication, we have some multiplication going on there, we have some division going on here, and a multiplication there. So we should do these next, before we do this addition right there."}, {"video_title": "Order of operations PEMDAS   Arithmetic properties   Pre-Algebra   Khan Academy (2).mp3", "Sentence": "Now, we don't have any parentheses left, so we don't have to worry about the p in PEMDAS. We have no e, no exponents in this, so then we go straight to multiplication and division. We have a multiplication, we have some multiplication going on there, we have some division going on here, and a multiplication there. So we should do these next, before we do this addition right there. So we could do this multiplication, we could do that multiplication. 7 times 2 is 14. We're going to wait to do that addition."}, {"video_title": "Order of operations PEMDAS   Arithmetic properties   Pre-Algebra   Khan Academy (2).mp3", "Sentence": "So we should do these next, before we do this addition right there. So we could do this multiplication, we could do that multiplication. 7 times 2 is 14. We're going to wait to do that addition. And then here we have a 16 divided by 4 times 2. That gets priority of the addition, so we're going to do that before we do the addition. But how do we evaluate that?"}, {"video_title": "Order of operations PEMDAS   Arithmetic properties   Pre-Algebra   Khan Academy (2).mp3", "Sentence": "We're going to wait to do that addition. And then here we have a 16 divided by 4 times 2. That gets priority of the addition, so we're going to do that before we do the addition. But how do we evaluate that? Do we do the division first, do we do the multiplication first? And remember, I told you in the last video, when you have multiple operations of the same level, in this case division and multiplication, they're at the same level, you're safest going left to right, or you should go left to right. So you do 16 divided by 4 is 4, so this thing right here simplifies 16 divided by 4 times 2, it simplifies to 4 times 2, that's this thing in green right there."}, {"video_title": "Order of operations PEMDAS   Arithmetic properties   Pre-Algebra   Khan Academy (2).mp3", "Sentence": "But how do we evaluate that? Do we do the division first, do we do the multiplication first? And remember, I told you in the last video, when you have multiple operations of the same level, in this case division and multiplication, they're at the same level, you're safest going left to right, or you should go left to right. So you do 16 divided by 4 is 4, so this thing right here simplifies 16 divided by 4 times 2, it simplifies to 4 times 2, that's this thing in green right there. And then we're going to want to do the multiplication next. So this is going to simplify to, because multiplication takes priority of addition, this simplifies to 8, and so you get 14, this 14 right here, plus 8. And what's 14 plus 8?"}, {"video_title": "Order of operations PEMDAS   Arithmetic properties   Pre-Algebra   Khan Academy (2).mp3", "Sentence": "So you do 16 divided by 4 is 4, so this thing right here simplifies 16 divided by 4 times 2, it simplifies to 4 times 2, that's this thing in green right there. And then we're going to want to do the multiplication next. So this is going to simplify to, because multiplication takes priority of addition, this simplifies to 8, and so you get 14, this 14 right here, plus 8. And what's 14 plus 8? That is 22. That is equal to 22. And we are done."}, {"video_title": "Solving two-step equations   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "We actually have two of them. So now we're going to try to figure out what x is. But before we even do that, what I want you to think about is a mathematical equation that can represent what is going on right here, that equates what we have on the left hand here to what we have on the right side of the scale right over there. And I'll give you a few seconds to think about it. So let's think about what we have on the left side here. We have three masses with mass x. So you could say that we have three x over here."}, {"video_title": "Solving two-step equations   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "And I'll give you a few seconds to think about it. So let's think about what we have on the left side here. We have three masses with mass x. So you could say that we have three x over here. We have three x's. And then we have two masses of one kilogram. So in total we have two kilograms, so plus two."}, {"video_title": "Solving two-step equations   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So you could say that we have three x over here. We have three x's. And then we have two masses of one kilogram. So in total we have two kilograms, so plus two. So one way to think about it, the total mass on the left hand side is three x plus two. Three masses with mass x plus two kilograms. That's what we have on the left hand side."}, {"video_title": "Solving two-step equations   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So in total we have two kilograms, so plus two. So one way to think about it, the total mass on the left hand side is three x plus two. Three masses with mass x plus two kilograms. That's what we have on the left hand side. Now let's think about what we have on the right hand side. We just have to count these. We have one, two, three, four, five, six, seven, eight, nine, 10, 11, 12, 13, 14."}, {"video_title": "Solving two-step equations   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "That's what we have on the left hand side. Now let's think about what we have on the right hand side. We just have to count these. We have one, two, three, four, five, six, seven, eight, nine, 10, 11, 12, 13, 14. 14 blocks each have a mass of one kilogram. So the total mass right over here is going to be 14 kilograms. So we get, and we see that the scale is balanced."}, {"video_title": "Solving two-step equations   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "We have one, two, three, four, five, six, seven, eight, nine, 10, 11, 12, 13, 14. 14 blocks each have a mass of one kilogram. So the total mass right over here is going to be 14 kilograms. So we get, and we see that the scale is balanced. It's not tilting down or upwards. So the scale is balanced. So this mass, this total mass right over here, must be equal to this total mass."}, {"video_title": "Solving two-step equations   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So we get, and we see that the scale is balanced. It's not tilting down or upwards. So the scale is balanced. So this mass, this total mass right over here, must be equal to this total mass. The scale is balanced. So we can write an equal sign. Maybe we'll do that in that white color."}, {"video_title": "Solving two-step equations   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So this mass, this total mass right over here, must be equal to this total mass. The scale is balanced. So we can write an equal sign. Maybe we'll do that in that white color. I don't like that brown. We can do it in that white color. Now, what I want you to think about, and you can think about it either through the symbols or think about it through the scale, is how would you go about, let's think about a few things."}, {"video_title": "Solving two-step equations   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Maybe we'll do that in that white color. I don't like that brown. We can do it in that white color. Now, what I want you to think about, and you can think about it either through the symbols or think about it through the scale, is how would you go about, let's think about a few things. How would you first go about at least getting rid of these little one kilogram blocks over here? And I'll give you a second to think about that. Well, the simplest thing is, well, you could take these one kilogram blocks off of the left-hand side."}, {"video_title": "Solving two-step equations   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Now, what I want you to think about, and you can think about it either through the symbols or think about it through the scale, is how would you go about, let's think about a few things. How would you first go about at least getting rid of these little one kilogram blocks over here? And I'll give you a second to think about that. Well, the simplest thing is, well, you could take these one kilogram blocks off of the left-hand side. But remember, if you just took these one kilogram blocks off the left-hand side and it was balanced before, now the left-hand side will be lighter and it'll move up. But we want to keep it balanced so that we can keep saying equal, that this mass is equal to that mass. So if we're going to remove two blocks from the left-hand side, we need to remove two blocks from the right-hand side."}, {"video_title": "Solving two-step equations   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "Well, the simplest thing is, well, you could take these one kilogram blocks off of the left-hand side. But remember, if you just took these one kilogram blocks off the left-hand side and it was balanced before, now the left-hand side will be lighter and it'll move up. But we want to keep it balanced so that we can keep saying equal, that this mass is equal to that mass. So if we're going to remove two blocks from the left-hand side, we need to remove two blocks from the right-hand side. So we can remove two there, and then we can remove two right over there. And mathematically, what we're essentially doing is we're subtracting two kilograms from each side. So we're subtracting two from this side."}, {"video_title": "Solving two-step equations   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So if we're going to remove two blocks from the left-hand side, we need to remove two blocks from the right-hand side. So we can remove two there, and then we can remove two right over there. And mathematically, what we're essentially doing is we're subtracting two kilograms from each side. So we're subtracting two from this side. So on the left-hand side, we now have 3x plus 2 minus 2. We're just left with 3x. And on the right-hand side, we had 14, and we took away 2."}, {"video_title": "Solving two-step equations   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So we're subtracting two from this side. So on the left-hand side, we now have 3x plus 2 minus 2. We're just left with 3x. And on the right-hand side, we had 14, and we took away 2. Let me write this. We took away 2, so we're going to be left with 12. We're going to be left with 12 blocks."}, {"video_title": "Solving two-step equations   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "And on the right-hand side, we had 14, and we took away 2. Let me write this. We took away 2, so we're going to be left with 12. We're going to be left with 12 blocks. And you see that there. The ones that I haven't crossed out, there's 12 left. And here, you only have three of those x blocks."}, {"video_title": "Solving two-step equations   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "We're going to be left with 12 blocks. And you see that there. The ones that I haven't crossed out, there's 12 left. And here, you only have three of those x blocks. And since we removed the exact same amount from both sides, our scale is still balanced. And our equation, 3x is now equal to 12. And now this turns into a problem very similar to what we saw in the last video."}, {"video_title": "Solving two-step equations   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "And here, you only have three of those x blocks. And since we removed the exact same amount from both sides, our scale is still balanced. And our equation, 3x is now equal to 12. And now this turns into a problem very similar to what we saw in the last video. So I will ask you, what can we do to isolate 1x? To only have 1x on the scale while keeping the scale, or 1x on the left-hand side of the scale, while keeping the scale balanced? So the easiest way to think about it is, if I want 1x on this left-hand side, that's 1 third of the total x is here."}, {"video_title": "Solving two-step equations   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "And now this turns into a problem very similar to what we saw in the last video. So I will ask you, what can we do to isolate 1x? To only have 1x on the scale while keeping the scale, or 1x on the left-hand side of the scale, while keeping the scale balanced? So the easiest way to think about it is, if I want 1x on this left-hand side, that's 1 third of the total x is here. So what if I were to essentially multiply the left-hand side by 1 third? But if I want to keep the scale balanced, I have to multiply the right-hand side by 1 third. And so if we can do that mathematically, so right over here, I can multiply the left-hand side by 1 third."}, {"video_title": "Solving two-step equations   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So the easiest way to think about it is, if I want 1x on this left-hand side, that's 1 third of the total x is here. So what if I were to essentially multiply the left-hand side by 1 third? But if I want to keep the scale balanced, I have to multiply the right-hand side by 1 third. And so if we can do that mathematically, so right over here, I can multiply the left-hand side by 1 third. But if I want to keep my scale balanced, I also have to multiply the right-hand side by 1 third. And multiplying it physically, that literally means just keeping 1 third of what we had here originally. So we would get rid of 2 of these."}, {"video_title": "Solving two-step equations   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "And so if we can do that mathematically, so right over here, I can multiply the left-hand side by 1 third. But if I want to keep my scale balanced, I also have to multiply the right-hand side by 1 third. And multiplying it physically, that literally means just keeping 1 third of what we had here originally. So we would get rid of 2 of these. And if we want to keep 1 third of what we had here originally, there are 12 blocks left over after removing those first 2. So 1 third of 12, we're only going to keep 4 of these little 1 kilogram boxes left. So let me remove all but 4."}, {"video_title": "Solving two-step equations   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So we would get rid of 2 of these. And if we want to keep 1 third of what we had here originally, there are 12 blocks left over after removing those first 2. So 1 third of 12, we're only going to keep 4 of these little 1 kilogram boxes left. So let me remove all but 4. So we're going to move those and remove those. I have left 1, 2, 3, 4 here. And so what you're left with, the only thing you have left is this x. I'll shade it in to show this is the one that we actually have left."}, {"video_title": "Solving two-step equations   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So let me remove all but 4. So we're going to move those and remove those. I have left 1, 2, 3, 4 here. And so what you're left with, the only thing you have left is this x. I'll shade it in to show this is the one that we actually have left. And then we have these boxes. We have these 1 kilogram boxes. And you see it mathematically right over here, 1 third times 3x."}, {"video_title": "Solving two-step equations   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "And so what you're left with, the only thing you have left is this x. I'll shade it in to show this is the one that we actually have left. And then we have these boxes. We have these 1 kilogram boxes. And you see it mathematically right over here, 1 third times 3x. Or you could have said 3x divided by 3. Either way, that gives us that. You could say that these 3's cancel out."}, {"video_title": "Solving two-step equations   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "And you see it mathematically right over here, 1 third times 3x. Or you could have said 3x divided by 3. Either way, that gives us that. You could say that these 3's cancel out. That'll give you just an x. And then on the right-hand side, 12 times 1 third, which is the same thing as 12 divided by 3, is equal to 4. And since we did the same thing to both sides, the scale is still balanced."}, {"video_title": "Example 3 Factoring quadratics as a perfect square of a sum (a+b)^2   Algebra I   Khan Academy.mp3", "Sentence": "We need to factor negative 4t squared minus 12t minus 9. And a good place to start is to say, well, are there any common factors for all of these terms? And when you look at them, well, these first two are divisible by 4, these last two are divisible by 3, but not all of them are divisible by any one number. Well, but you could factor out a negative 1. But even if you factor out a negative 1, so you say this is the same thing as negative 1, times positive 4t squared plus 12t plus 9, you still end up with a non-1 coefficient out here and on the second degree term, on the t squared term. So you might want to immediately start grouping this. And if you did factor it by grouping, it would work."}, {"video_title": "Example 3 Factoring quadratics as a perfect square of a sum (a+b)^2   Algebra I   Khan Academy.mp3", "Sentence": "Well, but you could factor out a negative 1. But even if you factor out a negative 1, so you say this is the same thing as negative 1, times positive 4t squared plus 12t plus 9, you still end up with a non-1 coefficient out here and on the second degree term, on the t squared term. So you might want to immediately start grouping this. And if you did factor it by grouping, it would work. You would get the right answer. But there is something you might be able to see, or there is something about this equation that might pop out at you that might make it a little bit simpler to solve. And to understand that, let's take a little bit of a break here on the right-hand side and just think about what happens if you take a plus b times a plus b, if you just have a binomial squared."}, {"video_title": "Example 3 Factoring quadratics as a perfect square of a sum (a+b)^2   Algebra I   Khan Academy.mp3", "Sentence": "And if you did factor it by grouping, it would work. You would get the right answer. But there is something you might be able to see, or there is something about this equation that might pop out at you that might make it a little bit simpler to solve. And to understand that, let's take a little bit of a break here on the right-hand side and just think about what happens if you take a plus b times a plus b, if you just have a binomial squared. Well, you have a times a, which is a squared. Then you have a times that b, which is plus ab. Then you have b times a, which is the same thing as ab."}, {"video_title": "Example 3 Factoring quadratics as a perfect square of a sum (a+b)^2   Algebra I   Khan Academy.mp3", "Sentence": "And to understand that, let's take a little bit of a break here on the right-hand side and just think about what happens if you take a plus b times a plus b, if you just have a binomial squared. Well, you have a times a, which is a squared. Then you have a times that b, which is plus ab. Then you have b times a, which is the same thing as ab. And then you have b times b, or you have b squared. And so if you add these middle two terms right here, you're left with a squared plus 2ab plus b squared. This is the square of a binomial."}, {"video_title": "Example 3 Factoring quadratics as a perfect square of a sum (a+b)^2   Algebra I   Khan Academy.mp3", "Sentence": "Then you have b times a, which is the same thing as ab. And then you have b times b, or you have b squared. And so if you add these middle two terms right here, you're left with a squared plus 2ab plus b squared. This is the square of a binomial. Now, does this right here, does 4t squared plus 12t plus 9 fit this pattern? Well, if 4t squared is a squared, so if this right here is a squared, if that is a squared right there, then what does a have to be? If this is a squared, then a would be equal to the square root of this."}, {"video_title": "Example 3 Factoring quadratics as a perfect square of a sum (a+b)^2   Algebra I   Khan Academy.mp3", "Sentence": "This is the square of a binomial. Now, does this right here, does 4t squared plus 12t plus 9 fit this pattern? Well, if 4t squared is a squared, so if this right here is a squared, if that is a squared right there, then what does a have to be? If this is a squared, then a would be equal to the square root of this. It would be 2t. And if this is b squared, let me do that in a different color. If this right here is b squared, if the 9 is b squared right there, then that means that b is equal to 3, equal to the positive square root of the 9."}, {"video_title": "Example 3 Factoring quadratics as a perfect square of a sum (a+b)^2   Algebra I   Khan Academy.mp3", "Sentence": "If this is a squared, then a would be equal to the square root of this. It would be 2t. And if this is b squared, let me do that in a different color. If this right here is b squared, if the 9 is b squared right there, then that means that b is equal to 3, equal to the positive square root of the 9. Now, this number right here, and actually it doesn't have to just be equal to 3. It might have been negative 3 as well. It could be plus or minus 3."}, {"video_title": "Example 3 Factoring quadratics as a perfect square of a sum (a+b)^2   Algebra I   Khan Academy.mp3", "Sentence": "If this right here is b squared, if the 9 is b squared right there, then that means that b is equal to 3, equal to the positive square root of the 9. Now, this number right here, and actually it doesn't have to just be equal to 3. It might have been negative 3 as well. It could be plus or minus 3. But this number here, is it 2 times ab? That's the middle term that we care about. Is it 2 times ab?"}, {"video_title": "Example 3 Factoring quadratics as a perfect square of a sum (a+b)^2   Algebra I   Khan Academy.mp3", "Sentence": "It could be plus or minus 3. But this number here, is it 2 times ab? That's the middle term that we care about. Is it 2 times ab? Well, if we multiply 2t times 3, we get 6t. And then we multiply that times 2, you get 12t. This right here, 12t, is equal to 2 times 2t times 3."}, {"video_title": "Example 3 Factoring quadratics as a perfect square of a sum (a+b)^2   Algebra I   Khan Academy.mp3", "Sentence": "Is it 2 times ab? Well, if we multiply 2t times 3, we get 6t. And then we multiply that times 2, you get 12t. This right here, 12t, is equal to 2 times 2t times 3. It is 2 times ab. And if this was a negative 3, we would look to see if this was a negative 12, but this does work for a positive 3. So this does fit the pattern of a perfect square."}, {"video_title": "Example 3 Factoring quadratics as a perfect square of a sum (a+b)^2   Algebra I   Khan Academy.mp3", "Sentence": "This right here, 12t, is equal to 2 times 2t times 3. It is 2 times ab. And if this was a negative 3, we would look to see if this was a negative 12, but this does work for a positive 3. So this does fit the pattern of a perfect square. This is a special type of, or you could view this as a square of a binomial. So if you wanted to factor this, the stuff on the inside, you still have that negative 1 out there, the 4t squared plus 12t plus 9, you could immediately say, well that's going to be a plus b times a plus b. Or 2t plus 3 times 2t plus 3."}, {"video_title": "Example 3 Factoring quadratics as a perfect square of a sum (a+b)^2   Algebra I   Khan Academy.mp3", "Sentence": "So this does fit the pattern of a perfect square. This is a special type of, or you could view this as a square of a binomial. So if you wanted to factor this, the stuff on the inside, you still have that negative 1 out there, the 4t squared plus 12t plus 9, you could immediately say, well that's going to be a plus b times a plus b. Or 2t plus 3 times 2t plus 3. Or you could just say it's 2t plus 3 squared. It fits this pattern. And of course, you can't forget about this negative 1 out here."}, {"video_title": "Example 3 Factoring quadratics as a perfect square of a sum (a+b)^2   Algebra I   Khan Academy.mp3", "Sentence": "Or 2t plus 3 times 2t plus 3. Or you could just say it's 2t plus 3 squared. It fits this pattern. And of course, you can't forget about this negative 1 out here. You could have also solved it by grouping, but this might be a quicker thing to recognize. This is a number squared. That's another number squared."}, {"video_title": "How to find two function inputs with the same output given graph (example)   Khan Academy.mp3", "Sentence": "The graph of the function f is shown below. What is the input value other than negative five for which f of x is equal to f of negative five? So we have our x-axis, we have our y-axis, and then in blue, they've graphed y equals f of x. So for example, when x is equal to one, f of x, when y is going to be equal to f of x, that's what this graph is, f of x is equal to one. When x is equal to seven, f of seven, we see, is equal to five. When x is equal to nine, we see that f of nine is equal to six. So what is the input value other than negative five for which f of x is equal to f of negative five?"}, {"video_title": "How to find two function inputs with the same output given graph (example)   Khan Academy.mp3", "Sentence": "So for example, when x is equal to one, f of x, when y is going to be equal to f of x, that's what this graph is, f of x is equal to one. When x is equal to seven, f of seven, we see, is equal to five. When x is equal to nine, we see that f of nine is equal to six. So what is the input value other than negative five for which f of x is equal to f of negative five? So let's see. If x is equal to negative five, f of negative five is, we move up here to the graph, it is equal to four, because this, once again, this is the graph y is equal to f of x. So the y-coordinate here, this is what f of x, this is what f of negative five is equal to."}, {"video_title": "How to find two function inputs with the same output given graph (example)   Khan Academy.mp3", "Sentence": "So what is the input value other than negative five for which f of x is equal to f of negative five? So let's see. If x is equal to negative five, f of negative five is, we move up here to the graph, it is equal to four, because this, once again, this is the graph y is equal to f of x. So the y-coordinate here, this is what f of x, this is what f of negative five is equal to. So f of negative five is equal to four. So where else does that happen? So let's see, let's just move horizontally to the right."}, {"video_title": "How to find two function inputs with the same output given graph (example)   Khan Academy.mp3", "Sentence": "So the y-coordinate here, this is what f of x, this is what f of negative five is equal to. So f of negative five is equal to four. So where else does that happen? So let's see, let's just move horizontally to the right. That happens here as well. So what input do we have to give? What's the x-coordinate here to get y is equal to f of x, or f of x is equal to four right over here?"}, {"video_title": "How to find two function inputs with the same output given graph (example)   Khan Academy.mp3", "Sentence": "So let's see, let's just move horizontally to the right. That happens here as well. So what input do we have to give? What's the x-coordinate here to get y is equal to f of x, or f of x is equal to four right over here? Well, we see the x-coordinate is also four. So this tells us that f of four is equal to four, which is the same thing as f of negative five. So when x is four, the function takes on the same value as when x is negative five."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "And in particular, I want to focus on quadratics that don't have a 1 as a leading coefficient. For example, if I wanted to factor 4x squared plus 25x minus 21. Everything we've factored so far, or all of the quadratics we've factored so far, had either a 1 or a negative 1 where this 4 is sitting. All of a sudden now we have this 4 here. So what I'm going to teach you is a technique called factoring by grouping. And it's a little bit more involved than what we've learned before, but it's a neat trick. But to some degree, it'll become obsolete once you learn the quadratic formula, because frankly, the quadratic formula is a lot easier."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "All of a sudden now we have this 4 here. So what I'm going to teach you is a technique called factoring by grouping. And it's a little bit more involved than what we've learned before, but it's a neat trick. But to some degree, it'll become obsolete once you learn the quadratic formula, because frankly, the quadratic formula is a lot easier. But this is how it goes. I'll show you the technique, and then at the end of this video, I'll actually show you why it works. So what we need to do here is we need to think of two numbers."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "But to some degree, it'll become obsolete once you learn the quadratic formula, because frankly, the quadratic formula is a lot easier. But this is how it goes. I'll show you the technique, and then at the end of this video, I'll actually show you why it works. So what we need to do here is we need to think of two numbers. We're going to think of two numbers, a and b, where a times b is equal to 4 times negative 21. So a times b is going to be equal to 4 times negative 21, which is equal to negative 84. And those same two numbers, a and b, a plus b, need to be equal to 25."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "So what we need to do here is we need to think of two numbers. We're going to think of two numbers, a and b, where a times b is equal to 4 times negative 21. So a times b is going to be equal to 4 times negative 21, which is equal to negative 84. And those same two numbers, a and b, a plus b, need to be equal to 25. They need to be equal to 25. Let me be very clear. This is the 25, so they need to be equal to 25."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "And those same two numbers, a and b, a plus b, need to be equal to 25. They need to be equal to 25. Let me be very clear. This is the 25, so they need to be equal to 25. This is where the 4 is, so they need to be equal to 4 times negative 21. That's a negative 21. So what two numbers are there that would do this?"}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "This is the 25, so they need to be equal to 25. This is where the 4 is, so they need to be equal to 4 times negative 21. That's a negative 21. So what two numbers are there that would do this? Well, we have to look at the factors of negative 84. And once again, one of these are going to have to be positive. The other ones are going to have to be negative, because their product is negative."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "So what two numbers are there that would do this? Well, we have to look at the factors of negative 84. And once again, one of these are going to have to be positive. The other ones are going to have to be negative, because their product is negative. So let's think about the different factors that might work. 4 and negative 21 look tantalizing, but when you add them, you get negative 17. Or if you had negative 4 and 21, you'd get positive 17."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "The other ones are going to have to be negative, because their product is negative. So let's think about the different factors that might work. 4 and negative 21 look tantalizing, but when you add them, you get negative 17. Or if you had negative 4 and 21, you'd get positive 17. It doesn't work. Let's try some other combinations. 1 and 84, too far apart when you take their difference, because that's essentially what you're going to do if one is negative and one is positive."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "Or if you had negative 4 and 21, you'd get positive 17. It doesn't work. Let's try some other combinations. 1 and 84, too far apart when you take their difference, because that's essentially what you're going to do if one is negative and one is positive. Too far apart. Let's see. You could do 2 and 42."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "1 and 84, too far apart when you take their difference, because that's essentially what you're going to do if one is negative and one is positive. Too far apart. Let's see. You could do 2 and 42. Once again, too far apart. Negative 2 plus 42 is 40. 2 plus negative 2 is negative 40."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "You could do 2 and 42. Once again, too far apart. Negative 2 plus 42 is 40. 2 plus negative 2 is negative 40. Too far apart. 3 goes into 84. 3 goes into 8."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "2 plus negative 2 is negative 40. Too far apart. 3 goes into 84. 3 goes into 8. 2 times 3 is 6. 8 minus 6 is 2. Bring down the 4."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "3 goes into 8. 2 times 3 is 6. 8 minus 6 is 2. Bring down the 4. It goes exactly 8 times. So 3 and 28. This seems interesting."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "Bring down the 4. It goes exactly 8 times. So 3 and 28. This seems interesting. 3 and 28. And remember, one of these has to be negative. So if we have negative 3 plus 28, that is equal to 25."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "This seems interesting. 3 and 28. And remember, one of these has to be negative. So if we have negative 3 plus 28, that is equal to 25. Now, we found our two numbers, but it's not going to be quite as simple of an operation as what we did when this wasn't a 1 or when this was a 1 or a negative 1. What we're going to do now is split up this term right here. We're going to split this up into negative... We're going to split it up into positive 28x minus 3x."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "So if we have negative 3 plus 28, that is equal to 25. Now, we found our two numbers, but it's not going to be quite as simple of an operation as what we did when this wasn't a 1 or when this was a 1 or a negative 1. What we're going to do now is split up this term right here. We're going to split this up into negative... We're going to split it up into positive 28x minus 3x. We're just going to split that term. That term is that term right there. And of course, you have your minus 21 there."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "We're going to split this up into negative... We're going to split it up into positive 28x minus 3x. We're just going to split that term. That term is that term right there. And of course, you have your minus 21 there. And you have your 4x squared over here. Now, you might say, how did you pick the 28 to go here and the negative 3 to go there? And it actually does matter."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "And of course, you have your minus 21 there. And you have your 4x squared over here. Now, you might say, how did you pick the 28 to go here and the negative 3 to go there? And it actually does matter. The way I thought about it is 3 or negative 3 and 21 or negative 21, they have some common factors in particular. They have the factor 3 in common. And 28 and 4 have some common factors."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "And it actually does matter. The way I thought about it is 3 or negative 3 and 21 or negative 21, they have some common factors in particular. They have the factor 3 in common. And 28 and 4 have some common factors. So I kind of grouped the 28 on the side of the 4. And you're going to see what I mean in a second. If we literally group these, so that term becomes 4x squared plus 28x."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "And 28 and 4 have some common factors. So I kind of grouped the 28 on the side of the 4. And you're going to see what I mean in a second. If we literally group these, so that term becomes 4x squared plus 28x. And then this side over here in pink, well, I could say it's plus negative 3x minus 21. Once again, I picked these. I grouped the negative 3 with the 21 or the negative 21 because they're both divisible by 3."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "If we literally group these, so that term becomes 4x squared plus 28x. And then this side over here in pink, well, I could say it's plus negative 3x minus 21. Once again, I picked these. I grouped the negative 3 with the 21 or the negative 21 because they're both divisible by 3. And I grouped the 28 with the 4 because they're both divisible by 4. And now, in each of these groups, we factor as much out as we can. So both of these terms are divisible by 4x."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "I grouped the negative 3 with the 21 or the negative 21 because they're both divisible by 3. And I grouped the 28 with the 4 because they're both divisible by 4. And now, in each of these groups, we factor as much out as we can. So both of these terms are divisible by 4x. So this orange term is equal to 4x times x. 4x squared divided by 4x is just x. Plus 28x divided by 4x is just 7."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "So both of these terms are divisible by 4x. So this orange term is equal to 4x times x. 4x squared divided by 4x is just x. Plus 28x divided by 4x is just 7. Now, this second term, remember, you factor out everything that you can factor out. Well, both of these terms are divisible by 3 or negative 3. So let's factor out a negative 3."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "Plus 28x divided by 4x is just 7. Now, this second term, remember, you factor out everything that you can factor out. Well, both of these terms are divisible by 3 or negative 3. So let's factor out a negative 3. And this becomes x plus 7. And now, something might pop out at you. We have x plus 7 times 4x plus x plus 7 times negative 3."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "So let's factor out a negative 3. And this becomes x plus 7. And now, something might pop out at you. We have x plus 7 times 4x plus x plus 7 times negative 3. So we can factor out an x plus 7. We can factor out an x plus 7. This might not be completely obvious."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "We have x plus 7 times 4x plus x plus 7 times negative 3. So we can factor out an x plus 7. We can factor out an x plus 7. This might not be completely obvious. You're probably not used to factoring out an entire binomial, but you could view this as like a. Or if you have 4xa minus 3a, you would be able to factor out an a. And I could just leave this as a minus sign."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "This might not be completely obvious. You're probably not used to factoring out an entire binomial, but you could view this as like a. Or if you have 4xa minus 3a, you would be able to factor out an a. And I could just leave this as a minus sign. Let me delete this plus right here because it's just minus 3. It's just minus 3. Plus negative 3, same thing as minus 3."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "And I could just leave this as a minus sign. Let me delete this plus right here because it's just minus 3. It's just minus 3. Plus negative 3, same thing as minus 3. So what can we do here? We have x plus 7 times 4x. We have an x plus 7 times negative 3."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "Plus negative 3, same thing as minus 3. So what can we do here? We have x plus 7 times 4x. We have an x plus 7 times negative 3. Let's factor out the x plus 7. We get x plus 7 times 4x minus 3. Minus that 3 right there."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "We have an x plus 7 times negative 3. Let's factor out the x plus 7. We get x plus 7 times 4x minus 3. Minus that 3 right there. And we've factored our binomial. Sorry, we've factored our quadratic by grouping, and we factored it into two binomials. Let's do another example of that and it's a little bit involved, but once you get the hang of it, it's kind of fun."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "Minus that 3 right there. And we've factored our binomial. Sorry, we've factored our quadratic by grouping, and we factored it into two binomials. Let's do another example of that and it's a little bit involved, but once you get the hang of it, it's kind of fun. So let's say we want to factor 6x squared plus 7x plus 1. Same drill. We want to find a times b that is equal to 1 times 6, and we want to find an a plus b needs to be equal to 7."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "Let's do another example of that and it's a little bit involved, but once you get the hang of it, it's kind of fun. So let's say we want to factor 6x squared plus 7x plus 1. Same drill. We want to find a times b that is equal to 1 times 6, and we want to find an a plus b needs to be equal to 7. This is a little bit more straightforward. The obvious one is 1 and 6. 1 times 6 is 6."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "We want to find a times b that is equal to 1 times 6, and we want to find an a plus b needs to be equal to 7. This is a little bit more straightforward. The obvious one is 1 and 6. 1 times 6 is 6. 1 plus 6 is 7. So we have a is equal to 1. Let me not even assign them."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "1 times 6 is 6. 1 plus 6 is 7. So we have a is equal to 1. Let me not even assign them. The numbers here are 1 and 6. Now we want to split this into a 1x and a 6x, but we want to group it so it's on the side of something that it shares a factor with. So we're going to have a 6x squared here plus, and so I'm going to put the 6x first because 6 and 6 share a factor."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "Let me not even assign them. The numbers here are 1 and 6. Now we want to split this into a 1x and a 6x, but we want to group it so it's on the side of something that it shares a factor with. So we're going to have a 6x squared here plus, and so I'm going to put the 6x first because 6 and 6 share a factor. Then we're going to have plus 1x. 6x plus 1x is 7x. That was the whole point."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "So we're going to have a 6x squared here plus, and so I'm going to put the 6x first because 6 and 6 share a factor. Then we're going to have plus 1x. 6x plus 1x is 7x. That was the whole point. They had to add up to 7. Then we have the final plus 1 there. Now in each of these groups, we can factor out as much as we like."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "That was the whole point. They had to add up to 7. Then we have the final plus 1 there. Now in each of these groups, we can factor out as much as we like. So in this first group, let's factor out a 6x. So this first group becomes 6x times 6x squared divided by 6x is just an x. 6x divided by 6x is just a 1."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "Now in each of these groups, we can factor out as much as we like. So in this first group, let's factor out a 6x. So this first group becomes 6x times 6x squared divided by 6x is just an x. 6x divided by 6x is just a 1. Then in the second group, we're going to have a plus here, but this second group, we just literally have an x plus 1, or we could even write a 1 times an x plus 1. You can imagine I just factored out a 1, so to speak. Now I have 6x times x plus 1 plus 1 times x plus 1."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "6x divided by 6x is just a 1. Then in the second group, we're going to have a plus here, but this second group, we just literally have an x plus 1, or we could even write a 1 times an x plus 1. You can imagine I just factored out a 1, so to speak. Now I have 6x times x plus 1 plus 1 times x plus 1. Well, I can factor out the x plus 1. If I factor out an x plus 1, that's equal to x plus 1 times 6x plus that 1. I'm just doing the distributive property in reverse."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "Now I have 6x times x plus 1 plus 1 times x plus 1. Well, I can factor out the x plus 1. If I factor out an x plus 1, that's equal to x plus 1 times 6x plus that 1. I'm just doing the distributive property in reverse. So hopefully you didn't find that too bad. Now I'm going to actually explain why this little magical system actually works. Why it actually works."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "I'm just doing the distributive property in reverse. So hopefully you didn't find that too bad. Now I'm going to actually explain why this little magical system actually works. Why it actually works. Let me take an example. Let's say I have, well, I'll do it in very general terms. Let's add ax plus b times cx."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "Why it actually works. Let me take an example. Let's say I have, well, I'll do it in very general terms. Let's add ax plus b times cx. Actually, I don't want to use, well, I'm afraid to use the a's and the b's. I think that'll confuse you because I use a's and b's here, and they won't be the same thing. Let me use completely different letters."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "Let's add ax plus b times cx. Actually, I don't want to use, well, I'm afraid to use the a's and the b's. I think that'll confuse you because I use a's and b's here, and they won't be the same thing. Let me use completely different letters. Let's say I have fx plus g times hx plus, I'll use j instead of i. You'll learn in the future why I don't like using i as a variable. So what is this going to be equal to?"}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "Let me use completely different letters. Let's say I have fx plus g times hx plus, I'll use j instead of i. You'll learn in the future why I don't like using i as a variable. So what is this going to be equal to? Well, it's going to be fx times hx, which is fhx, and then fx times j, so plus fjx. Then we're going to have g times hx, so plus ghx, and then g times j, plus gj. Or if we add these two middle terms, if you add the two middle terms, you have fh times x plus, add these two terms, fj plus ghx, plus gj."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "So what is this going to be equal to? Well, it's going to be fx times hx, which is fhx, and then fx times j, so plus fjx. Then we're going to have g times hx, so plus ghx, and then g times j, plus gj. Or if we add these two middle terms, if you add the two middle terms, you have fh times x plus, add these two terms, fj plus ghx, plus gj. Now, what did I do here? Well, remember, in all of these problems where you have a non-1 or non-negative-1 coefficient, we look for two numbers that add up to this whose product is equal to the product of that times that. Well, here we have two numbers that add up."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "Or if we add these two middle terms, if you add the two middle terms, you have fh times x plus, add these two terms, fj plus ghx, plus gj. Now, what did I do here? Well, remember, in all of these problems where you have a non-1 or non-negative-1 coefficient, we look for two numbers that add up to this whose product is equal to the product of that times that. Well, here we have two numbers that add up. Let's say that a is equal to fj. Let's say that a is equal to fj. That is a, and b is equal to gh."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "Well, here we have two numbers that add up. Let's say that a is equal to fj. Let's say that a is equal to fj. That is a, and b is equal to gh. So a plus b is going to be equal to that middle coefficient. a plus b is going to be equal to that middle coefficient there. And then what is a times b?"}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "That is a, and b is equal to gh. So a plus b is going to be equal to that middle coefficient. a plus b is going to be equal to that middle coefficient there. And then what is a times b? a times b is going to be equal to fj times gh, which we could just reorder these terms. We're just multiplying a bunch of terms, so that could be rewritten as f times h times g times j. These are all the same things."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "And then what is a times b? a times b is going to be equal to fj times gh, which we could just reorder these terms. We're just multiplying a bunch of terms, so that could be rewritten as f times h times g times j. These are all the same things. Well, what is fh times gj? This is equal to fh times gj. Well, this is equal to the first coefficient times the constant term."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "These are all the same things. Well, what is fh times gj? This is equal to fh times gj. Well, this is equal to the first coefficient times the constant term. So if a plus b will be equal to the middle coefficient, then a times b will equal the first coefficient times the constant term. So that's why this whole factoring by grouping even works, or how we're able to figure out what a and b even are. Now I'm going to close up with something slightly different, but just to make sure that you have a well-rounded education in factoring things."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "Well, this is equal to the first coefficient times the constant term. So if a plus b will be equal to the middle coefficient, then a times b will equal the first coefficient times the constant term. So that's why this whole factoring by grouping even works, or how we're able to figure out what a and b even are. Now I'm going to close up with something slightly different, but just to make sure that you have a well-rounded education in factoring things. What I want to do is teach you to factor things a little bit more completely. This is a little bit of an add-on. I was going to make a whole video on this, but I think on some level it might be a little obvious for you."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "Now I'm going to close up with something slightly different, but just to make sure that you have a well-rounded education in factoring things. What I want to do is teach you to factor things a little bit more completely. This is a little bit of an add-on. I was going to make a whole video on this, but I think on some level it might be a little obvious for you. So let's say we had 2... Let me get a good one here. Let's say we had negative x to the third plus 17x squared minus 70. Now, I have 70x."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "I was going to make a whole video on this, but I think on some level it might be a little obvious for you. So let's say we had 2... Let me get a good one here. Let's say we had negative x to the third plus 17x squared minus 70. Now, I have 70x. Now, immediately you say, gee, this isn't even a quadratic. I don't know how to solve something like this. It has an x to the third power."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "Now, I have 70x. Now, immediately you say, gee, this isn't even a quadratic. I don't know how to solve something like this. It has an x to the third power. The first thing you should realize is that every term here is divisible by x. So let's factor out an x, or even better, let's factor out a negative x. So if you factor out a negative x, this is equal to negative x times..."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "It has an x to the third power. The first thing you should realize is that every term here is divisible by x. So let's factor out an x, or even better, let's factor out a negative x. So if you factor out a negative x, this is equal to negative x times... Negative x to the third divided by negative x is x squared. 17x squared divided by negative x is negative 17x. Negative 70x divided by negative x is positive 70."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "So if you factor out a negative x, this is equal to negative x times... Negative x to the third divided by negative x is x squared. 17x squared divided by negative x is negative 17x. Negative 70x divided by negative x is positive 70. These cancel out. And now you have something that might look a little bit familiar. We have just a standard quadratic where the leading coefficient is a 1, so we just have to find 2 numbers whose product is 70 and that add up to negative 17."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "Negative 70x divided by negative x is positive 70. These cancel out. And now you have something that might look a little bit familiar. We have just a standard quadratic where the leading coefficient is a 1, so we just have to find 2 numbers whose product is 70 and that add up to negative 17. And the numbers that immediately jumped into my head are negative 10 and negative 7. You take their product, you get 70, you add them up, you get negative 17. So this part right here is going to be x minus 10 times x minus 7."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "We have just a standard quadratic where the leading coefficient is a 1, so we just have to find 2 numbers whose product is 70 and that add up to negative 17. And the numbers that immediately jumped into my head are negative 10 and negative 7. You take their product, you get 70, you add them up, you get negative 17. So this part right here is going to be x minus 10 times x minus 7. And of course, you have that leading negative x. The general idea here is just see if there's anything you can factor out and then it will get into a form that you might recognize. Hopefully you found this helpful."}, {"video_title": "More examples of factoring by grouping   Algebra I   Khan Academy.mp3", "Sentence": "So this part right here is going to be x minus 10 times x minus 7. And of course, you have that leading negative x. The general idea here is just see if there's anything you can factor out and then it will get into a form that you might recognize. Hopefully you found this helpful. Now, I want to reiterate what I showed you at the beginning of this video. I think it's a really cool trick, so to speak, to be able to factor things that have a non-1 or non-negative 1 leading coefficient. But to some degree, you're going to find out easier ways to do this, especially with the quadratic formula, in not too long."}, {"video_title": "Reporting measurements   Working with units   Algebra 1   Khan Academy.mp3", "Sentence": "You're like, well, no, of course we can measure the dimensions of something. Let's say I have some type of a gear over here. So let me draw my gear. And if I were to ask you, that's not the best drawn gear, but if I were to ask you, what's the inner diameter of the hole of the gear right over here? Maybe you take a ruler out right over here. So this is my ruler. And that you are able to see when you measure it that it is one centimeter in diameter."}, {"video_title": "Reporting measurements   Working with units   Algebra 1   Khan Academy.mp3", "Sentence": "And if I were to ask you, that's not the best drawn gear, but if I were to ask you, what's the inner diameter of the hole of the gear right over here? Maybe you take a ruler out right over here. So this is my ruler. And that you are able to see when you measure it that it is one centimeter in diameter. But then I say, is it exactly one centimeter? And then you realize, well, yeah, let me get a little bit more precise. Maybe you get a magnifying glass out here."}, {"video_title": "Reporting measurements   Working with units   Algebra 1   Khan Academy.mp3", "Sentence": "And that you are able to see when you measure it that it is one centimeter in diameter. But then I say, is it exactly one centimeter? And then you realize, well, yeah, let me get a little bit more precise. Maybe you get a magnifying glass out here. So this is the lens of my magnifying glass. And you zoom in a little bit. Maybe you get a better ruler that marks off the millimeters."}, {"video_title": "Reporting measurements   Working with units   Algebra 1   Khan Academy.mp3", "Sentence": "Maybe you get a magnifying glass out here. So this is the lens of my magnifying glass. And you zoom in a little bit. Maybe you get a better ruler that marks off the millimeters. And you actually say, oh, well, when I look a little bit closer, it actually turns out it's not exactly one centimeter. It's actually closer to 1.1 centimeters. And then I ask you, is that exactly the inner diameter of this gear here?"}, {"video_title": "Reporting measurements   Working with units   Algebra 1   Khan Academy.mp3", "Sentence": "Maybe you get a better ruler that marks off the millimeters. And you actually say, oh, well, when I look a little bit closer, it actually turns out it's not exactly one centimeter. It's actually closer to 1.1 centimeters. And then I ask you, is that exactly the inner diameter of this gear here? And you're like, okay, well, let me get out a microscope. And then you realize, oh, you're right. It's actually 1.089 centimeters."}, {"video_title": "Reporting measurements   Working with units   Algebra 1   Khan Academy.mp3", "Sentence": "And then I ask you, is that exactly the inner diameter of this gear here? And you're like, okay, well, let me get out a microscope. And then you realize, oh, you're right. It's actually 1.089 centimeters. And then I ask you, is that exactly right? And then you're like, yeah, I guess you're right. I haven't measured to the nearest, to the height or the width of an atom."}, {"video_title": "Reporting measurements   Working with units   Algebra 1   Khan Academy.mp3", "Sentence": "It's actually 1.089 centimeters. And then I ask you, is that exactly right? And then you're like, yeah, I guess you're right. I haven't measured to the nearest, to the height or the width of an atom. To do that, I would need a lot more precision right over here. And so maybe I need some type of an electron microscope. But even if you're able to do that, and that would be many decimal places to the right of the decimal point here if you're measuring in centimeters, you could still ask, well, is that exactly right?"}, {"video_title": "Reporting measurements   Working with units   Algebra 1   Khan Academy.mp3", "Sentence": "I haven't measured to the nearest, to the height or the width of an atom. To do that, I would need a lot more precision right over here. And so maybe I need some type of an electron microscope. But even if you're able to do that, and that would be many decimal places to the right of the decimal point here if you're measuring in centimeters, you could still ask, well, is that exactly right? Maybe you could measure the parts of an atom or to a measurement even smaller than an atom. And later on, you might study quantum physics and there are some levels of granularity where you can't get a true measurement below that. But as you can see, it is somewhat arbitrary for our everyday life."}, {"video_title": "Reporting measurements   Working with units   Algebra 1   Khan Academy.mp3", "Sentence": "But even if you're able to do that, and that would be many decimal places to the right of the decimal point here if you're measuring in centimeters, you could still ask, well, is that exactly right? Maybe you could measure the parts of an atom or to a measurement even smaller than an atom. And later on, you might study quantum physics and there are some levels of granularity where you can't get a true measurement below that. But as you can see, it is somewhat arbitrary for our everyday life. And so the question is, which one do you pick? Or how much trouble do you get? Or how much trouble do you take to get to these different levels of precision?"}, {"video_title": "Reporting measurements   Working with units   Algebra 1   Khan Academy.mp3", "Sentence": "But as you can see, it is somewhat arbitrary for our everyday life. And so the question is, which one do you pick? Or how much trouble do you get? Or how much trouble do you take to get to these different levels of precision? And the answer is, it just depends. If the goal was, hey, we just wanna make multiple copies of maybe jewelry of this little cog here. So we're gonna wanna put some type of, I don't know, gold chain through it."}, {"video_title": "Reporting measurements   Working with units   Algebra 1   Khan Academy.mp3", "Sentence": "Or how much trouble do you take to get to these different levels of precision? And the answer is, it just depends. If the goal was, hey, we just wanna make multiple copies of maybe jewelry of this little cog here. So we're gonna wanna put some type of, I don't know, gold chain through it. And we say, hey, you know, we need at least three quarters of a centimeter in order to get the rope or the chain through it. Well then, this first measurement, that's enough precision. But if I told you this gear is going to be an essential part of the space shuttle or some type of really important machinery that has really fine tolerances, I guess people aren't using the space shuttle anymore, but some finely engineered automobile or something that's going to have a lot of, needs really close tolerances, it needs to be really, really precise, well then, even this 1.089 centimeters might not be enough."}, {"video_title": "Reporting measurements   Working with units   Algebra 1   Khan Academy.mp3", "Sentence": "So we're gonna wanna put some type of, I don't know, gold chain through it. And we say, hey, you know, we need at least three quarters of a centimeter in order to get the rope or the chain through it. Well then, this first measurement, that's enough precision. But if I told you this gear is going to be an essential part of the space shuttle or some type of really important machinery that has really fine tolerances, I guess people aren't using the space shuttle anymore, but some finely engineered automobile or something that's going to have a lot of, needs really close tolerances, it needs to be really, really precise, well then, even this 1.089 centimeters might not be enough. You might have to get to something like it's 1.089203 centimeters to be able to be really, really finely crafted. We're nowhere close with our everyday tools to get anywhere close to, say, the width or the height of an atom. And you could even, in theory, measure within the atom."}, {"video_title": "Reporting measurements   Working with units   Algebra 1   Khan Academy.mp3", "Sentence": "But if I told you this gear is going to be an essential part of the space shuttle or some type of really important machinery that has really fine tolerances, I guess people aren't using the space shuttle anymore, but some finely engineered automobile or something that's going to have a lot of, needs really close tolerances, it needs to be really, really precise, well then, even this 1.089 centimeters might not be enough. You might have to get to something like it's 1.089203 centimeters to be able to be really, really finely crafted. We're nowhere close with our everyday tools to get anywhere close to, say, the width or the height of an atom. And you could even, in theory, measure within the atom. And so you just have to think about what the measurement is for. I'll give another example. This right over here is a picture of Mount Everest."}, {"video_title": "Reporting measurements   Working with units   Algebra 1   Khan Academy.mp3", "Sentence": "And you could even, in theory, measure within the atom. And so you just have to think about what the measurement is for. I'll give another example. This right over here is a picture of Mount Everest. You might know it as the tallest mountain in the world. And if you were to ask someone, how tall is Mount Everest, if you were to do a web search for it right now, you would find that it is 8,848 meters tall. Now, this is clearly rounded to the nearest meter."}, {"video_title": "Reporting measurements   Working with units   Algebra 1   Khan Academy.mp3", "Sentence": "This right over here is a picture of Mount Everest. You might know it as the tallest mountain in the world. And if you were to ask someone, how tall is Mount Everest, if you were to do a web search for it right now, you would find that it is 8,848 meters tall. Now, this is clearly rounded to the nearest meter. Because if you were to go to the top of Mount Everest, you'll see little pebbles. In fact, those pebbles might move around. And so the actual precise height of Mount Everest might change actually second by second, depending if rain is falling, snow is falling, how the wind is moving different pebbles around."}, {"video_title": "Reporting measurements   Working with units   Algebra 1   Khan Academy.mp3", "Sentence": "Now, this is clearly rounded to the nearest meter. Because if you were to go to the top of Mount Everest, you'll see little pebbles. In fact, those pebbles might move around. And so the actual precise height of Mount Everest might change actually second by second, depending if rain is falling, snow is falling, how the wind is moving different pebbles around. But for most of our daily purposes, this is sufficient. In fact, for a lot of us, we might not even need this level of precision. You might say, hey, it's roughly or it's approximately, we'd estimate that it's about 9,000 meters."}, {"video_title": "Reporting measurements   Working with units   Algebra 1   Khan Academy.mp3", "Sentence": "And so the actual precise height of Mount Everest might change actually second by second, depending if rain is falling, snow is falling, how the wind is moving different pebbles around. But for most of our daily purposes, this is sufficient. In fact, for a lot of us, we might not even need this level of precision. You might say, hey, it's roughly or it's approximately, we'd estimate that it's about 9,000 meters. But there are applications where you would need at least this level of precision or maybe something even more precise. For example, if you wanted to compare it to another mountain, say K2, which is the second tallest mountain in the world, and let's say they are close in height. And actually, if you were to do a Google search, you would see that K2 has a height of 8,611 meters, rounded to the nearest meter."}, {"video_title": "Reporting measurements   Working with units   Algebra 1   Khan Academy.mp3", "Sentence": "You might say, hey, it's roughly or it's approximately, we'd estimate that it's about 9,000 meters. But there are applications where you would need at least this level of precision or maybe something even more precise. For example, if you wanted to compare it to another mountain, say K2, which is the second tallest mountain in the world, and let's say they are close in height. And actually, if you were to do a Google search, you would see that K2 has a height of 8,611 meters, rounded to the nearest meter. You would see that that 9,000-meter approximation wouldn't be enough. If you round to the nearest kilometer, I guess, that wouldn't be enough to be able to compare Mount Everest to K2, because rounded to the nearest kilometer, they're both approximately nine kilometers. So this is approximately 9,000 meters as well."}, {"video_title": "Reporting measurements   Working with units   Algebra 1   Khan Academy.mp3", "Sentence": "And actually, if you were to do a Google search, you would see that K2 has a height of 8,611 meters, rounded to the nearest meter. You would see that that 9,000-meter approximation wouldn't be enough. If you round to the nearest kilometer, I guess, that wouldn't be enough to be able to compare Mount Everest to K2, because rounded to the nearest kilometer, they're both approximately nine kilometers. So this is approximately 9,000 meters as well. So you would need more precision. If you wanted to answer which one is taller, you'd have to get at least to the closest 100-meter. And then there's reasons why you might wanna get even more precise."}, {"video_title": "Reporting measurements   Working with units   Algebra 1   Khan Academy.mp3", "Sentence": "So this is approximately 9,000 meters as well. So you would need more precision. If you wanted to answer which one is taller, you'd have to get at least to the closest 100-meter. And then there's reasons why you might wanna get even more precise. Maybe you wanna create a slide from the top of K2 to the bottom of K2. And so you can imagine if your slide is too long by, let's say, three meters, well, it's going to be hard to get on that slide on the top, or it's going to dig into the snow at the bottom. And if your slide is too short by three meters, that's a pretty unpleasant thing to have you go on this seemingly super fun slide."}, {"video_title": "Reporting measurements   Working with units   Algebra 1   Khan Academy.mp3", "Sentence": "And then there's reasons why you might wanna get even more precise. Maybe you wanna create a slide from the top of K2 to the bottom of K2. And so you can imagine if your slide is too long by, let's say, three meters, well, it's going to be hard to get on that slide on the top, or it's going to dig into the snow at the bottom. And if your slide is too short by three meters, that's a pretty unpleasant thing to have you go on this seemingly super fun slide. You have to drop nine feet at the end, or really, what if you're off by 10 meters? Then you're gonna drop 30 feet off the end, which could really break some bones and be unpleasant. So the big takeaway is it's very hard to measure anything perfectly precisely, and you have to think about what's the application?"}, {"video_title": "Example 4 Factoring quadratics by taking a negative common factor and grouping   Khan Academy.mp3", "Sentence": "We need to factor negative 12 f squared minus 38f plus 22. So a good place to start is to see if, is there any common factor for all three of these terms? When you look at them, they're all even. And we don't like a negative number out here. So let's divide everything, or let's factor out a negative 2. So this expression right here is the same thing as negative 2 times negative 12 f squared divided by negative 2. So it's positive 6 f squared."}, {"video_title": "Example 4 Factoring quadratics by taking a negative common factor and grouping   Khan Academy.mp3", "Sentence": "And we don't like a negative number out here. So let's divide everything, or let's factor out a negative 2. So this expression right here is the same thing as negative 2 times negative 12 f squared divided by negative 2. So it's positive 6 f squared. Negative 38 divided by negative 2 is positive 19. So it'll be positive 19f. And then 22 divided by negative 2 is negative 11."}, {"video_title": "Example 4 Factoring quadratics by taking a negative common factor and grouping   Khan Academy.mp3", "Sentence": "So it's positive 6 f squared. Negative 38 divided by negative 2 is positive 19. So it'll be positive 19f. And then 22 divided by negative 2 is negative 11. So we've simplified it a bit. We have the 6 f squared plus 19f minus 11. We'll just focus on that part right now."}, {"video_title": "Example 4 Factoring quadratics by taking a negative common factor and grouping   Khan Academy.mp3", "Sentence": "And then 22 divided by negative 2 is negative 11. So we've simplified it a bit. We have the 6 f squared plus 19f minus 11. We'll just focus on that part right now. And the best way to factor this thing, since we don't have a 1 here as the coefficient on the f squared, is to factor it by grouping. So we need to look for two numbers whose products is 6 times negative 11. So two numbers."}, {"video_title": "Example 4 Factoring quadratics by taking a negative common factor and grouping   Khan Academy.mp3", "Sentence": "We'll just focus on that part right now. And the best way to factor this thing, since we don't have a 1 here as the coefficient on the f squared, is to factor it by grouping. So we need to look for two numbers whose products is 6 times negative 11. So two numbers. So a times b needs to be equal to 6 times negative 11, or negative 66. And a plus b needs to be equal to 19. So let's try a few numbers here."}, {"video_title": "Example 4 Factoring quadratics by taking a negative common factor and grouping   Khan Academy.mp3", "Sentence": "So two numbers. So a times b needs to be equal to 6 times negative 11, or negative 66. And a plus b needs to be equal to 19. So let's try a few numbers here. So let's see, 22, 22. I'm just thinking of numbers that are roughly 19 apart, because they're going to be of different signs. So 22 and 3, I think, will work."}, {"video_title": "Example 4 Factoring quadratics by taking a negative common factor and grouping   Khan Academy.mp3", "Sentence": "So let's try a few numbers here. So let's see, 22, 22. I'm just thinking of numbers that are roughly 19 apart, because they're going to be of different signs. So 22 and 3, I think, will work. Right. If we take 22 times negative 3, that is negative 66. And 22 plus negative 3 is equal to 19."}, {"video_title": "Example 4 Factoring quadratics by taking a negative common factor and grouping   Khan Academy.mp3", "Sentence": "So 22 and 3, I think, will work. Right. If we take 22 times negative 3, that is negative 66. And 22 plus negative 3 is equal to 19. And the way I kind of got pretty close to this number is, well, they're going to be of different signs. So the positive versions of them have to be about 19 apart. And that worked out."}, {"video_title": "Example 4 Factoring quadratics by taking a negative common factor and grouping   Khan Academy.mp3", "Sentence": "And 22 plus negative 3 is equal to 19. And the way I kind of got pretty close to this number is, well, they're going to be of different signs. So the positive versions of them have to be about 19 apart. And that worked out. So 22 and negative 3. So now we can rewrite this 19f right here as the sum of negative 3f and 22f. So it's negative 3f plus 22f."}, {"video_title": "Example 4 Factoring quadratics by taking a negative common factor and grouping   Khan Academy.mp3", "Sentence": "And that worked out. So 22 and negative 3. So now we can rewrite this 19f right here as the sum of negative 3f and 22f. So it's negative 3f plus 22f. That's the same thing as 19f. I just kind of broke it apart. And of course, we have the 6f squared."}, {"video_title": "Example 4 Factoring quadratics by taking a negative common factor and grouping   Khan Academy.mp3", "Sentence": "So it's negative 3f plus 22f. That's the same thing as 19f. I just kind of broke it apart. And of course, we have the 6f squared. And we have the minus 11 here. Now, you're probably saying, hey, Sal, why did you put the 22 here and the negative 3 there? Why didn't you do it the other way around?"}, {"video_title": "Example 4 Factoring quadratics by taking a negative common factor and grouping   Khan Academy.mp3", "Sentence": "And of course, we have the 6f squared. And we have the minus 11 here. Now, you're probably saying, hey, Sal, why did you put the 22 here and the negative 3 there? Why didn't you do it the other way around? Why didn't you put the 22 and then the negative 3 there? And my main motivation for doing it, I like to put the negative 3 on the same side with the 6, because they have the common factor of the 3. I like to put the 22 with the negative 11."}, {"video_title": "Example 4 Factoring quadratics by taking a negative common factor and grouping   Khan Academy.mp3", "Sentence": "Why didn't you do it the other way around? Why didn't you put the 22 and then the negative 3 there? And my main motivation for doing it, I like to put the negative 3 on the same side with the 6, because they have the common factor of the 3. I like to put the 22 with the negative 11. They have the same factor of 11. So that's why I decided to do it that way. So now let's do the grouping."}, {"video_title": "Example 4 Factoring quadratics by taking a negative common factor and grouping   Khan Academy.mp3", "Sentence": "I like to put the 22 with the negative 11. They have the same factor of 11. So that's why I decided to do it that way. So now let's do the grouping. And of course, you can't forget this negative 2 that we have sitting out here the whole time. So let me put that negative 2 out there. But that'll just kind of hang out for a while."}, {"video_title": "Example 4 Factoring quadratics by taking a negative common factor and grouping   Khan Academy.mp3", "Sentence": "So now let's do the grouping. And of course, you can't forget this negative 2 that we have sitting out here the whole time. So let me put that negative 2 out there. But that'll just kind of hang out for a while. But let's do some grouping. So let's group these first two. And then we're going to group this."}, {"video_title": "Example 4 Factoring quadratics by taking a negative common factor and grouping   Khan Academy.mp3", "Sentence": "But that'll just kind of hang out for a while. But let's do some grouping. So let's group these first two. And then we're going to group this. Let me get a nice color here. And then we're going to group this second 2. That's almost an identical color."}, {"video_title": "Example 4 Factoring quadratics by taking a negative common factor and grouping   Khan Academy.mp3", "Sentence": "And then we're going to group this. Let me get a nice color here. And then we're going to group this second 2. That's almost an identical color. Let me do it in this purple color. And then we can group that second 2 right there. So these first two, we could factor out a negative 3f."}, {"video_title": "Example 4 Factoring quadratics by taking a negative common factor and grouping   Khan Academy.mp3", "Sentence": "That's almost an identical color. Let me do it in this purple color. And then we can group that second 2 right there. So these first two, we could factor out a negative 3f. So it's negative 3f times 6f squared divided by negative 3f is negative 2f. And the negative 3f divided by negative 3f is just positive f. Actually, a better way to start, instead of factoring out a negative 3f, let's just factor out 3f. So we don't have a negative out here."}, {"video_title": "Example 4 Factoring quadratics by taking a negative common factor and grouping   Khan Academy.mp3", "Sentence": "So these first two, we could factor out a negative 3f. So it's negative 3f times 6f squared divided by negative 3f is negative 2f. And the negative 3f divided by negative 3f is just positive f. Actually, a better way to start, instead of factoring out a negative 3f, let's just factor out 3f. So we don't have a negative out here. We could do it either way. But if we just factor out a 3f, 6f squared divided by 3f is 2f. And then negative 3f divided by 3f is negative 1."}, {"video_title": "Example 4 Factoring quadratics by taking a negative common factor and grouping   Khan Academy.mp3", "Sentence": "So we don't have a negative out here. We could do it either way. But if we just factor out a 3f, 6f squared divided by 3f is 2f. And then negative 3f divided by 3f is negative 1. So that's what that factors into. And then that second part, in that dark purple color, we can factor out an 11. So we factor out an 11."}, {"video_title": "Example 4 Factoring quadratics by taking a negative common factor and grouping   Khan Academy.mp3", "Sentence": "And then negative 3f divided by 3f is negative 1. So that's what that factors into. And then that second part, in that dark purple color, we can factor out an 11. So we factor out an 11. And if we factor that out, 22f divided by 11 is 2f. And negative 11 divided by 11 is negative 1. And of course, once again, you have that negative 2 hanging out there."}, {"video_title": "Example 4 Factoring quadratics by taking a negative common factor and grouping   Khan Academy.mp3", "Sentence": "So we factor out an 11. And if we factor that out, 22f divided by 11 is 2f. And negative 11 divided by 11 is negative 1. And of course, once again, you have that negative 2 hanging out there. You have that negative 2. Now, inside the parentheses, we have two terms, both of which have 2f minus 1 as a factor. So we can factor that out."}, {"video_title": "Example 4 Factoring quadratics by taking a negative common factor and grouping   Khan Academy.mp3", "Sentence": "And of course, once again, you have that negative 2 hanging out there. You have that negative 2. Now, inside the parentheses, we have two terms, both of which have 2f minus 1 as a factor. So we can factor that out. This whole thing is just an exercise in doing the reverse distributive property, if you will. So let's factor that out. So you have 2f minus 1 times this 3f, times that 3f, and then times that plus 11."}, {"video_title": "Example 4 Factoring quadratics by taking a negative common factor and grouping   Khan Academy.mp3", "Sentence": "So we can factor that out. This whole thing is just an exercise in doing the reverse distributive property, if you will. So let's factor that out. So you have 2f minus 1 times this 3f, times that 3f, and then times that plus 11. Let me do that in the same shade of purple right over there. And you can distribute, if you like. 2f minus 1 times 3f will give you this term."}, {"video_title": "Example 4 Factoring quadratics by taking a negative common factor and grouping   Khan Academy.mp3", "Sentence": "So you have 2f minus 1 times this 3f, times that 3f, and then times that plus 11. Let me do that in the same shade of purple right over there. And you can distribute, if you like. 2f minus 1 times 3f will give you this term. 2f minus 1 times 11 will give you that term. And we can't forget that we still have that negative 2 hanging out outside. I don't want to change the colors on it."}, {"video_title": "Example 4 Factoring quadratics by taking a negative common factor and grouping   Khan Academy.mp3", "Sentence": "2f minus 1 times 3f will give you this term. 2f minus 1 times 11 will give you that term. And we can't forget that we still have that negative 2 hanging out outside. I don't want to change the colors on it. We have the negative 2 hanging out, that same negative 2 over there. And we're done factoring it. Negative 12f squared minus 38."}, {"video_title": "Independent and dependent systems   Algebra II   Khan Academy.mp3", "Sentence": "And they give us two equations right here. Before I tackle this specific problem, let's just do a little bit of a review of what dependent or independent means. And actually I'll compare that to consistent and inconsistent. So just to start off with, if we're dealing with systems of linear equations in two dimensions, there's only three possibilities that the lines or the equations can have relative to each other. So let me draw the three possibilities. So let me draw three coordinate axes. So that's my first x-axis and y-axis."}, {"video_title": "Independent and dependent systems   Algebra II   Khan Academy.mp3", "Sentence": "So just to start off with, if we're dealing with systems of linear equations in two dimensions, there's only three possibilities that the lines or the equations can have relative to each other. So let me draw the three possibilities. So let me draw three coordinate axes. So that's my first x-axis and y-axis. So x and y. Let me draw another one. That is x and that is y."}, {"video_title": "Independent and dependent systems   Algebra II   Khan Academy.mp3", "Sentence": "So that's my first x-axis and y-axis. So x and y. Let me draw another one. That is x and that is y. Let me draw one more because there's only three possibilities in two dimensions. x and y, if we're dealing with linear equations. x and y."}, {"video_title": "Independent and dependent systems   Algebra II   Khan Academy.mp3", "Sentence": "That is x and that is y. Let me draw one more because there's only three possibilities in two dimensions. x and y, if we're dealing with linear equations. x and y. So you can have the situation where the lines just intersect in one point. So you could have one line like that and maybe the other line does something like that. And they intersect at one point."}, {"video_title": "Independent and dependent systems   Algebra II   Khan Academy.mp3", "Sentence": "x and y. So you can have the situation where the lines just intersect in one point. So you could have one line like that and maybe the other line does something like that. And they intersect at one point. You could have the situation where the two lines are parallel. So you could have a situation, actually let me draw it over here, where you have one line that goes like that and the other line has the same slope but it's shifted. It has a different y-intercept so maybe it looks like this."}, {"video_title": "Independent and dependent systems   Algebra II   Khan Academy.mp3", "Sentence": "And they intersect at one point. You could have the situation where the two lines are parallel. So you could have a situation, actually let me draw it over here, where you have one line that goes like that and the other line has the same slope but it's shifted. It has a different y-intercept so maybe it looks like this. And you have no points of intersection. And then you could have the situation where they're actually the same line. So that both lines have the same slope and the same y-intercept."}, {"video_title": "Independent and dependent systems   Algebra II   Khan Academy.mp3", "Sentence": "It has a different y-intercept so maybe it looks like this. And you have no points of intersection. And then you could have the situation where they're actually the same line. So that both lines have the same slope and the same y-intercept. So really they are the same line. They intersect on an infinite number of points. Every point on either of those lines is also a point on the other line."}, {"video_title": "Independent and dependent systems   Algebra II   Khan Academy.mp3", "Sentence": "So that both lines have the same slope and the same y-intercept. So really they are the same line. They intersect on an infinite number of points. Every point on either of those lines is also a point on the other line. So just to give you a little bit of the terminology here, and we learned this in the last video, this type of system where they don't intersect, where you have no solutions, this is an inconsistent system. And by definition, or I guess just taking the opposite of inconsistent, both of these would be considered consistent. But then within consistent, there's obviously a difference."}, {"video_title": "Independent and dependent systems   Algebra II   Khan Academy.mp3", "Sentence": "Every point on either of those lines is also a point on the other line. So just to give you a little bit of the terminology here, and we learned this in the last video, this type of system where they don't intersect, where you have no solutions, this is an inconsistent system. And by definition, or I guess just taking the opposite of inconsistent, both of these would be considered consistent. But then within consistent, there's obviously a difference. Here we only have one solution. These are two different lines that intersect in one place. They're essentially the same exact line."}, {"video_title": "Independent and dependent systems   Algebra II   Khan Academy.mp3", "Sentence": "But then within consistent, there's obviously a difference. Here we only have one solution. These are two different lines that intersect in one place. They're essentially the same exact line. And so we differentiate between these two scenarios by calling this one over here independent. And this one over here dependent. So independent, both lines are doing their own thing."}, {"video_title": "Independent and dependent systems   Algebra II   Khan Academy.mp3", "Sentence": "They're essentially the same exact line. And so we differentiate between these two scenarios by calling this one over here independent. And this one over here dependent. So independent, both lines are doing their own thing. They're not dependent on each other. They're not the same line. They will intersect at one place."}, {"video_title": "Independent and dependent systems   Algebra II   Khan Academy.mp3", "Sentence": "So independent, both lines are doing their own thing. They're not dependent on each other. They're not the same line. They will intersect at one place. Dependent, they're the exact same line. Any point that satisfies one line will satisfy the other. Any point that satisfies one equation will satisfy the other."}, {"video_title": "Independent and dependent systems   Algebra II   Khan Academy.mp3", "Sentence": "They will intersect at one place. Dependent, they're the exact same line. Any point that satisfies one line will satisfy the other. Any point that satisfies one equation will satisfy the other. So with that said, let's see if this system of linear equations right here is dependent or independent. So they're kind of having us assume that it's going to be consistent, that we're either going to intersect in one place or we're going to intersect in an infinite number of places. And the easiest way to do this, we already have this second equation here."}, {"video_title": "Independent and dependent systems   Algebra II   Khan Academy.mp3", "Sentence": "Any point that satisfies one equation will satisfy the other. So with that said, let's see if this system of linear equations right here is dependent or independent. So they're kind of having us assume that it's going to be consistent, that we're either going to intersect in one place or we're going to intersect in an infinite number of places. And the easiest way to do this, we already have this second equation here. It's already in slope-intercept form. We know the slope is negative 2. The y-intercept is 8."}, {"video_title": "Independent and dependent systems   Algebra II   Khan Academy.mp3", "Sentence": "And the easiest way to do this, we already have this second equation here. It's already in slope-intercept form. We know the slope is negative 2. The y-intercept is 8. So let's put this first equation up here in slope-intercept form and see if it has a different slope or a different intercept or maybe it's the same line. So we have 4x plus 2y is equal to 16. We can subtract 4x from both sides."}, {"video_title": "Independent and dependent systems   Algebra II   Khan Academy.mp3", "Sentence": "The y-intercept is 8. So let's put this first equation up here in slope-intercept form and see if it has a different slope or a different intercept or maybe it's the same line. So we have 4x plus 2y is equal to 16. We can subtract 4x from both sides. What we want to do is isolate the y on the left-hand side. So let's subtract 4x from both sides. The left-hand side, we are just left with a 2y."}, {"video_title": "Independent and dependent systems   Algebra II   Khan Academy.mp3", "Sentence": "We can subtract 4x from both sides. What we want to do is isolate the y on the left-hand side. So let's subtract 4x from both sides. The left-hand side, we are just left with a 2y. The right-hand side, we have a negative 4x plus 16. I just wrote the negative 4 in front of the 16 just so that we have it in the traditional slope-intercept form. And now we can divide both sides of this equation by 2 so that we can isolate the y on the left-hand side."}, {"video_title": "Independent and dependent systems   Algebra II   Khan Academy.mp3", "Sentence": "The left-hand side, we are just left with a 2y. The right-hand side, we have a negative 4x plus 16. I just wrote the negative 4 in front of the 16 just so that we have it in the traditional slope-intercept form. And now we can divide both sides of this equation by 2 so that we can isolate the y on the left-hand side. Divide both sides by 2. We are left with y is equal to negative 4 divided by 2 is negative 2x plus 16 over 2, plus 8. So all I did is algebraically manipulated this top equation up here."}, {"video_title": "Independent and dependent systems   Algebra II   Khan Academy.mp3", "Sentence": "And now we can divide both sides of this equation by 2 so that we can isolate the y on the left-hand side. Divide both sides by 2. We are left with y is equal to negative 4 divided by 2 is negative 2x plus 16 over 2, plus 8. So all I did is algebraically manipulated this top equation up here. And when I did that, when I solved essentially for y, I got this right over here, which is the exact same thing as the second equation. We have the exact same slope, negative 2, negative 2, and we have the exact same y-intercept, 8 and 8. If I were to graph these equations, that's my x-axis and that is my y-axis, both of them have a y-intercept at 8 and then have a slope of negative 2."}, {"video_title": "Independent and dependent systems   Algebra II   Khan Academy.mp3", "Sentence": "So all I did is algebraically manipulated this top equation up here. And when I did that, when I solved essentially for y, I got this right over here, which is the exact same thing as the second equation. We have the exact same slope, negative 2, negative 2, and we have the exact same y-intercept, 8 and 8. If I were to graph these equations, that's my x-axis and that is my y-axis, both of them have a y-intercept at 8 and then have a slope of negative 2. So they look something... I'm just drawing an approximation of it, but they would look something like that. So maybe this is the graph of this equation right here, this first equation."}, {"video_title": "Independent and dependent systems   Algebra II   Khan Academy.mp3", "Sentence": "If I were to graph these equations, that's my x-axis and that is my y-axis, both of them have a y-intercept at 8 and then have a slope of negative 2. So they look something... I'm just drawing an approximation of it, but they would look something like that. So maybe this is the graph of this equation right here, this first equation. And then the second equation will be the exact same graph. It has the exact same y-intercept and the exact same slope. So clearly these two lines are dependent."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "On both sides of the scale, we have our mystery mass. And now I'm calling the mystery mass having a mass of y, just to show you that it doesn't always have to be x. It can be any symbol, as long as you can keep track of that symbol. But all of these have the same mass. That's why I wrote y on all of them. And we also have the little one kilogram boxes on both sides of this scale. So the first thing I wanna do, we're gonna go step by step and try to figure out what this mystery mass is."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "But all of these have the same mass. That's why I wrote y on all of them. And we also have the little one kilogram boxes on both sides of this scale. So the first thing I wanna do, we're gonna go step by step and try to figure out what this mystery mass is. But the first thing I wanna do is have you think about whether you can represent this algebraically, whether with a little bit of mathematic symbolry, you can represent what's going on in this scale. Over here, I have three y's and three of these boxes, and their total mass is equal to this one y. And I think I have about, let's see, I have seven boxes right over here."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "So the first thing I wanna do, we're gonna go step by step and try to figure out what this mystery mass is. But the first thing I wanna do is have you think about whether you can represent this algebraically, whether with a little bit of mathematic symbolry, you can represent what's going on in this scale. Over here, I have three y's and three of these boxes, and their total mass is equal to this one y. And I think I have about, let's see, I have seven boxes right over here. So I'll give you a few seconds to do that. So let's think about the total mass over here. We have three boxes of mass y."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "And I think I have about, let's see, I have seven boxes right over here. So I'll give you a few seconds to do that. So let's think about the total mass over here. We have three boxes of mass y. So they're going to have a mass of three y. And then you have three boxes with a mass of one kilogram. So they're going to have a mass of three kilograms."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "We have three boxes of mass y. So they're going to have a mass of three y. And then you have three boxes with a mass of one kilogram. So they're going to have a mass of three kilograms. Now over here, I have one box with a mass of y kilograms. So that's going to be my one y right over there. I could have written one y, but I don't need to."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "So they're going to have a mass of three kilograms. Now over here, I have one box with a mass of y kilograms. So that's going to be my one y right over there. I could have written one y, but I don't need to. A y is the same thing as one y. So I have the y kilograms right there. And then I have seven of these, right?"}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "I could have written one y, but I don't need to. A y is the same thing as one y. So I have the y kilograms right there. And then I have seven of these, right? One, two, three, four, five, six, seven. Yep, seven of these. So I have y plus seven kilograms on the right-hand side."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "And then I have seven of these, right? One, two, three, four, five, six, seven. Yep, seven of these. So I have y plus seven kilograms on the right-hand side. And once again, it's balanced. The scale is balanced. This total mass is equal to this total mass."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "So I have y plus seven kilograms on the right-hand side. And once again, it's balanced. The scale is balanced. This total mass is equal to this total mass. So we can write an equal sign right over there. So that's a good starting point. We were able to represent this situation, this real-life situation, you know, back in the day when people actually had to figure out the mass of things, if you were to go to the jewelry store or whatever."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "This total mass is equal to this total mass. So we can write an equal sign right over there. So that's a good starting point. We were able to represent this situation, this real-life situation, you know, back in the day when people actually had to figure out the mass of things, if you were to go to the jewelry store or whatever. They actually did have problems like this. But we were able to represent it mathematically. Now, the next thing to do is, what are some reasonable next steps?"}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "We were able to represent this situation, this real-life situation, you know, back in the day when people actually had to figure out the mass of things, if you were to go to the jewelry store or whatever. They actually did have problems like this. But we were able to represent it mathematically. Now, the next thing to do is, what are some reasonable next steps? How can we start to simplify this a little bit? And once again, I'll give you a few seconds to think about that. Well, the neat thing about algebra is there's actually multiple paths that you could go down."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "Now, the next thing to do is, what are some reasonable next steps? How can we start to simplify this a little bit? And once again, I'll give you a few seconds to think about that. Well, the neat thing about algebra is there's actually multiple paths that you could go down. You might say, well, why don't we remove three of these yellow blocks from both sides? That would be completely legitimate. You might say, well, why don't we remove one of these y's from both sides?"}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "Well, the neat thing about algebra is there's actually multiple paths that you could go down. You might say, well, why don't we remove three of these yellow blocks from both sides? That would be completely legitimate. You might say, well, why don't we remove one of these y's from both sides? That also would be legitimate. And we can do it in either order. So let's just pick one of them."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "You might say, well, why don't we remove one of these y's from both sides? That also would be legitimate. And we can do it in either order. So let's just pick one of them. Let's say that we first want to remove, let's say that we first want to remove a y from either side, just so that we feel a little bit more comfortable with all of our y's sitting on only one side. And so the best way, if we want all of our y's to sit on one side, we can remove a y from each side. Remember, if we removed a y from only one side, that would unbalance the scale."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "So let's just pick one of them. Let's say that we first want to remove, let's say that we first want to remove a y from either side, just so that we feel a little bit more comfortable with all of our y's sitting on only one side. And so the best way, if we want all of our y's to sit on one side, we can remove a y from each side. Remember, if we removed a y from only one side, that would unbalance the scale. The scale was already balanced. Whatever I have to do to one side, I have to do to the other. So I'm going to remove a y. I'm going to remove y mass from both sides."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "Remember, if we removed a y from only one side, that would unbalance the scale. The scale was already balanced. Whatever I have to do to one side, I have to do to the other. So I'm going to remove a y. I'm going to remove y mass from both sides. Now, what would that look like algebraically? Well, I removed a y from both sides. So I subtracted y from the left-hand side, and I subtracted y from the right-hand side."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "So I'm going to remove a y. I'm going to remove y mass from both sides. Now, what would that look like algebraically? Well, I removed a y from both sides. So I subtracted y from the left-hand side, and I subtracted y from the right-hand side. That's exactly what I did. It had a mass of y. I don't know what that is, but I did take it away. I lifted that little block."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "So I subtracted y from the left-hand side, and I subtracted y from the right-hand side. That's exactly what I did. It had a mass of y. I don't know what that is, but I did take it away. I lifted that little block. And so on the left-hand side, what am I left with? And you could think of it mathematically, but you could even look up here and see what you're left with. If I have three of something, and I take away one of them, I'm left with two of that something."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "I lifted that little block. And so on the left-hand side, what am I left with? And you could think of it mathematically, but you could even look up here and see what you're left with. If I have three of something, and I take away one of them, I'm left with two of that something. So I'm left with two y right over here. And you see it. I had three."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "If I have three of something, and I take away one of them, I'm left with two of that something. So I'm left with two y right over here. And you see it. I had three. I got rid of one, so I'm left with two. And I still have those three yellow blocks. On the right-hand side, I had a y. I took away the y."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "I had three. I got rid of one, so I'm left with two. And I still have those three yellow blocks. On the right-hand side, I had a y. I took away the y. And so now I have no y's left. And we see it visually right over here. But I still have seven of the yellow blocks."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "On the right-hand side, I had a y. I took away the y. And so now I have no y's left. And we see it visually right over here. But I still have seven of the yellow blocks. So I still have seven of the yellow blocks. And since I took away the exact same mass from both sides of the scale, the scale is still going to be balanced. It was balanced before."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "But I still have seven of the yellow blocks. So I still have seven of the yellow blocks. And since I took away the exact same mass from both sides of the scale, the scale is still going to be balanced. It was balanced before. I took away the same thing from both sides. And so the scale is still balanced. So this is going to be equal to that."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "It was balanced before. I took away the same thing from both sides. And so the scale is still balanced. So this is going to be equal to that. Now this is starting to look a little bit similar to what we saw in the last video. But I will ask you, what can we do from this point to simplify it further? Or so even better, think of it so we can isolate these y's on the left-hand side."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "So this is going to be equal to that. Now this is starting to look a little bit similar to what we saw in the last video. But I will ask you, what can we do from this point to simplify it further? Or so even better, think of it so we can isolate these y's on the left-hand side. And I'll give you a few seconds to think about that. Well, if we want to isolate these y's on the left-hand side, these two y's, the best way is to get rid of this three, to get rid of these three blocks. So why don't we do that?"}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "Or so even better, think of it so we can isolate these y's on the left-hand side. And I'll give you a few seconds to think about that. Well, if we want to isolate these y's on the left-hand side, these two y's, the best way is to get rid of this three, to get rid of these three blocks. So why don't we do that? Let's take three blocks from this side. But we can't just take it from that side if we want to keep it balanced. We have to do it to this side, too."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "So why don't we do that? Let's take three blocks from this side. But we can't just take it from that side if we want to keep it balanced. We have to do it to this side, too. We've got to take away three blocks. So we're subtracting three from that side and subtracting three from the right side. So on the left-hand side, we're going to be left with just these two blocks of mass y."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "We have to do it to this side, too. We've got to take away three blocks. So we're subtracting three from that side and subtracting three from the right side. So on the left-hand side, we're going to be left with just these two blocks of mass y. So our total mass is now going to be 2y. These 3 minus 3 is 0. And you see that here."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "So on the left-hand side, we're going to be left with just these two blocks of mass y. So our total mass is now going to be 2y. These 3 minus 3 is 0. And you see that here. We're just left with two y's right over here. And on the right-hand side, we got rid of three of the blocks. So we only have four of them left."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "And you see that here. We're just left with two y's right over here. And on the right-hand side, we got rid of three of the blocks. So we only have four of them left. So you have two of these y masses is equal to 4 kilograms. Because we did the same thing to both sides, the scale is still balanced. And now, how do we solve this?"}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "So we only have four of them left. So you have two of these y masses is equal to 4 kilograms. Because we did the same thing to both sides, the scale is still balanced. And now, how do we solve this? And you might be able to solve this in your head. I have 2 times something is equal to 4. You could kind of think about what that is."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "And now, how do we solve this? And you might be able to solve this in your head. I have 2 times something is equal to 4. You could kind of think about what that is. But if we want to stay true to what we've been doing before, let's think about it. I have 2 of something is equal to something else. What if I multiplied both sides by 2?"}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "You could kind of think about what that is. But if we want to stay true to what we've been doing before, let's think about it. I have 2 of something is equal to something else. What if I multiplied both sides by 2? Sorry, what if I multiplied both sides by 1 half? Or in other ways, dividing both sides by 2. If I multiply this side by 1 half, if I essentially take away half of the mass, or I only leave half of the mass, then I'm only going to have one block here."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "What if I multiplied both sides by 2? Sorry, what if I multiplied both sides by 1 half? Or in other ways, dividing both sides by 2. If I multiply this side by 1 half, if I essentially take away half of the mass, or I only leave half of the mass, then I'm only going to have one block here. And if I take away half of the mass over here, I'm going to have to take away two of these blocks right over there. And what I just did, you could say I multiplied both sides by 1 half. Or just for the sake of a little change, you could say I divided both sides by 2."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "If I multiply this side by 1 half, if I essentially take away half of the mass, or I only leave half of the mass, then I'm only going to have one block here. And if I take away half of the mass over here, I'm going to have to take away two of these blocks right over there. And what I just did, you could say I multiplied both sides by 1 half. Or just for the sake of a little change, you could say I divided both sides by 2. And on the left-hand side, I'm left with a mass of y. And on the right-hand side, I'm left with a mass of 4 divided by 2 is 2. And once again, I can still write this equal sign because the scale is balanced."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "Or just for the sake of a little change, you could say I divided both sides by 2. And on the left-hand side, I'm left with a mass of y. And on the right-hand side, I'm left with a mass of 4 divided by 2 is 2. And once again, I can still write this equal sign because the scale is balanced. I did the exact same thing to both sides. I left half of what was on the left-hand side and half of what was on the right-hand side. It was balanced before, half of each side, so it's going to be balanced again."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "And once again, I can still write this equal sign because the scale is balanced. I did the exact same thing to both sides. I left half of what was on the left-hand side and half of what was on the right-hand side. It was balanced before, half of each side, so it's going to be balanced again. But there we've done it. We've solved something that's actually not so easy to solve, or might not look so easy at first. We figured out that our mystery mass, y, is 2 kilograms."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "It was balanced before, half of each side, so it's going to be balanced again. But there we've done it. We've solved something that's actually not so easy to solve, or might not look so easy at first. We figured out that our mystery mass, y, is 2 kilograms. And you can verify this. This is the really fun thing about algebra, is that once you get to this point, you can go back and think about whether the original problem we saw made sense. Let's do that."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "We figured out that our mystery mass, y, is 2 kilograms. And you can verify this. This is the really fun thing about algebra, is that once you get to this point, you can go back and think about whether the original problem we saw made sense. Let's do that. Let's think about whether the original problem made sense. And to do that, I want you to calculate, now that we know what the mass, y, is, is 2 kilograms, what was the total mass on each side? Well, let's calculate it."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "Let's do that. Let's think about whether the original problem made sense. And to do that, I want you to calculate, now that we know what the mass, y, is, is 2 kilograms, what was the total mass on each side? Well, let's calculate it. You have 2, I'll write it right over here. This is 2 kilograms. I'll do it in purple color."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "Well, let's calculate it. You have 2, I'll write it right over here. This is 2 kilograms. I'll do it in purple color. So this is a 2. So we had 6 kilograms plus these 3. You had 9 kilograms on the left-hand side."}, {"video_title": "Why we do the same thing to both sides Multi-step equations   Algebra I   Khan Academy.mp3", "Sentence": "I'll do it in purple color. So this is a 2. So we had 6 kilograms plus these 3. You had 9 kilograms on the left-hand side. And on the right-hand side, I had these 7 plus 2 here. 7 plus 2 is 9 kilograms. That's why it was balanced, our mystery mass."}, {"video_title": "Fraction to decimal with rounding   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "This is also 16 divided by 21, so we can literally just divide 21 into 16. And because 21 is larger than 16, we're going to get something less than one. So let's just literally divide 21 into 16. And we're gonna have something less than one, so let's add some decimal places here, and we're gonna round to the nearest thousandths in case our digits keep going on and on and on. And let's start dividing. 21 goes into one zero times. 21 goes into 16 zero times."}, {"video_title": "Fraction to decimal with rounding   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "And we're gonna have something less than one, so let's add some decimal places here, and we're gonna round to the nearest thousandths in case our digits keep going on and on and on. And let's start dividing. 21 goes into one zero times. 21 goes into 16 zero times. 21 goes into 160. Well, 20 would go into 160 eight times, so let's try seven. 21, let's see if seven is the right thing."}, {"video_title": "Fraction to decimal with rounding   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "21 goes into 16 zero times. 21 goes into 160. Well, 20 would go into 160 eight times, so let's try seven. 21, let's see if seven is the right thing. So seven times one is seven. Seven times two is 14. And then when we subtract it, we should get a remainder less than 21."}, {"video_title": "Fraction to decimal with rounding   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "21, let's see if seven is the right thing. So seven times one is seven. Seven times two is 14. And then when we subtract it, we should get a remainder less than 21. If we pick the largest number here that goes into the largest number here, where if I multiply it by 21, I get close to 160 without going over. And so if we subtract, we do get 13. We do get 13, so that worked."}, {"video_title": "Fraction to decimal with rounding   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "And then when we subtract it, we should get a remainder less than 21. If we pick the largest number here that goes into the largest number here, where if I multiply it by 21, I get close to 160 without going over. And so if we subtract, we do get 13. We do get 13, so that worked. 13 is less than 21. And you could just subtract it. I did it in my head right there, but you could regroup."}, {"video_title": "Fraction to decimal with rounding   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "We do get 13, so that worked. 13 is less than 21. And you could just subtract it. I did it in my head right there, but you could regroup. You could say this is a 10, and then this would be a five. 10 minus seven is three. Five minus four is one."}, {"video_title": "Fraction to decimal with rounding   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "I did it in my head right there, but you could regroup. You could say this is a 10, and then this would be a five. 10 minus seven is three. Five minus four is one. One minus one is zero. Now let's bring down a zero. 21 goes into 130."}, {"video_title": "Fraction to decimal with rounding   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "Five minus four is one. One minus one is zero. Now let's bring down a zero. 21 goes into 130. So let's see, would six work? It looks like six would work. Six times 21 is 126, so that looks like it works."}, {"video_title": "Fraction to decimal with rounding   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "21 goes into 130. So let's see, would six work? It looks like six would work. Six times 21 is 126, so that looks like it works. So let's put a six there. Six times one is six. Six times two is 120."}, {"video_title": "Fraction to decimal with rounding   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "Six times 21 is 126, so that looks like it works. So let's put a six there. Six times one is six. Six times two is 120. There's a little bit of an art to this. All right, now let's subtract. And once again, we can regroup."}, {"video_title": "Fraction to decimal with rounding   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "Six times two is 120. There's a little bit of an art to this. All right, now let's subtract. And once again, we can regroup. This would be a 10. We've taken 10 from essentially this 30, so now this becomes a two. 10 minus six is four."}, {"video_title": "Fraction to decimal with rounding   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "And once again, we can regroup. This would be a 10. We've taken 10 from essentially this 30, so now this becomes a two. 10 minus six is four. Two minus two is zero. One minus one is zero. Now let's bring down another zero."}, {"video_title": "Fraction to decimal with rounding   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "10 minus six is four. Two minus two is zero. One minus one is zero. Now let's bring down another zero. 21 goes into 40, well, almost two times, but not quite. So only one time. One times 21 is 21."}, {"video_title": "Fraction to decimal with rounding   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "Now let's bring down another zero. 21 goes into 40, well, almost two times, but not quite. So only one time. One times 21 is 21. And now let's subtract. This is a 10. This becomes a three."}, {"video_title": "Fraction to decimal with rounding   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "One times 21 is 21. And now let's subtract. This is a 10. This becomes a three. 10 minus one is nine. Three minus two is one. And we're gonna have to get this digit, because we want to round to the nearest thousand."}, {"video_title": "Fraction to decimal with rounding   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "This becomes a three. 10 minus one is nine. Three minus two is one. And we're gonna have to get this digit, because we want to round to the nearest thousand. So if this is five or over, we're gonna round up. If this is less than five, we're gonna round down. So let's bring another zero."}, {"video_title": "Fraction to decimal with rounding   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "And we're gonna have to get this digit, because we want to round to the nearest thousand. So if this is five or over, we're gonna round up. If this is less than five, we're gonna round down. So let's bring another zero. Let's bring another zero down here. And 21 goes into 190. Let's see, I think nine will work."}, {"video_title": "Fraction to decimal with rounding   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "So let's bring another zero. Let's bring another zero down here. And 21 goes into 190. Let's see, I think nine will work. Let's try nine. Nine times one is nine. Nine times two is 18."}, {"video_title": "Fraction to decimal with rounding   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "Let's see, I think nine will work. Let's try nine. Nine times one is nine. Nine times two is 18. When you subtract, 190 minus 189 is one. And we could keep going on and on and on, but we already have enough digits to round to the nearest thousandth. This digit right over here is greater than."}, {"video_title": "Fraction to decimal with rounding   Decimals   Pre-Algebra   Khan Academy.mp3", "Sentence": "Nine times two is 18. When you subtract, 190 minus 189 is one. And we could keep going on and on and on, but we already have enough digits to round to the nearest thousandth. This digit right over here is greater than. This is greater than or equal to five. So we will round up in the thousandths place. So if we round to the nearest thousandths, we can say that this is 0.76, and then this is where we're gonna round up, 762."}, {"video_title": "Multiplying & dividing powers (integer exponents)   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "And I encourage you to pause the video and think about it on your own. Well, there's a couple of ways to do this. One, you say, oh look, I'm multiplying two things that have the same base, so this is going to be that base, four, and then I add the exponents. Four to the negative three plus five power, which is equal to four to the second power. And that's just a straightforward exponent property, but you can also think about why does that actually make sense? Four to the negative three power, that is one over four to the third power. Or you could view that as one over four times four times four and then four to the fifth, that's five fours being multiplied together, so it's times four times four times four times four times four."}, {"video_title": "Multiplying & dividing powers (integer exponents)   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "Four to the negative three plus five power, which is equal to four to the second power. And that's just a straightforward exponent property, but you can also think about why does that actually make sense? Four to the negative three power, that is one over four to the third power. Or you could view that as one over four times four times four and then four to the fifth, that's five fours being multiplied together, so it's times four times four times four times four times four. And so notice, when you multiply this out, you're gonna have five fours in the numerator and three fours in the denominator. And so three of these in the denominator are gonna cancel out with three of these in the numerator. And so you're gonna be left with five minus three, or negative three plus five fours."}, {"video_title": "Multiplying & dividing powers (integer exponents)   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "Or you could view that as one over four times four times four and then four to the fifth, that's five fours being multiplied together, so it's times four times four times four times four times four. And so notice, when you multiply this out, you're gonna have five fours in the numerator and three fours in the denominator. And so three of these in the denominator are gonna cancel out with three of these in the numerator. And so you're gonna be left with five minus three, or negative three plus five fours. So this four times four is the same thing as four squared. Now let's do one with variables. So let's say that you have a to the negative four power times a to the, let's say a squared."}, {"video_title": "Multiplying & dividing powers (integer exponents)   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "And so you're gonna be left with five minus three, or negative three plus five fours. So this four times four is the same thing as four squared. Now let's do one with variables. So let's say that you have a to the negative four power times a to the, let's say a squared. What is that going to be? Well, once again, you have the same base. In this case, it's a."}, {"video_title": "Multiplying & dividing powers (integer exponents)   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "So let's say that you have a to the negative four power times a to the, let's say a squared. What is that going to be? Well, once again, you have the same base. In this case, it's a. And so, and since I'm multiplying them, you can just add the exponents. So it's gonna be a to the negative four plus two power, which is equal to a to the negative two power. And once again, it should make sense."}, {"video_title": "Multiplying & dividing powers (integer exponents)   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "In this case, it's a. And so, and since I'm multiplying them, you can just add the exponents. So it's gonna be a to the negative four plus two power, which is equal to a to the negative two power. And once again, it should make sense. This right over here, that is one over a times a times a times a. And then this is times a times a. So that cancels with that, that cancels with that, and you're still left with one over a times a, which is the same thing as a to the negative two power."}, {"video_title": "Multiplying & dividing powers (integer exponents)   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "And once again, it should make sense. This right over here, that is one over a times a times a times a. And then this is times a times a. So that cancels with that, that cancels with that, and you're still left with one over a times a, which is the same thing as a to the negative two power. Now let's do it with some quotients. So what if I were to ask you, what is 12 to the negative seven divided by 12 to the negative five power? Well, when you're dividing, you subtract exponents if you have the same base."}, {"video_title": "Multiplying & dividing powers (integer exponents)   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "So that cancels with that, that cancels with that, and you're still left with one over a times a, which is the same thing as a to the negative two power. Now let's do it with some quotients. So what if I were to ask you, what is 12 to the negative seven divided by 12 to the negative five power? Well, when you're dividing, you subtract exponents if you have the same base. So this is going to be equal to 12 to the negative seven minus negative five power. You're subtracting the bottom exponent. And so this is going to be equal to 12 to the, well, subtracting a negative is the same thing as adding the positive, a 12 to the negative two power."}, {"video_title": "Multiplying & dividing powers (integer exponents)   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "Well, when you're dividing, you subtract exponents if you have the same base. So this is going to be equal to 12 to the negative seven minus negative five power. You're subtracting the bottom exponent. And so this is going to be equal to 12 to the, well, subtracting a negative is the same thing as adding the positive, a 12 to the negative two power. And once again, we just have to think about why does this actually make sense? Well, you can actually rewrite this. 12 to the negative seven divided by 12 to the negative five, that's the same thing as 12 to the negative seven times 12 to the fifth power."}, {"video_title": "Multiplying & dividing powers (integer exponents)   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "And so this is going to be equal to 12 to the, well, subtracting a negative is the same thing as adding the positive, a 12 to the negative two power. And once again, we just have to think about why does this actually make sense? Well, you can actually rewrite this. 12 to the negative seven divided by 12 to the negative five, that's the same thing as 12 to the negative seven times 12 to the fifth power. If we take the reciprocal of, if we take the reciprocal of this right over here, you would make the exponent positive. And then you get exactly what we were doing in those previous examples with products. And so let's just do one more with variables for good measure."}, {"video_title": "Multiplying & dividing powers (integer exponents)   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "12 to the negative seven divided by 12 to the negative five, that's the same thing as 12 to the negative seven times 12 to the fifth power. If we take the reciprocal of, if we take the reciprocal of this right over here, you would make the exponent positive. And then you get exactly what we were doing in those previous examples with products. And so let's just do one more with variables for good measure. Let's say I have x to the negative 20th power divided by x to the fifth power. Well, once again, we have the same base and we're taking a quotient. So this is going to be x to the negative 20 minus five because we have this one right over here in the denominator."}, {"video_title": "Multiplying & dividing powers (integer exponents)   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "And so let's just do one more with variables for good measure. Let's say I have x to the negative 20th power divided by x to the fifth power. Well, once again, we have the same base and we're taking a quotient. So this is going to be x to the negative 20 minus five because we have this one right over here in the denominator. So this is going to be equal to x to the negative 25th power. And once again, you could view our original expression as x to the negative 20th. And having an x to the fifth in the denominator, dividing by x to the fifth, is the same thing as multiplying by x to the negative five."}, {"video_title": "Absolute value and number lines   Negative numbers and absolute value   Pre-Algebra   Khan Academy.mp3", "Sentence": "And you see every one of these values have an absolute value sign, so let's take a little bit of a review of what absolute value even is. Absolute value. The way I think about it, there's two ways to think about it. The first way to think about it is how far is something from 0? So let me plot this negative 3 here. So let me do a number line. This isn't the number line for our actual answer to this command, plot these values on a number line."}, {"video_title": "Absolute value and number lines   Negative numbers and absolute value   Pre-Algebra   Khan Academy.mp3", "Sentence": "The first way to think about it is how far is something from 0? So let me plot this negative 3 here. So let me do a number line. This isn't the number line for our actual answer to this command, plot these values on a number line. I'm just first going to plot the numbers inside the absolute value sign, and then we're going to take the absolute value and plot those, just like they're asking us to do. So on this number line, if this is 0, if we go to the negative, we're going to go to the left of 0. So this is negative 1, negative 2, negative 3."}, {"video_title": "Absolute value and number lines   Negative numbers and absolute value   Pre-Algebra   Khan Academy.mp3", "Sentence": "This isn't the number line for our actual answer to this command, plot these values on a number line. I'm just first going to plot the numbers inside the absolute value sign, and then we're going to take the absolute value and plot those, just like they're asking us to do. So on this number line, if this is 0, if we go to the negative, we're going to go to the left of 0. So this is negative 1, negative 2, negative 3. Negative 3 sits right over there. So this is negative 3 right there. The absolute value of negative 3 is essentially saying how far are you away from 0?"}, {"video_title": "Absolute value and number lines   Negative numbers and absolute value   Pre-Algebra   Khan Academy.mp3", "Sentence": "So this is negative 1, negative 2, negative 3. Negative 3 sits right over there. So this is negative 3 right there. The absolute value of negative 3 is essentially saying how far are you away from 0? How far is negative 3 from 0? And you say, well, it's 1, 2, 3 away from 0. So you'd say that the absolute value of negative 3 is equal to positive 3."}, {"video_title": "Absolute value and number lines   Negative numbers and absolute value   Pre-Algebra   Khan Academy.mp3", "Sentence": "The absolute value of negative 3 is essentially saying how far are you away from 0? How far is negative 3 from 0? And you say, well, it's 1, 2, 3 away from 0. So you'd say that the absolute value of negative 3 is equal to positive 3. Now, that's really the conceptual way to imagine absolute value. How far are you away from 0? But the easy way to calculate absolute value signs, if you don't care too much about the concept, is whether it's negative or positive, the absolute value of it's always going to be positive."}, {"video_title": "Absolute value and number lines   Negative numbers and absolute value   Pre-Algebra   Khan Academy.mp3", "Sentence": "So you'd say that the absolute value of negative 3 is equal to positive 3. Now, that's really the conceptual way to imagine absolute value. How far are you away from 0? But the easy way to calculate absolute value signs, if you don't care too much about the concept, is whether it's negative or positive, the absolute value of it's always going to be positive. Absolute value of negative 3 is positive 3. Absolute value of positive 3 is still positive 3. So you're always going to get the positive version of the number, so to speak."}, {"video_title": "Absolute value and number lines   Negative numbers and absolute value   Pre-Algebra   Khan Academy.mp3", "Sentence": "But the easy way to calculate absolute value signs, if you don't care too much about the concept, is whether it's negative or positive, the absolute value of it's always going to be positive. Absolute value of negative 3 is positive 3. Absolute value of positive 3 is still positive 3. So you're always going to get the positive version of the number, so to speak. But conceptually, you're just saying how far away are you from 0? So let's do what they asked. So that first value on this number line, so all of these are absolute values."}, {"video_title": "Absolute value and number lines   Negative numbers and absolute value   Pre-Algebra   Khan Academy.mp3", "Sentence": "So you're always going to get the positive version of the number, so to speak. But conceptually, you're just saying how far away are you from 0? So let's do what they asked. So that first value on this number line, so all of these are absolute values. So they're all going to be positive values. So they're all going to be greater than 0. So let me draw my number line like this."}, {"video_title": "Absolute value and number lines   Negative numbers and absolute value   Pre-Algebra   Khan Academy.mp3", "Sentence": "So that first value on this number line, so all of these are absolute values. So they're all going to be positive values. So they're all going to be greater than 0. So let me draw my number line like this. I could do a straighter number line than that. Let's see. Well, that's a little bit straighter."}, {"video_title": "Absolute value and number lines   Negative numbers and absolute value   Pre-Algebra   Khan Academy.mp3", "Sentence": "So let me draw my number line like this. I could do a straighter number line than that. Let's see. Well, that's a little bit straighter. And let's say if this is 0, this would be negative 1, then you'd have 1, 2, 3, 4, 5, 6, 7. I think that'll do the trick. So this first quantity here, I'll do it in orange, the absolute value of negative 3 we just figured out."}, {"video_title": "Absolute value and number lines   Negative numbers and absolute value   Pre-Algebra   Khan Academy.mp3", "Sentence": "Well, that's a little bit straighter. And let's say if this is 0, this would be negative 1, then you'd have 1, 2, 3, 4, 5, 6, 7. I think that'll do the trick. So this first quantity here, I'll do it in orange, the absolute value of negative 3 we just figured out. That is positive 3. So I'll plot it right over there. Positive 3."}, {"video_title": "Absolute value and number lines   Negative numbers and absolute value   Pre-Algebra   Khan Academy.mp3", "Sentence": "So this first quantity here, I'll do it in orange, the absolute value of negative 3 we just figured out. That is positive 3. So I'll plot it right over there. Positive 3. Then this next value right here, the absolute value of 7. If we look over here, 1, 2, 3, 4, 5, 6, 7, 7 is how far away from 0? It is 7 away from 0."}, {"video_title": "Absolute value and number lines   Negative numbers and absolute value   Pre-Algebra   Khan Academy.mp3", "Sentence": "Positive 3. Then this next value right here, the absolute value of 7. If we look over here, 1, 2, 3, 4, 5, 6, 7, 7 is how far away from 0? It is 7 away from 0. So the absolute value of 7 is equal to 7. So you already see the pattern there. If it's negative, it just becomes positive."}, {"video_title": "Absolute value and number lines   Negative numbers and absolute value   Pre-Algebra   Khan Academy.mp3", "Sentence": "It is 7 away from 0. So the absolute value of 7 is equal to 7. So you already see the pattern there. If it's negative, it just becomes positive. If it's already positive, it just equals itself. So plotting this value, we'll just place it right over there. So the absolute value of 7 is 7."}, {"video_title": "Absolute value and number lines   Negative numbers and absolute value   Pre-Algebra   Khan Academy.mp3", "Sentence": "If it's negative, it just becomes positive. If it's already positive, it just equals itself. So plotting this value, we'll just place it right over there. So the absolute value of 7 is 7. Absolute value of negative 3 is positive 3. Let me mark out the 0 a little bit better so you see relative to 0. Now we have the absolute value of 8 minus 12."}, {"video_title": "Absolute value and number lines   Negative numbers and absolute value   Pre-Algebra   Khan Academy.mp3", "Sentence": "So the absolute value of 7 is 7. Absolute value of negative 3 is positive 3. Let me mark out the 0 a little bit better so you see relative to 0. Now we have the absolute value of 8 minus 12. The absolute value of 8 minus 12. Well, first of all, let's figure out what 8 minus 12 is. So if you take 12 away from 8, you're at negative 4."}, {"video_title": "Absolute value and number lines   Negative numbers and absolute value   Pre-Algebra   Khan Academy.mp3", "Sentence": "Now we have the absolute value of 8 minus 12. The absolute value of 8 minus 12. Well, first of all, let's figure out what 8 minus 12 is. So if you take 12 away from 8, you're at negative 4. 12 less than 8 is negative 4. And you can do that on the number line if this is a little bit, if you don't quite remember how to do this. But if you take 8 away from 8, you're at 0."}, {"video_title": "Absolute value and number lines   Negative numbers and absolute value   Pre-Algebra   Khan Academy.mp3", "Sentence": "So if you take 12 away from 8, you're at negative 4. 12 less than 8 is negative 4. And you can do that on the number line if this is a little bit, if you don't quite remember how to do this. But if you take 8 away from 8, you're at 0. And then you take another one, you're at negative 1, then at negative 2, negative 3, all the way to negative 4. So this is equal to the absolute value of negative 4. If we just plot negative 4, we go 1, 2, 3, negative 4 is right over there."}, {"video_title": "Absolute value and number lines   Negative numbers and absolute value   Pre-Algebra   Khan Academy.mp3", "Sentence": "But if you take 8 away from 8, you're at 0. And then you take another one, you're at negative 1, then at negative 2, negative 3, all the way to negative 4. So this is equal to the absolute value of negative 4. If we just plot negative 4, we go 1, 2, 3, negative 4 is right over there. But if we're taking its absolute value, we're saying how far is negative 4 from 0? Well, it's 4 away from 0. 1, 2, 3, 4."}, {"video_title": "Absolute value and number lines   Negative numbers and absolute value   Pre-Algebra   Khan Academy.mp3", "Sentence": "If we just plot negative 4, we go 1, 2, 3, negative 4 is right over there. But if we're taking its absolute value, we're saying how far is negative 4 from 0? Well, it's 4 away from 0. 1, 2, 3, 4. So this is equal to positive 4. So we'll plot it right here. This number line is the answer to this command up here."}, {"video_title": "Absolute value and number lines   Negative numbers and absolute value   Pre-Algebra   Khan Academy.mp3", "Sentence": "1, 2, 3, 4. So this is equal to positive 4. So we'll plot it right here. This number line is the answer to this command up here. So the absolute value of 8 minus 12, which is negative 4, is positive 4. Then we have the absolute value of 0. So how far is 0 from 0?"}, {"video_title": "Absolute value and number lines   Negative numbers and absolute value   Pre-Algebra   Khan Academy.mp3", "Sentence": "This number line is the answer to this command up here. So the absolute value of 8 minus 12, which is negative 4, is positive 4. Then we have the absolute value of 0. So how far is 0 from 0? Well, it's 0 away from 0. The absolute value of 0 is 0. So you can just plot it right over there."}, {"video_title": "Absolute value and number lines   Negative numbers and absolute value   Pre-Algebra   Khan Academy.mp3", "Sentence": "So how far is 0 from 0? Well, it's 0 away from 0. The absolute value of 0 is 0. So you can just plot it right over there. And we have one left. Let me pick a suitable color here. The absolute value of 7 minus 2."}, {"video_title": "Absolute value and number lines   Negative numbers and absolute value   Pre-Algebra   Khan Academy.mp3", "Sentence": "So you can just plot it right over there. And we have one left. Let me pick a suitable color here. The absolute value of 7 minus 2. The absolute value of 7 minus 2. Well, 7 minus 2 is 5. So this is the same thing as the absolute value of 5."}, {"video_title": "Absolute value and number lines   Negative numbers and absolute value   Pre-Algebra   Khan Academy.mp3", "Sentence": "The absolute value of 7 minus 2. The absolute value of 7 minus 2. Well, 7 minus 2 is 5. So this is the same thing as the absolute value of 5. How far is 5 away from 0? Well, it's just 5 away. It's almost too easy."}, {"video_title": "Absolute value and number lines   Negative numbers and absolute value   Pre-Algebra   Khan Academy.mp3", "Sentence": "So this is the same thing as the absolute value of 5. How far is 5 away from 0? Well, it's just 5 away. It's almost too easy. That's what makes it confusing. If I were to plot 5, it's 1, 2, 3, 4, 5. It is 1, 2, 3, 4, 5 spaces from 0."}, {"video_title": "Absolute value and number lines   Negative numbers and absolute value   Pre-Algebra   Khan Academy.mp3", "Sentence": "It's almost too easy. That's what makes it confusing. If I were to plot 5, it's 1, 2, 3, 4, 5. It is 1, 2, 3, 4, 5 spaces from 0. So the absolute value of 5 is 5. So you plot it just like that. So conceptually, it's how far you are away from 0."}, {"video_title": "Checking solutions of systems of inequalities example   Algebra I   Khan Academy.mp3", "Sentence": "is 2,5, a solution of this system. And we have a system of inequalities right over here. We have y is greater than or equal to 2x plus 1, and x is greater than 1. In order for 2,5 to be a solution of the system, it just has to satisfy both inequalities. So just try it out. So when x is equal to 2 and y is equal to 5, it has to satisfy both of these. So let's try it with the first one."}, {"video_title": "Checking solutions of systems of inequalities example   Algebra I   Khan Academy.mp3", "Sentence": "In order for 2,5 to be a solution of the system, it just has to satisfy both inequalities. So just try it out. So when x is equal to 2 and y is equal to 5, it has to satisfy both of these. So let's try it with the first one. So if we assume x is 2 and y is 5, we would get an inequality that says 5 is greater than or equal to 2 times 2 plus 1. x is 2, y is 5. This gives us 5 is greater than or equal to 2 times 2 is 4, plus 1 is 5. y is greater than or equal to 5. That's true."}, {"video_title": "Checking solutions of systems of inequalities example   Algebra I   Khan Academy.mp3", "Sentence": "So let's try it with the first one. So if we assume x is 2 and y is 5, we would get an inequality that says 5 is greater than or equal to 2 times 2 plus 1. x is 2, y is 5. This gives us 5 is greater than or equal to 2 times 2 is 4, plus 1 is 5. y is greater than or equal to 5. That's true. 5 is equal to 5. So that equal part of the greater than or equal saves us. So it definitely satisfies the first inequality."}, {"video_title": "Checking solutions of systems of inequalities example   Algebra I   Khan Academy.mp3", "Sentence": "That's true. 5 is equal to 5. So that equal part of the greater than or equal saves us. So it definitely satisfies the first inequality. Let's see the second one. x needs to be greater than 1. So in 2,5, x is 2."}, {"video_title": "Checking solutions of systems of inequalities example   Algebra I   Khan Academy.mp3", "Sentence": "So it definitely satisfies the first inequality. Let's see the second one. x needs to be greater than 1. So in 2,5, x is 2. So 2 is greater than 1. So it actually satisfies both of these inequalities. So 2,5 is a solution for this system."}, {"video_title": "Understanding inverse functions   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "You'd say, okay, if we input negative nine into our function, if x is negative nine, this table tells us that f of x is going to be equal to five. And you might already have experience with doing composite functions, where you say f of, let's say actually f of, f of negative nine plus one. So this is interesting. It seems very daunting, but you say, oh well, we know what f of negative nine is. This is going to be five. So it's gonna be f of five plus one. So this is going to be equal to f of six."}, {"video_title": "Understanding inverse functions   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "It seems very daunting, but you say, oh well, we know what f of negative nine is. This is going to be five. So it's gonna be f of five plus one. So this is going to be equal to f of six. And if we look at our table, f of six is equal to negative seven. So all of that is a review so far. But what I wanna now do is start evaluating the inverse of functions."}, {"video_title": "Understanding inverse functions   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "So this is going to be equal to f of six. And if we look at our table, f of six is equal to negative seven. So all of that is a review so far. But what I wanna now do is start evaluating the inverse of functions. And this function f is invertible, because it's a one-to-one mapping between the x's and the f of x's. No two x's map to the same f of x. And so this is an invertible function."}, {"video_title": "Understanding inverse functions   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "But what I wanna now do is start evaluating the inverse of functions. And this function f is invertible, because it's a one-to-one mapping between the x's and the f of x's. No two x's map to the same f of x. And so this is an invertible function. So with that in mind, let's see if we can evaluate something like f inverse, f inverse of eight. What is that going to be? I encourage you to pause the video and try to think about it."}, {"video_title": "Understanding inverse functions   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "And so this is an invertible function. So with that in mind, let's see if we can evaluate something like f inverse, f inverse of eight. What is that going to be? I encourage you to pause the video and try to think about it. All right, so f of x, just as a reminder of what functions do, f of x is going to map from this domain, from a value in its domain, to a corresponding value in the range. In the range. So this is what f does."}, {"video_title": "Understanding inverse functions   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "I encourage you to pause the video and try to think about it. All right, so f of x, just as a reminder of what functions do, f of x is going to map from this domain, from a value in its domain, to a corresponding value in the range. In the range. So this is what f does. So this is domain, domain, and this right over here is going to be the range. Now f inverse, if you pass it the value in the range, it'll map it back to the corresponding value in the domain. But how do we think about it like this?"}, {"video_title": "Understanding inverse functions   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "So this is what f does. So this is domain, domain, and this right over here is going to be the range. Now f inverse, if you pass it the value in the range, it'll map it back to the corresponding value in the domain. But how do we think about it like this? Well f inverse of eight, this is whatever maps to eight. So if this was eight, we have to say, well what mapped to eight? Well we see here f of nine is eight."}, {"video_title": "Understanding inverse functions   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "But how do we think about it like this? Well f inverse of eight, this is whatever maps to eight. So if this was eight, we have to say, well what mapped to eight? Well we see here f of nine is eight. f of nine is eight. So f inverse of eight is going to be, is going to be equal to, and actually we do this in that same color, is going to be equal to nine. And if it makes it easier, we could actually construct a table here."}, {"video_title": "Understanding inverse functions   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "Well we see here f of nine is eight. f of nine is eight. So f inverse of eight is going to be, is going to be equal to, and actually we do this in that same color, is going to be equal to nine. And if it makes it easier, we could actually construct a table here. And this is actually what I probably would do, just to make sure I'm not doing something strange. Where I could say x and f inverse of x, and what I'd essentially do is I'd swap these two columns. So f of x goes from negative nine to five, f inverse of x is going to go from five to negative nine."}, {"video_title": "Understanding inverse functions   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "And if it makes it easier, we could actually construct a table here. And this is actually what I probably would do, just to make sure I'm not doing something strange. Where I could say x and f inverse of x, and what I'd essentially do is I'd swap these two columns. So f of x goes from negative nine to five, f inverse of x is going to go from five to negative nine. All I did is I swapped these two. Now we're mapping from this to that. So f inverse of x is gonna map from seven to negative seven."}, {"video_title": "Understanding inverse functions   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "So f of x goes from negative nine to five, f inverse of x is going to go from five to negative nine. All I did is I swapped these two. Now we're mapping from this to that. So f inverse of x is gonna map from seven to negative seven. Notice instead of going from this, mapping from this thing to that thing, we're now gonna map from that thing to that thing. So f inverse is gonna map from 13 to five. 13 to five is gonna map from negative seven to six."}, {"video_title": "Understanding inverse functions   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "So f inverse of x is gonna map from seven to negative seven. Notice instead of going from this, mapping from this thing to that thing, we're now gonna map from that thing to that thing. So f inverse is gonna map from 13 to five. 13 to five is gonna map from negative seven to six. Negative seven to six is gonna map from eight to nine. Eight to nine. And it's gonna map from 12 to 11."}, {"video_title": "Understanding inverse functions   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "13 to five is gonna map from negative seven to six. Negative seven to six is gonna map from eight to nine. Eight to nine. And it's gonna map from 12 to 11. From 12 to 11. So let's see, did I do that right? Yeah, it looks like I got all of, yep."}, {"video_title": "Understanding inverse functions   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "And it's gonna map from 12 to 11. From 12 to 11. So let's see, did I do that right? Yeah, it looks like I got all of, yep. So all I did is really I just swapped these columns. It's mapping, the f inverse maps from this column to that column. So I just swapped them out."}, {"video_title": "Understanding inverse functions   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "Yeah, it looks like I got all of, yep. So all I did is really I just swapped these columns. It's mapping, the f inverse maps from this column to that column. So I just swapped them out. So now it becomes a little bit clearer that way. And you see it right over here. F inverse of eight, if you input eight into f inverse, you get nine."}, {"video_title": "Understanding inverse functions   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "So I just swapped them out. So now it becomes a little bit clearer that way. And you see it right over here. F inverse of eight, if you input eight into f inverse, you get nine. So now we can use that to start doing fancier things. We can evaluate something like f inverse, f inverse of, actually, let's evaluate this. Let's evaluate f of, f of f inverse of, let's see, f of f inverse of seven."}, {"video_title": "Understanding inverse functions   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "F inverse of eight, if you input eight into f inverse, you get nine. So now we can use that to start doing fancier things. We can evaluate something like f inverse, f inverse of, actually, let's evaluate this. Let's evaluate f of, f of f inverse of, let's see, f of f inverse of seven. What is this going to be? Well, let's first evaluate f inverse of seven. F inverse of seven maps from seven to negative seven."}, {"video_title": "Understanding inverse functions   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "Let's evaluate f of, f of f inverse of, let's see, f of f inverse of seven. What is this going to be? Well, let's first evaluate f inverse of seven. F inverse of seven maps from seven to negative seven. Maps from seven to negative seven. So this is going to be f of this stuff in here. I think we do it in yellow."}, {"video_title": "Understanding inverse functions   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "F inverse of seven maps from seven to negative seven. Maps from seven to negative seven. So this is going to be f of this stuff in here. I think we do it in yellow. This stuff in here, f inverse of seven, we see is negative seven. So it's going to be f of negative seven. And then to evaluate the function, well, f of negative seven, that's just going to be seven again."}, {"video_title": "Understanding inverse functions   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "I think we do it in yellow. This stuff in here, f inverse of seven, we see is negative seven. So it's going to be f of negative seven. And then to evaluate the function, well, f of negative seven, that's just going to be seven again. And that makes complete sense. We essentially went, we mapped from seven, f inverse of seven went from seven to negative seven, and then evaluating the function of that went back to, went back to seven. So let's do one more of these, just to really feel comfortable with mapping back and forth between these two sets, between applying the function and the inverse of the function."}, {"video_title": "Understanding inverse functions   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "And then to evaluate the function, well, f of negative seven, that's just going to be seven again. And that makes complete sense. We essentially went, we mapped from seven, f inverse of seven went from seven to negative seven, and then evaluating the function of that went back to, went back to seven. So let's do one more of these, just to really feel comfortable with mapping back and forth between these two sets, between applying the function and the inverse of the function. So let's evaluate what, do it in purple, f, I mean, no, that wasn't purple. So it's going, let's try to evaluate f, f inverse of, f inverse of 13, of f inverse of 13. What is that going to be?"}, {"video_title": "Understanding inverse functions   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "So let's do one more of these, just to really feel comfortable with mapping back and forth between these two sets, between applying the function and the inverse of the function. So let's evaluate what, do it in purple, f, I mean, no, that wasn't purple. So it's going, let's try to evaluate f, f inverse of, f inverse of 13, of f inverse of 13. What is that going to be? I encourage you to pause the video and try to figure it out. Well, what's f inverse of 13? Well, that's looking at this table right here."}, {"video_title": "Understanding inverse functions   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "What is that going to be? I encourage you to pause the video and try to figure it out. Well, what's f inverse of 13? Well, that's looking at this table right here. f inverse goes from 13 to five. And you see it over here. f went from five to 13, so f inverse is going to go from 13 to five."}, {"video_title": "Understanding inverse functions   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "Well, that's looking at this table right here. f inverse goes from 13 to five. And you see it over here. f went from five to 13, so f inverse is going to go from 13 to five. So this right over here, f inverse of 13 is just going to be five. So this whole thing is the same thing as f inverse of five. And f inverse of five, well, f inverse goes from five to negative nine."}, {"video_title": "Understanding inverse functions   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "f went from five to 13, so f inverse is going to go from 13 to five. So this right over here, f inverse of 13 is just going to be five. So this whole thing is the same thing as f inverse of five. And f inverse of five, well, f inverse goes from five to negative nine. So this is going to be equal to negative nine. Once again, f inverse goes from five, f goes from negative nine to five, so f inverse is gonna go from five to negative nine. So at first, when you start doing these kind of inverse and function and inverse of a function, it looks a little confusing."}, {"video_title": "Understanding inverse functions   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "And f inverse of five, well, f inverse goes from five to negative nine. So this is going to be equal to negative nine. Once again, f inverse goes from five, f goes from negative nine to five, so f inverse is gonna go from five to negative nine. So at first, when you start doing these kind of inverse and function and inverse of a function, it looks a little confusing. Hey, I'm going back and forth. But you just have to remember, a function maps from one set of numbers to another set of numbers, and the inverse of that function goes the other way. So if the function goes from nine to eight, the inverse is gonna go from eight to nine."}, {"video_title": "Combining like terms, but more complicated   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Now we have a very, very, very hairy expression. And once again, I'm going to see if you can simplify this. And I'll give you a little time to do it. So this one is even crazier than the last few we've looked at. We've got y's and xy's and x squared's and x's and, well, more just xy's and y squared's and on and on and on. And there will be a temptation, because you see a y here and a y here, to say, oh, maybe I can add this negative 3y plus this 4xy somehow, since I see a y and a y. But the important thing to realize here is that a y is different than an xy."}, {"video_title": "Combining like terms, but more complicated   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "So this one is even crazier than the last few we've looked at. We've got y's and xy's and x squared's and x's and, well, more just xy's and y squared's and on and on and on. And there will be a temptation, because you see a y here and a y here, to say, oh, maybe I can add this negative 3y plus this 4xy somehow, since I see a y and a y. But the important thing to realize here is that a y is different than an xy. Think about if they were numbers. If y was 3 and an x was a 2, then a y would be a 3, while an xy would have been a 6. And a y is very different than a y squared."}, {"video_title": "Combining like terms, but more complicated   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "But the important thing to realize here is that a y is different than an xy. Think about if they were numbers. If y was 3 and an x was a 2, then a y would be a 3, while an xy would have been a 6. And a y is very different than a y squared. Once again, if the y took on the value 3, then the y squared would be the value 9. So even though you see the same letter here, they aren't the same. I guess you cannot add these two or subtract these two terms."}, {"video_title": "Combining like terms, but more complicated   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And a y is very different than a y squared. Once again, if the y took on the value 3, then the y squared would be the value 9. So even though you see the same letter here, they aren't the same. I guess you cannot add these two or subtract these two terms. A y is different than a y squared is different than an xy. Now with that said, let's see if we can, if there is anything that we can simplify. So first let's think about the y terms."}, {"video_title": "Combining like terms, but more complicated   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "I guess you cannot add these two or subtract these two terms. A y is different than a y squared is different than an xy. Now with that said, let's see if we can, if there is anything that we can simplify. So first let's think about the y terms. So you have a negative 3y there. Do we have any more y terms? Yes, we do."}, {"video_title": "Combining like terms, but more complicated   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "So first let's think about the y terms. So you have a negative 3y there. Do we have any more y terms? Yes, we do. We have this 2y right over there. So I'll just write it out. I'll just reorder it."}, {"video_title": "Combining like terms, but more complicated   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Yes, we do. We have this 2y right over there. So I'll just write it out. I'll just reorder it. So we have negative 3y plus 2y. Now let's think about, I'm just going in an arbitrary order, but since our next term is an xy term, let's think about all of the xy terms. So we have plus 4xy right over here."}, {"video_title": "Combining like terms, but more complicated   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "I'll just reorder it. So we have negative 3y plus 2y. Now let's think about, I'm just going in an arbitrary order, but since our next term is an xy term, let's think about all of the xy terms. So we have plus 4xy right over here. So let me just write it down. I'm just reordering the whole expression. Plus 4xy and then I have minus 4xy right over here."}, {"video_title": "Combining like terms, but more complicated   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "So we have plus 4xy right over here. So let me just write it down. I'm just reordering the whole expression. Plus 4xy and then I have minus 4xy right over here. So minus 4xy. Then let's go to the x squared terms. I have negative 2 times x squared or minus 2x squared."}, {"video_title": "Combining like terms, but more complicated   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Plus 4xy and then I have minus 4xy right over here. So minus 4xy. Then let's go to the x squared terms. I have negative 2 times x squared or minus 2x squared. So let's look at this. So I have minus 2x squared. Do I have any other x squareds?"}, {"video_title": "Combining like terms, but more complicated   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "I have negative 2 times x squared or minus 2x squared. So let's look at this. So I have minus 2x squared. Do I have any other x squareds? Yes, I do. I have this 3x squared right over there. So plus 3x squareds."}, {"video_title": "Combining like terms, but more complicated   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Do I have any other x squareds? Yes, I do. I have this 3x squared right over there. So plus 3x squareds. And then let's see. I have an x term right over here. And that actually looks like the only x term."}, {"video_title": "Combining like terms, but more complicated   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "So plus 3x squareds. And then let's see. I have an x term right over here. And that actually looks like the only x term. So that's plus 2x. And then I only have one y squared term. I'll circle that in orange."}, {"video_title": "Combining like terms, but more complicated   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And that actually looks like the only x term. So that's plus 2x. And then I only have one y squared term. I'll circle that in orange. So plus y squared. So all I have done is I've reordered the statement and I've color coded it based on the type of term we have. And now it should be a little bit simpler."}, {"video_title": "Combining like terms, but more complicated   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "I'll circle that in orange. So plus y squared. So all I have done is I've reordered the statement and I've color coded it based on the type of term we have. And now it should be a little bit simpler. So let's try it out. If I have negative 3 of something plus 2 of that something, what do I have? Or another way to say it, if I have 2 of something and I subtract 3 of that, what am I left with?"}, {"video_title": "Combining like terms, but more complicated   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And now it should be a little bit simpler. So let's try it out. If I have negative 3 of something plus 2 of that something, what do I have? Or another way to say it, if I have 2 of something and I subtract 3 of that, what am I left with? I'm left with negative 1 of that something. So I could write negative 1y or I could just write negative y. And another way to think about it, but I like to think about it intuitively more, is what's the coefficient here?"}, {"video_title": "Combining like terms, but more complicated   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Or another way to say it, if I have 2 of something and I subtract 3 of that, what am I left with? I'm left with negative 1 of that something. So I could write negative 1y or I could just write negative y. And another way to think about it, but I like to think about it intuitively more, is what's the coefficient here? It is negative 3. What's the coefficient here? It's 2."}, {"video_title": "Combining like terms, but more complicated   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And another way to think about it, but I like to think about it intuitively more, is what's the coefficient here? It is negative 3. What's the coefficient here? It's 2. We're obviously both dealing with y terms. Not xy terms, not y squared terms, just y. And so negative 3 plus 2 is negative 1."}, {"video_title": "Combining like terms, but more complicated   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "It's 2. We're obviously both dealing with y terms. Not xy terms, not y squared terms, just y. And so negative 3 plus 2 is negative 1. Or negative 1y is the same thing as negative y. So those simplify to this right over here. Now let's look at the xy terms."}, {"video_title": "Combining like terms, but more complicated   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "And so negative 3 plus 2 is negative 1. Or negative 1y is the same thing as negative y. So those simplify to this right over here. Now let's look at the xy terms. If I have 4 of this, 4 xy's, and I were to take away 4 xy's, how many xy's am I left with? Well, I'm left with no xy's. Or you could say, add the coefficients, 4 plus negative 4 gives you 0 xy's."}, {"video_title": "Combining like terms, but more complicated   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Now let's look at the xy terms. If I have 4 of this, 4 xy's, and I were to take away 4 xy's, how many xy's am I left with? Well, I'm left with no xy's. Or you could say, add the coefficients, 4 plus negative 4 gives you 0 xy's. Either way, these two cancel out. If I have 4 of something and I take away those 4 of that something, I'm left with none of them. So I'm left with no xy's."}, {"video_title": "Combining like terms, but more complicated   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Or you could say, add the coefficients, 4 plus negative 4 gives you 0 xy's. Either way, these two cancel out. If I have 4 of something and I take away those 4 of that something, I'm left with none of them. So I'm left with no xy's. And then I have right over here, I could have written 0 xy, but that seems unnecessary. Then right over here, I have my x squared terms. Negative 2 plus 3 is 1."}, {"video_title": "Combining like terms, but more complicated   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "So I'm left with no xy's. And then I have right over here, I could have written 0 xy, but that seems unnecessary. Then right over here, I have my x squared terms. Negative 2 plus 3 is 1. Or another way of saying, if I have 3 x squared's and I were to take away 2 of those x squared's, I'm left with 1 x squared. So this right over here simplifies to 1 x squared. Or I could literally just write x squared."}, {"video_title": "Combining like terms, but more complicated   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Negative 2 plus 3 is 1. Or another way of saying, if I have 3 x squared's and I were to take away 2 of those x squared's, I'm left with 1 x squared. So this right over here simplifies to 1 x squared. Or I could literally just write x squared. 1 x squared is the same thing as x squared. So plus x squared. And then these, there's nothing really left to simplify."}, {"video_title": "Combining like terms, but more complicated   Introduction to algebra   Algebra I   Khan Academy.mp3", "Sentence": "Or I could literally just write x squared. 1 x squared is the same thing as x squared. So plus x squared. And then these, there's nothing really left to simplify. So plus 2x plus y squared. And we're done. And obviously, you might have gotten an answer in some other order, but the order in which I write these terms don't matter."}, {"video_title": "Recursive formulas for arithmetic sequences   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "Here are the first few terms of the sequence. So they say the first term is four, second term is three and four fifths, third term is three and three fifths, fourth term is three and two fifths. Find the values of the missing parameters a and b in the following recursive definition of the sequence. So they say the nth term is going to be equal to a if n is equal to one, and it's going to be equal to g of n minus one plus b if n is greater than one. And so I encourage you to pause this video and see if you can figure out what a and b are going to be. Well, the first one to figure out a is actually pretty straightforward. If n is equal to one, if n is equal to one, the first term when n equals one is four."}, {"video_title": "Recursive formulas for arithmetic sequences   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "So they say the nth term is going to be equal to a if n is equal to one, and it's going to be equal to g of n minus one plus b if n is greater than one. And so I encourage you to pause this video and see if you can figure out what a and b are going to be. Well, the first one to figure out a is actually pretty straightforward. If n is equal to one, if n is equal to one, the first term when n equals one is four. So a is equal to four. So we could write this as g of n is equal to four if n is equal to one. And now let's think about the second line."}, {"video_title": "Recursive formulas for arithmetic sequences   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "If n is equal to one, if n is equal to one, the first term when n equals one is four. So a is equal to four. So we could write this as g of n is equal to four if n is equal to one. And now let's think about the second line. The second line is interesting. It's saying it's going to be equal to the previous term, g of n minus one. This means the n minus oneth term plus b will give you the nth term."}, {"video_title": "Recursive formulas for arithmetic sequences   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "And now let's think about the second line. The second line is interesting. It's saying it's going to be equal to the previous term, g of n minus one. This means the n minus oneth term plus b will give you the nth term. Let's just think about what's happening with this arithmetic sequence. When I go from the first term to the second term, what have I done? I have, looks like I've subtracted 1 5th."}, {"video_title": "Recursive formulas for arithmetic sequences   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "This means the n minus oneth term plus b will give you the nth term. Let's just think about what's happening with this arithmetic sequence. When I go from the first term to the second term, what have I done? I have, looks like I've subtracted 1 5th. So minus 1 5th. And then it's an arithmetic sequence, so I should subtract or add the same amount every time. And I am."}, {"video_title": "Recursive formulas for arithmetic sequences   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "I have, looks like I've subtracted 1 5th. So minus 1 5th. And then it's an arithmetic sequence, so I should subtract or add the same amount every time. And I am. I'm subtracting 1 5th. And so I am subtracting 1 5th. And so one way to think about it, if we were to go the other way, we could say, for example, that g of four is equal to g of three minus 1 5th."}, {"video_title": "Recursive formulas for arithmetic sequences   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "And I am. I'm subtracting 1 5th. And so I am subtracting 1 5th. And so one way to think about it, if we were to go the other way, we could say, for example, that g of four is equal to g of three minus 1 5th. Minus 1 5th. You see that right over there. G of three is this."}, {"video_title": "Recursive formulas for arithmetic sequences   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "And so one way to think about it, if we were to go the other way, we could say, for example, that g of four is equal to g of three minus 1 5th. Minus 1 5th. You see that right over there. G of three is this. You subtract 1 5th, you get g of four. You see that right over there. And of course I could have written this like g of four is equal to g of four minus one minus 1 5th."}, {"video_title": "Recursive formulas for arithmetic sequences   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "G of three is this. You subtract 1 5th, you get g of four. You see that right over there. And of course I could have written this like g of four is equal to g of four minus one minus 1 5th. So when you look at it this way, you could see that if I'm trying to find the nth term, it's going to be the n minus 1th term plus negative 1 5th. So b is negative 1 5th. Once again, if I'm trying to find the fourth term, if n is equal to four, I'm not going to use this first case because this has to be for n equals one."}, {"video_title": "Recursive formulas for arithmetic sequences   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "And of course I could have written this like g of four is equal to g of four minus one minus 1 5th. So when you look at it this way, you could see that if I'm trying to find the nth term, it's going to be the n minus 1th term plus negative 1 5th. So b is negative 1 5th. Once again, if I'm trying to find the fourth term, if n is equal to four, I'm not going to use this first case because this has to be for n equals one. So if n equals four, I would use the second case. So then it would be g of four minus one. It would be g of three minus 1 5th."}, {"video_title": "Recursive formulas for arithmetic sequences   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "Once again, if I'm trying to find the fourth term, if n is equal to four, I'm not going to use this first case because this has to be for n equals one. So if n equals four, I would use the second case. So then it would be g of four minus one. It would be g of three minus 1 5th. And so we could say g of n is equal to g of n minus one. So the term right before that minus 1 5th if n is greater than one. But for the sake of this problem, we see that a is equal to four and b is equal to negative 1 5th."}, {"video_title": "Worked example rational vs. irrational expressions (unknowns)   High School Math   Khan Academy.mp3", "Sentence": "Is a over b rational or irrational? Well, let's think about it. They're both rational numbers, so that means that a, since it's rational, can be expressed as the ratio of two integers. So I could write a is equal to m over n, and same thing about b. I could write b as being equal to p over q, where m, n, p, and q are integers. By definition of what a rational number is, they're telling us these numbers are rational so I can express them as these type of ratios. So what is a over b going to be? a over b is going to be m over n over p over q, which is equal to m over n. If I divide by a fraction, it's the same thing as multiplying by the reciprocal."}, {"video_title": "Worked example rational vs. irrational expressions (unknowns)   High School Math   Khan Academy.mp3", "Sentence": "So I could write a is equal to m over n, and same thing about b. I could write b as being equal to p over q, where m, n, p, and q are integers. By definition of what a rational number is, they're telling us these numbers are rational so I can express them as these type of ratios. So what is a over b going to be? a over b is going to be m over n over p over q, which is equal to m over n. If I divide by a fraction, it's the same thing as multiplying by the reciprocal. q over p, let me write that a little bit, q over p, which is equal to mq over np. Well, mq is going to be an integer if the product of two integers is going to be an integer, and np is going to be another integer. The product of two integers is an integer."}, {"video_title": "Worked example rational vs. irrational expressions (unknowns)   High School Math   Khan Academy.mp3", "Sentence": "a over b is going to be m over n over p over q, which is equal to m over n. If I divide by a fraction, it's the same thing as multiplying by the reciprocal. q over p, let me write that a little bit, q over p, which is equal to mq over np. Well, mq is going to be an integer if the product of two integers is going to be an integer, and np is going to be another integer. The product of two integers is an integer. So I've just shown that a over b can be expressed as the ratio of two integers. So a over b is for sure. In fact, I've just proven it to you."}, {"video_title": "Worked example rational vs. irrational expressions (unknowns)   High School Math   Khan Academy.mp3", "Sentence": "The product of two integers is an integer. So I've just shown that a over b can be expressed as the ratio of two integers. So a over b is for sure. In fact, I've just proven it to you. a over b is for sure going to be rational. Let's do a few more of these. This is interesting."}, {"video_title": "Worked example rational vs. irrational expressions (unknowns)   High School Math   Khan Academy.mp3", "Sentence": "In fact, I've just proven it to you. a over b is for sure going to be rational. Let's do a few more of these. This is interesting. All right. So now we're saying let a and b be irrational numbers. Is a over b, let a and b be irrational numbers."}, {"video_title": "Worked example rational vs. irrational expressions (unknowns)   High School Math   Khan Academy.mp3", "Sentence": "This is interesting. All right. So now we're saying let a and b be irrational numbers. Is a over b, let a and b be irrational numbers. Is a over b rational or irrational? And like always, pause the video and try to think this through. And you might want to do some examples of some irrational numbers and see if you can get, when you divide them, you can get rational or irrational numbers."}, {"video_title": "Worked example rational vs. irrational expressions (unknowns)   High School Math   Khan Academy.mp3", "Sentence": "Is a over b, let a and b be irrational numbers. Is a over b rational or irrational? And like always, pause the video and try to think this through. And you might want to do some examples of some irrational numbers and see if you can get, when you divide them, you can get rational or irrational numbers. Well, let's just imagine a world where let's say that a is equal to, I don't know, two square roots of two, and b is equal to the square root of two. Well, in that world, a over b, a over b would be two square roots of two over the square root of two, which would be two, which is very much a rational number. I can express that as a ratio of integers."}, {"video_title": "Worked example rational vs. irrational expressions (unknowns)   High School Math   Khan Academy.mp3", "Sentence": "And you might want to do some examples of some irrational numbers and see if you can get, when you divide them, you can get rational or irrational numbers. Well, let's just imagine a world where let's say that a is equal to, I don't know, two square roots of two, and b is equal to the square root of two. Well, in that world, a over b, a over b would be two square roots of two over the square root of two, which would be two, which is very much a rational number. I can express that as a ratio of integers. I can write that as two over one. And there's actually an infinite number of ways I can express that as a ratio of two integers. So in this case, I was able to get a over b to be rational, based on a and b being irrational."}, {"video_title": "Worked example rational vs. irrational expressions (unknowns)   High School Math   Khan Academy.mp3", "Sentence": "I can express that as a ratio of integers. I can write that as two over one. And there's actually an infinite number of ways I can express that as a ratio of two integers. So in this case, I was able to get a over b to be rational, based on a and b being irrational. But what if, what if, instead of, what if a was equal to the square root of two and b is equal to the square root of, let's say b is equal to the square root of seven. Well, then a over b would be equal to the square root of two over the square root of seven, which is still going to be irrational. I mean, another way to think about it, and I'm not proving it here, but you could think about this is going to be the square root of 2 7ths."}, {"video_title": "Worked example rational vs. irrational expressions (unknowns)   High School Math   Khan Academy.mp3", "Sentence": "So in this case, I was able to get a over b to be rational, based on a and b being irrational. But what if, what if, instead of, what if a was equal to the square root of two and b is equal to the square root of, let's say b is equal to the square root of seven. Well, then a over b would be equal to the square root of two over the square root of seven, which is still going to be irrational. I mean, another way to think about it, and I'm not proving it here, but you could think about this is going to be the square root of 2 7ths. So we have something that's not a perfect square under the radical, so we're going to end up with an irrational number. So we can, we can show one example where a over b is rational, and we showed one example where it is irrational. So it can be either way, either way."}, {"video_title": "Worked example rational vs. irrational expressions (unknowns)   High School Math   Khan Academy.mp3", "Sentence": "I mean, another way to think about it, and I'm not proving it here, but you could think about this is going to be the square root of 2 7ths. So we have something that's not a perfect square under the radical, so we're going to end up with an irrational number. So we can, we can show one example where a over b is rational, and we showed one example where it is irrational. So it can be either way, either way. Let's do a few more of these. All right. Let a be a non-zero rational number."}, {"video_title": "Worked example rational vs. irrational expressions (unknowns)   High School Math   Khan Academy.mp3", "Sentence": "So it can be either way, either way. Let's do a few more of these. All right. Let a be a non-zero rational number. Is a times square root of eight rational or irrational? Well, the key here is if you multiply an irrational number, and why is this an irrational number? It has a perfect square in it, but it's not a perfect square in and of itself."}, {"video_title": "Worked example rational vs. irrational expressions (unknowns)   High School Math   Khan Academy.mp3", "Sentence": "Let a be a non-zero rational number. Is a times square root of eight rational or irrational? Well, the key here is if you multiply an irrational number, and why is this an irrational number? It has a perfect square in it, but it's not a perfect square in and of itself. Square root of eight is, square root of eight is equal to the square root of four times two, which is equal to the square root of four times square root of two, which is equal to two square roots of two. And this is kind of getting to the punchline of this problem, but if I multiply a rational times an irrational, I am going to get an irrational. So square root of eight is an irrational, and if I multiply that times a rational number, I'm still going to get an irrational number."}, {"video_title": "Worked example rational vs. irrational expressions (unknowns)   High School Math   Khan Academy.mp3", "Sentence": "It has a perfect square in it, but it's not a perfect square in and of itself. Square root of eight is, square root of eight is equal to the square root of four times two, which is equal to the square root of four times square root of two, which is equal to two square roots of two. And this is kind of getting to the punchline of this problem, but if I multiply a rational times an irrational, I am going to get an irrational. So square root of eight is an irrational, and if I multiply that times a rational number, I'm still going to get an irrational number. So this is going to be, for sure, irrational. Let's do one more of these. So we're told let a be an irrational number."}, {"video_title": "Worked example rational vs. irrational expressions (unknowns)   High School Math   Khan Academy.mp3", "Sentence": "So square root of eight is an irrational, and if I multiply that times a rational number, I'm still going to get an irrational number. So this is going to be, for sure, irrational. Let's do one more of these. So we're told let a be an irrational number. Is negative 24 plus a rational or irrational? And I won't give a formal proof here, but it'll give you more of an intuitive feel. It's nice to just try out some numbers, and I encourage you to pause the video and try to think through it yourself."}, {"video_title": "Worked example rational vs. irrational expressions (unknowns)   High School Math   Khan Academy.mp3", "Sentence": "So we're told let a be an irrational number. Is negative 24 plus a rational or irrational? And I won't give a formal proof here, but it'll give you more of an intuitive feel. It's nice to just try out some numbers, and I encourage you to pause the video and try to think through it yourself. Let's just imagine some values. Imagine if a is irrational. So what if a was equal to negative pi, which is approximately equal to negative 3.14159, and it keeps going on and on forever, never repeating?"}, {"video_title": "Worked example rational vs. irrational expressions (unknowns)   High School Math   Khan Academy.mp3", "Sentence": "It's nice to just try out some numbers, and I encourage you to pause the video and try to think through it yourself. Let's just imagine some values. Imagine if a is irrational. So what if a was equal to negative pi, which is approximately equal to negative 3.14159, and it keeps going on and on forever, never repeating? Well, then we would have negative 24 plus a would be equal to negative 24 minus pi, which would be approximately negative 27.14159. The decimal expansion, everything to the right of the decimal is going to be the exact same thing as pi. So this looks like, at least for this example, it's going to be irrational."}, {"video_title": "Worked example rational vs. irrational expressions (unknowns)   High School Math   Khan Academy.mp3", "Sentence": "So what if a was equal to negative pi, which is approximately equal to negative 3.14159, and it keeps going on and on forever, never repeating? Well, then we would have negative 24 plus a would be equal to negative 24 minus pi, which would be approximately negative 27.14159. The decimal expansion, everything to the right of the decimal is going to be the exact same thing as pi. So this looks like, at least for this example, it's going to be irrational. And let's see, if a was square root of two, negative 24 plus the square root of two, well, once again, I'm not doing a proof here, but intuitively, this is going to be a decimal, it's going to have a decimal expansion that's going to go on forever, never repeat. And so this would just change what's to the left of the decimal, but not really change what's, well, it would change what's to the right of the decimal, because this is negative, but it's still going to go on forever and never repeat. And if, in fact, this was, if this was this way, then to the right of the decimal, you would have the same thing as the square root of two."}, {"video_title": "Worked example rational vs. irrational expressions (unknowns)   High School Math   Khan Academy.mp3", "Sentence": "So this looks like, at least for this example, it's going to be irrational. And let's see, if a was square root of two, negative 24 plus the square root of two, well, once again, I'm not doing a proof here, but intuitively, this is going to be a decimal, it's going to have a decimal expansion that's going to go on forever, never repeat. And so this would just change what's to the left of the decimal, but not really change what's, well, it would change what's to the right of the decimal, because this is negative, but it's still going to go on forever and never repeat. And if, in fact, this was, if this was this way, then to the right of the decimal, you would have the same thing as the square root of two. To the left of the decimal, you would just have a different value. You would have, what, negative 25 point whatever, whatever, whatever. And so this is, when you add a rational number to an irrational number, we prove it in other videos, a rational plus an irrational is going to be irrational."}, {"video_title": "Dividing negative fractions   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "Let's do some examples dividing fractions. Let's say that I have negative 5 6 divided by 3 4ths, divided by positive 3 4ths. Well, we've already talked about when you divide by something, it's the exact same thing as multiplying by its reciprocal. So this is going to be the exact same thing as negative 5 6 times the reciprocal of 3 4ths, which is 4 over 3. I'm just swapping the numerator and the denominator. So it's going to be 4 over 3. And we've already seen lots of examples multiplying fractions."}, {"video_title": "Dividing negative fractions   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "So this is going to be the exact same thing as negative 5 6 times the reciprocal of 3 4ths, which is 4 over 3. I'm just swapping the numerator and the denominator. So it's going to be 4 over 3. And we've already seen lots of examples multiplying fractions. This is going to be the numerators times each other. So we're going to multiply negative 5 times 4. I'll give the negative sign to the 5 there."}, {"video_title": "Dividing negative fractions   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "And we've already seen lots of examples multiplying fractions. This is going to be the numerators times each other. So we're going to multiply negative 5 times 4. I'll give the negative sign to the 5 there. So negative 5 times 4. Let me do 4 in that yellow color. And then the denominator is 6 times 3."}, {"video_title": "Dividing negative fractions   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "I'll give the negative sign to the 5 there. So negative 5 times 4. Let me do 4 in that yellow color. And then the denominator is 6 times 3. 6 times 3. Now, in the numerator here, you see we have a negative number. You might already know that 5 times 4 is 20."}, {"video_title": "Dividing negative fractions   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "And then the denominator is 6 times 3. 6 times 3. Now, in the numerator here, you see we have a negative number. You might already know that 5 times 4 is 20. And you just have to remember that, look, we're multiplying a negative times a positive. We're essentially going to have negative 5 four times. So negative 5 plus negative 5 plus negative 5 plus negative 5 is negative 20."}, {"video_title": "Dividing negative fractions   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "You might already know that 5 times 4 is 20. And you just have to remember that, look, we're multiplying a negative times a positive. We're essentially going to have negative 5 four times. So negative 5 plus negative 5 plus negative 5 plus negative 5 is negative 20. So the numerator here is negative 20. And the denominator here is 18. So we get 20 over 18."}, {"video_title": "Dividing negative fractions   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "So negative 5 plus negative 5 plus negative 5 plus negative 5 is negative 20. So the numerator here is negative 20. And the denominator here is 18. So we get 20 over 18. But we can simplify this. Both the numerator and the denominator, they're both divisible by 2. So let's divide them both by 2."}, {"video_title": "Dividing negative fractions   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "So we get 20 over 18. But we can simplify this. Both the numerator and the denominator, they're both divisible by 2. So let's divide them both by 2. Let me give myself a little more space. So if we divide both the numerator and the denominator by 2, just to simplify this. And I picked 2 because that's the largest number that goes into both of these."}, {"video_title": "Dividing negative fractions   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "So let's divide them both by 2. Let me give myself a little more space. So if we divide both the numerator and the denominator by 2, just to simplify this. And I picked 2 because that's the largest number that goes into both of these. That's the greatest common divisor of 20 and 18. 20 divided by 2 is 10. And 18 divided by 2 is 9."}, {"video_title": "Dividing negative fractions   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "And I picked 2 because that's the largest number that goes into both of these. That's the greatest common divisor of 20 and 18. 20 divided by 2 is 10. And 18 divided by 2 is 9. So negative 5 6 divided by 3 4ths is, oh, I have to be very careful here. It's negative 10 9ths. Just how we always learned."}, {"video_title": "Dividing negative fractions   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "And 18 divided by 2 is 9. So negative 5 6 divided by 3 4ths is, oh, I have to be very careful here. It's negative 10 9ths. Just how we always learned. If you have a negative divided by a positive, if the signs are different, then you're going to get a negative value. Let's do another example. Let's say that I have negative 4 divided by negative 1 half."}, {"video_title": "Dividing negative fractions   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "Just how we always learned. If you have a negative divided by a positive, if the signs are different, then you're going to get a negative value. Let's do another example. Let's say that I have negative 4 divided by negative 1 half. So using the exact logic that we just said, we said, hey, look, dividing by something is equivalent to multiplying by its reciprocal. So this is going to be equal to negative 4. Instead of writing it as negative 4, let me just write it as a fraction so that we are clear what its numerator is and what its denominator is."}, {"video_title": "Dividing negative fractions   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "Let's say that I have negative 4 divided by negative 1 half. So using the exact logic that we just said, we said, hey, look, dividing by something is equivalent to multiplying by its reciprocal. So this is going to be equal to negative 4. Instead of writing it as negative 4, let me just write it as a fraction so that we are clear what its numerator is and what its denominator is. So negative 4 is the exact same thing as negative 4 over 1. And we're going to multiply that times the reciprocal of negative 1 half. The reciprocal of negative 1 half is negative 2 over 1."}, {"video_title": "Dividing negative fractions   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "Instead of writing it as negative 4, let me just write it as a fraction so that we are clear what its numerator is and what its denominator is. So negative 4 is the exact same thing as negative 4 over 1. And we're going to multiply that times the reciprocal of negative 1 half. The reciprocal of negative 1 half is negative 2 over 1. You could view it as negative 2 over 1, or you could use it as positive 2 over negative 1, or you could use it as negative 2. Either way, these are all the same value. And now we're ready to multiply."}, {"video_title": "Dividing negative fractions   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "The reciprocal of negative 1 half is negative 2 over 1. You could view it as negative 2 over 1, or you could use it as positive 2 over negative 1, or you could use it as negative 2. Either way, these are all the same value. And now we're ready to multiply. Notice, all I did here, I rewrote the negative 4 just as negative 4 over 1. Negative 4 divided by 1 is negative 4. And here, for the negative 1 half, since I'm multiplying now, I'm multiplying by its reciprocal."}, {"video_title": "Dividing negative fractions   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "And now we're ready to multiply. Notice, all I did here, I rewrote the negative 4 just as negative 4 over 1. Negative 4 divided by 1 is negative 4. And here, for the negative 1 half, since I'm multiplying now, I'm multiplying by its reciprocal. I've swapped the denominator and the numerator. Or I swapped the denominator and the numerator. What was the denominator is now the numerator."}, {"video_title": "Dividing negative fractions   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "And here, for the negative 1 half, since I'm multiplying now, I'm multiplying by its reciprocal. I've swapped the denominator and the numerator. Or I swapped the denominator and the numerator. What was the denominator is now the numerator. What was the numerator is now the denominator. And I'm ready to multiply. This is going to be equal to, I gave both the negative signs to the numerator, so it's going to be negative 4 times negative 2 in the numerator."}, {"video_title": "Dividing negative fractions   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "What was the denominator is now the numerator. What was the numerator is now the denominator. And I'm ready to multiply. This is going to be equal to, I gave both the negative signs to the numerator, so it's going to be negative 4 times negative 2 in the numerator. And then in the denominator, it's going to be 1 times 1. Let me write that down. 1 times 1."}, {"video_title": "Dividing negative fractions   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "This is going to be equal to, I gave both the negative signs to the numerator, so it's going to be negative 4 times negative 2 in the numerator. And then in the denominator, it's going to be 1 times 1. Let me write that down. 1 times 1. And so this gives us, so we have a negative 4 times a negative 2. So it's a negative times a negative. So we're going to get a positive value here."}, {"video_title": "Dividing negative fractions   Fractions   Pre-Algebra   Khan Academy.mp3", "Sentence": "1 times 1. And so this gives us, so we have a negative 4 times a negative 2. So it's a negative times a negative. So we're going to get a positive value here. And 4 times 2 is 8. So this is a positive 8 over 1. And 8 divided by 1 is just equal to 8."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "He says, my fault, my apologies. I realize now, I realize now what the mistake was. There was a slight, I guess, typing error or writing error. The first week when they went to the market and bought two pounds of apples and one pounds of bananas, it wasn't a $3 cost. It was a $5 cost. It was a $5 cost. Now surely, considering how smart you and this bird seem to be, you surely could figure out what is the per pound cost of apples and what is the per pound cost of bananas."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "The first week when they went to the market and bought two pounds of apples and one pounds of bananas, it wasn't a $3 cost. It was a $5 cost. It was a $5 cost. Now surely, considering how smart you and this bird seem to be, you surely could figure out what is the per pound cost of apples and what is the per pound cost of bananas. So you think for a little bit. Is there now going to be a solution? So let's break it down using the exact same variables."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "Now surely, considering how smart you and this bird seem to be, you surely could figure out what is the per pound cost of apples and what is the per pound cost of bananas. So you think for a little bit. Is there now going to be a solution? So let's break it down using the exact same variables. You say, well, if a is the cost of apples per pound and b is the cost of bananas, this first constraint tells us that two pounds of apples, so two pounds of apples are going to cost 2a, because it's $a per pound. And one pound of bananas is going to cost b dollars, because it's one pound times b dollars per pound, is now going to cost $5. This is the corrected."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "So let's break it down using the exact same variables. You say, well, if a is the cost of apples per pound and b is the cost of bananas, this first constraint tells us that two pounds of apples, so two pounds of apples are going to cost 2a, because it's $a per pound. And one pound of bananas is going to cost b dollars, because it's one pound times b dollars per pound, is now going to cost $5. This is the corrected. This is the corrected number. And we saw from the last scenario, this information hasn't changed. Six pounds of apples is going to cost 6a, six pounds times $a per pound."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "This is the corrected. This is the corrected number. And we saw from the last scenario, this information hasn't changed. Six pounds of apples is going to cost 6a, six pounds times $a per pound. And three pounds of bananas is going to cost 3b, three pounds times b dollars per pound. The total cost of the apples and bananas in this trip we are given is $15. So once again, you say, well, let me try to solve this maybe through elimination."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "Six pounds of apples is going to cost 6a, six pounds times $a per pound. And three pounds of bananas is going to cost 3b, three pounds times b dollars per pound. The total cost of the apples and bananas in this trip we are given is $15. So once again, you say, well, let me try to solve this maybe through elimination. And once again, you say, well, let me cancel out the a's. I have 2a here. I have 6a here."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "So once again, you say, well, let me try to solve this maybe through elimination. And once again, you say, well, let me cancel out the a's. I have 2a here. I have 6a here. If I multiply the 2a here by negative 3, then this will become a negative 6a, and it might be able to cancel out with all of this business. So you do that. You multiply this entire equation."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "I have 6a here. If I multiply the 2a here by negative 3, then this will become a negative 6a, and it might be able to cancel out with all of this business. So you do that. You multiply this entire equation. You can't just multiply one term. You have to multiply the entire equation times negative 3 if you want the equation to still hold. And so we're multiplying by negative 3."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "You multiply this entire equation. You can't just multiply one term. You have to multiply the entire equation times negative 3 if you want the equation to still hold. And so we're multiplying by negative 3. So 2a times negative 3 is negative 6a. b times negative 3 is negative 3b. And then 5 times negative 3 is negative 15."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "And so we're multiplying by negative 3. So 2a times negative 3 is negative 6a. b times negative 3 is negative 3b. And then 5 times negative 3 is negative 15. And now something fishy starts to look like it's about to happen. Because when you add the left-hand side of this blue equation or this purplish equation to the green one, you get 0. All of these things right over here just cancel out."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "And then 5 times negative 3 is negative 15. And now something fishy starts to look like it's about to happen. Because when you add the left-hand side of this blue equation or this purplish equation to the green one, you get 0. All of these things right over here just cancel out. And on the right-hand side, 15 minus 15, that is also equal to 0. And you get 0 equals 0, which seems a little bit better than the last time you worked through it. Last time we got 0 equals 6."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "All of these things right over here just cancel out. And on the right-hand side, 15 minus 15, that is also equal to 0. And you get 0 equals 0, which seems a little bit better than the last time you worked through it. Last time we got 0 equals 6. But 0 equals 0 doesn't really tell you anything about the x's and y's. This is true. This is absolutely true that 0 does definitely equal 0."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "Last time we got 0 equals 6. But 0 equals 0 doesn't really tell you anything about the x's and y's. This is true. This is absolutely true that 0 does definitely equal 0. But it doesn't tell you any information about x and y. And so then the bird whispers in the king's ear. And then the king says, well, the bird says you should graph it to figure out what's actually going on."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "This is absolutely true that 0 does definitely equal 0. But it doesn't tell you any information about x and y. And so then the bird whispers in the king's ear. And then the king says, well, the bird says you should graph it to figure out what's actually going on. And so you've learned that listening to the bird actually makes a lot of sense. So you try to graph these two constraints. So let's do it the same way."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "And then the king says, well, the bird says you should graph it to figure out what's actually going on. And so you've learned that listening to the bird actually makes a lot of sense. So you try to graph these two constraints. So let's do it the same way. We'll have a b-axis. That's our b-axis. And we will have our a-axis."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "So let's do it the same way. We'll have a b-axis. That's our b-axis. And we will have our a-axis. Let me mark off some markers here. 1, 2, 3, 4, 5, and 1, 2, 3, 4, 5. So this first equation right over here, if we subtract 2a from both sides, I'm just going to put it into slope-intercept form, you get b is equal to negative 2a plus 5."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "And we will have our a-axis. Let me mark off some markers here. 1, 2, 3, 4, 5, and 1, 2, 3, 4, 5. So this first equation right over here, if we subtract 2a from both sides, I'm just going to put it into slope-intercept form, you get b is equal to negative 2a plus 5. All I did is subtract 2a from both sides. And if we were to graph that, our b-intercept, when a is equal to 0, b is equal to 5. So that's right over here."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "So this first equation right over here, if we subtract 2a from both sides, I'm just going to put it into slope-intercept form, you get b is equal to negative 2a plus 5. All I did is subtract 2a from both sides. And if we were to graph that, our b-intercept, when a is equal to 0, b is equal to 5. So that's right over here. And our slope is negative 2. Every time you add 1 to a, so if a goes from 0 to 1, b is going to go down by 2. So go down by 2, go down by 2."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "So that's right over here. And our slope is negative 2. Every time you add 1 to a, so if a goes from 0 to 1, b is going to go down by 2. So go down by 2, go down by 2. So this first white equation looks like this if we graph the solution set. These are all of the prices for bananas and apples that meet this constraint. Now let's graph the second equation."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "So go down by 2, go down by 2. So this first white equation looks like this if we graph the solution set. These are all of the prices for bananas and apples that meet this constraint. Now let's graph the second equation. If we subtract 6a from both sides, we get 3b is equal to negative 6a plus 15. And now we can divide both sides by 3. Divide everything by 3."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "Now let's graph the second equation. If we subtract 6a from both sides, we get 3b is equal to negative 6a plus 15. And now we can divide both sides by 3. Divide everything by 3. We are left with b is equal to negative 2a plus 5. Well, this is interesting. This looks very similar, or it looks exactly the same."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "Divide everything by 3. We are left with b is equal to negative 2a plus 5. Well, this is interesting. This looks very similar, or it looks exactly the same. Our b-intercept is 5, and our slope is negative 2a. So this is essentially the same line. So these are essentially the same constraints."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "This looks very similar, or it looks exactly the same. Our b-intercept is 5, and our slope is negative 2a. So this is essentially the same line. So these are essentially the same constraints. And so you start to look at it a little bit confused. And you say, OK, I see why we got 0 equals 0. There's actually an infinite number of solutions."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "So these are essentially the same constraints. And so you start to look at it a little bit confused. And you say, OK, I see why we got 0 equals 0. There's actually an infinite number of solutions. You pick any x, and then the corresponding y for each of these could be a solution for either of these things. So there's an infinite number of solutions. But you start to wonder, why is this happening?"}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "There's actually an infinite number of solutions. You pick any x, and then the corresponding y for each of these could be a solution for either of these things. So there's an infinite number of solutions. But you start to wonder, why is this happening? And so the bird whispers again into the king's ear. And the king says, well, the bird says this is because in both trips to the market, the same ratio of apples and bananas was bought. In the green trip versus the white trip, bought three times as many apples, bought three times as many bananas, and you had three times the cost."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "But you start to wonder, why is this happening? And so the bird whispers again into the king's ear. And the king says, well, the bird says this is because in both trips to the market, the same ratio of apples and bananas was bought. In the green trip versus the white trip, bought three times as many apples, bought three times as many bananas, and you had three times the cost. So in any situation, for any prices, per pound prices of apples and bananas, if you buy exactly three times the number of apples, three times the number of bananas, and have three times the cost, that could be true for any prices. And so this is actually consistent. We can't say that our begla is lying to us."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "In the green trip versus the white trip, bought three times as many apples, bought three times as many bananas, and you had three times the cost. So in any situation, for any prices, per pound prices of apples and bananas, if you buy exactly three times the number of apples, three times the number of bananas, and have three times the cost, that could be true for any prices. And so this is actually consistent. We can't say that our begla is lying to us. But it's not giving us enough information. This is what we call a consistent system. It's consistent information here."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "We can't say that our begla is lying to us. But it's not giving us enough information. This is what we call a consistent system. It's consistent information here. So let me write this down. This is consistent. It is consistent."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "It's consistent information here. So let me write this down. This is consistent. It is consistent. 0 equals 0. There's no shadiness going on here. But it's not enough information."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "It is consistent. 0 equals 0. There's no shadiness going on here. But it's not enough information. This system of equations is dependent. It is dependent. And you have an infinite number of solutions."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "But it's not enough information. This system of equations is dependent. It is dependent. And you have an infinite number of solutions. Any point on this line represents a solution. So you tell our begla, well, if you really want us to figure this out, you need to give us more information. And preferably, we buy a different ratio of apples to bananas."}, {"video_title": "Simplifying square roots   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "So let's actually take its prime factorization and see if any of those prime factors show up more than once. So it's clearly an odd number. It's clearly not divisible by 2. To test whether it's divisible by 3, we can add up all of the digits. And we explained why this works in another place on Khan Academy. But if you add up all the digits, you get a 9. And 9 is divisible by 3."}, {"video_title": "Simplifying square roots   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "To test whether it's divisible by 3, we can add up all of the digits. And we explained why this works in another place on Khan Academy. But if you add up all the digits, you get a 9. And 9 is divisible by 3. So 117 is going to be divisible by 3. Now let's do a little aside here and figure out what 117 divided by 3 actually is. So 3 doesn't go into 1."}, {"video_title": "Simplifying square roots   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "And 9 is divisible by 3. So 117 is going to be divisible by 3. Now let's do a little aside here and figure out what 117 divided by 3 actually is. So 3 doesn't go into 1. It does go into 11 3 times. 3 times 3 is 9. Subtract."}, {"video_title": "Simplifying square roots   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "So 3 doesn't go into 1. It does go into 11 3 times. 3 times 3 is 9. Subtract. You got a remainder of 2. Bring down a 7. 3 goes into 27 9 times."}, {"video_title": "Simplifying square roots   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "Subtract. You got a remainder of 2. Bring down a 7. 3 goes into 27 9 times. 9 times 3 is 27. Subtract, and you're done. It goes in perfectly."}, {"video_title": "Simplifying square roots   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "3 goes into 27 9 times. 9 times 3 is 27. Subtract, and you're done. It goes in perfectly. And so we can factor 117 as 3 times 39. Now 39 we can factor as. That jumps out more at us."}, {"video_title": "Simplifying square roots   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "It goes in perfectly. And so we can factor 117 as 3 times 39. Now 39 we can factor as. That jumps out more at us. That's divisible by 3. That's equivalent to 3 times 13. And then all of these are now prime numbers."}, {"video_title": "Simplifying square roots   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "That jumps out more at us. That's divisible by 3. That's equivalent to 3 times 13. And then all of these are now prime numbers. So we could say that this thing is the same as 5 times the square root of 3 times 3 times 13. And this is going to be the same thing as, and we know this from our exponent properties, as 5 times the square root of 3 times 3 times the square root of 13. Now what's the square root of 3 times 3?"}, {"video_title": "Simplifying square roots   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "And then all of these are now prime numbers. So we could say that this thing is the same as 5 times the square root of 3 times 3 times 13. And this is going to be the same thing as, and we know this from our exponent properties, as 5 times the square root of 3 times 3 times the square root of 13. Now what's the square root of 3 times 3? Well, that's the square root of 9. That's the square root of 3 squared. Any of those, well, that's just going to give you 3."}, {"video_title": "Simplifying square roots   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "Now what's the square root of 3 times 3? Well, that's the square root of 9. That's the square root of 3 squared. Any of those, well, that's just going to give you 3. So this is just going to simplify to 3. So this whole thing is 5 times 3 times the square root of 13. So this part right over here would give us 15 times the square root of 13."}, {"video_title": "Simplifying square roots   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "Any of those, well, that's just going to give you 3. So this is just going to simplify to 3. So this whole thing is 5 times 3 times the square root of 13. So this part right over here would give us 15 times the square root of 13. Let's do one more example here. So let's try to simplify 3 times the square root of 26. Actually, I'm going to put 26 in yellow like I did in the previous problem."}, {"video_title": "Simplifying square roots   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "So this part right over here would give us 15 times the square root of 13. Let's do one more example here. So let's try to simplify 3 times the square root of 26. Actually, I'm going to put 26 in yellow like I did in the previous problem. 26. Well, 26 is clearly an even number. So it's going to be divisible by 2."}, {"video_title": "Simplifying square roots   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "Actually, I'm going to put 26 in yellow like I did in the previous problem. 26. Well, 26 is clearly an even number. So it's going to be divisible by 2. We can rewrite it as 2 times 13. And then we're done. 13 is a prime number."}, {"video_title": "Simplifying square roots   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "So it's going to be divisible by 2. We can rewrite it as 2 times 13. And then we're done. 13 is a prime number. We can't factor this anymore. And so 26 doesn't have any perfect squares in it. It's not like we can factor it out as a factor of some other numbers and some perfect squares like we did here."}, {"video_title": "Simplifying square roots   Exponents, radicals, and scientific notation   Pre-Algebra   Khan Academy.mp3", "Sentence": "13 is a prime number. We can't factor this anymore. And so 26 doesn't have any perfect squares in it. It's not like we can factor it out as a factor of some other numbers and some perfect squares like we did here. 117 is 13 times 9. It's the product of a perfect square and 13. 26 isn't."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "So this is an example from the Khan Academy exercise, graphing solutions to two variable linear equations. And they tell us to complete the table so each row represents a solution of the following equation. And they give us the equation, and then they want us to figure out what does y equal when x is equal to negative five, and what does x equal when y is equal to eight? And to figure this out, I've actually copied and pasted this part of the problem onto my scratch pad, so let me get that out. And so this is the exact same problem. And there's a couple of ways that we could try to tackle it. One way is you could try to simplify this more, get all your x's on one side and all your y's on the other side."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "And to figure this out, I've actually copied and pasted this part of the problem onto my scratch pad, so let me get that out. And so this is the exact same problem. And there's a couple of ways that we could try to tackle it. One way is you could try to simplify this more, get all your x's on one side and all your y's on the other side. Or we could just literally substitute when x equals negative five, what must y equal? Actually, let me do it the second way first. So if we take this equation and we substitute x with negative five, what do we get?"}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "One way is you could try to simplify this more, get all your x's on one side and all your y's on the other side. Or we could just literally substitute when x equals negative five, what must y equal? Actually, let me do it the second way first. So if we take this equation and we substitute x with negative five, what do we get? We get negative three times, well, we're gonna say x is negative five, times negative five plus seven y is equal to five times, x is once again, it's gonna be negative five, x is negative five, five times negative five plus two y. See, negative three times negative five is positive 15, plus seven y is equal to negative 25 plus two y. And now to solve for y, let's see, I could subtract two y from both sides so that I get rid of the two y here on the right."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "So if we take this equation and we substitute x with negative five, what do we get? We get negative three times, well, we're gonna say x is negative five, times negative five plus seven y is equal to five times, x is once again, it's gonna be negative five, x is negative five, five times negative five plus two y. See, negative three times negative five is positive 15, plus seven y is equal to negative 25 plus two y. And now to solve for y, let's see, I could subtract two y from both sides so that I get rid of the two y here on the right. So let me subtract two y, subtract two y from both sides. And then if I want all my constants on the right-hand side, I can subtract 15 from both sides. So let me subtract 15 from both sides."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "And now to solve for y, let's see, I could subtract two y from both sides so that I get rid of the two y here on the right. So let me subtract two y, subtract two y from both sides. And then if I want all my constants on the right-hand side, I can subtract 15 from both sides. So let me subtract 15 from both sides. And I am going to be left with 15 minus 15, that's zero. That's the whole point of subtracting 15 from both sides, so I get rid of this 15 here. Seven y minus two y, seven of something minus two of that same something is gonna be five of that something."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "So let me subtract 15 from both sides. And I am going to be left with 15 minus 15, that's zero. That's the whole point of subtracting 15 from both sides, so I get rid of this 15 here. Seven y minus two y, seven of something minus two of that same something is gonna be five of that something. It's gonna be equal to five y, is equal to negative 25 minus 15. Well, that's gonna be negative 40. And then two y minus two y, well, that's just gonna be zero."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "Seven y minus two y, seven of something minus two of that same something is gonna be five of that something. It's gonna be equal to five y, is equal to negative 25 minus 15. Well, that's gonna be negative 40. And then two y minus two y, well, that's just gonna be zero. That was the whole point of subtracting two y from both sides. So you have five times y is equal to negative 40, or if we divide both sides by five, we divide both sides by five, we would get y is equal to negative eight. So when x is equal to negative five, y is equal to negative eight."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "And then two y minus two y, well, that's just gonna be zero. That was the whole point of subtracting two y from both sides. So you have five times y is equal to negative 40, or if we divide both sides by five, we divide both sides by five, we would get y is equal to negative eight. So when x is equal to negative five, y is equal to negative eight. Y is equal to negative eight. And actually, we can fill that in. So this y is going to be equal to negative eight."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "So when x is equal to negative five, y is equal to negative eight. Y is equal to negative eight. And actually, we can fill that in. So this y is going to be equal to negative eight. And now we gotta figure this out. What does x equal when y is positive eight? Well, we can go back to our scratch pad here."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "So this y is going to be equal to negative eight. And now we gotta figure this out. What does x equal when y is positive eight? Well, we can go back to our scratch pad here. And now let's take the same equation, but let's make y equal to positive eight. So you have negative three x plus seven. Now y is going to be eight."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "Well, we can go back to our scratch pad here. And now let's take the same equation, but let's make y equal to positive eight. So you have negative three x plus seven. Now y is going to be eight. Y is eight. Seven times eight is equal to five times x plus two times, once again, y is eight. Two times eight."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "Now y is going to be eight. Y is eight. Seven times eight is equal to five times x plus two times, once again, y is eight. Two times eight. So we get negative three x plus 56, that's 56, is equal to five x plus 16. Now if we want to get all of our constants on one side and of all of our x terms on the other side, well, what could we do? Let's see, we could add three x to both sides."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "Two times eight. So we get negative three x plus 56, that's 56, is equal to five x plus 16. Now if we want to get all of our constants on one side and of all of our x terms on the other side, well, what could we do? Let's see, we could add three x to both sides. That would get rid of all of the x's on this side and put them all on this side. So we're gonna add three x to both sides. And let's see, if we want to get all the constants on the left-hand side, we'd want to get rid of the 16, so we could subtract 16 from the right-hand side."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "Let's see, we could add three x to both sides. That would get rid of all of the x's on this side and put them all on this side. So we're gonna add three x to both sides. And let's see, if we want to get all the constants on the left-hand side, we'd want to get rid of the 16, so we could subtract 16 from the right-hand side. If we do it from the right, we're gonna have to do it from the left as well. And we're going to be left with, these cancel out, 56 minus 16 is positive 40. And then, let's see, 16 minus 16 is zero."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "And let's see, if we want to get all the constants on the left-hand side, we'd want to get rid of the 16, so we could subtract 16 from the right-hand side. If we do it from the right, we're gonna have to do it from the left as well. And we're going to be left with, these cancel out, 56 minus 16 is positive 40. And then, let's see, 16 minus 16 is zero. Five x plus three x is equal to eight x. We get eight x is equal to 40. We could divide both sides by eight."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "And then, let's see, 16 minus 16 is zero. Five x plus three x is equal to eight x. We get eight x is equal to 40. We could divide both sides by eight. And we get five is equal to x. So this right over here is going to be equal to five. So let's go back."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "We could divide both sides by eight. And we get five is equal to x. So this right over here is going to be equal to five. So let's go back. Let's go back now. So when y is positive eight, x is positive five. Now they ask us, use your two solutions to graph the equation."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "So let's go back. Let's go back now. So when y is positive eight, x is positive five. Now they ask us, use your two solutions to graph the equation. So let's see if we can do, whoops, let me use my mouse now. So to graph the equation. So when x is negative five, y is negative eight."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "Now they ask us, use your two solutions to graph the equation. So let's see if we can do, whoops, let me use my mouse now. So to graph the equation. So when x is negative five, y is negative eight. So the point negative five comma negative eight. So that's right over there. Actually, let me move my browser up so you can see that."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "So when x is negative five, y is negative eight. So the point negative five comma negative eight. So that's right over there. Actually, let me move my browser up so you can see that. Negative five, when x is negative five, y is negative eight. And when x is positive five, we see that up here, when x is positive five, y is positive eight. When x is positive five, y is positive eight."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "Actually, let me move my browser up so you can see that. Negative five, when x is negative five, y is negative eight. And when x is positive five, we see that up here, when x is positive five, y is positive eight. When x is positive five, y is positive eight. And we're done. We can check our answer if we like. We got it right."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "When x is positive five, y is positive eight. And we're done. We can check our answer if we like. We got it right. Now I said there was two ways to tackle it. I kind of just did it, I guess you could say the naive way. I just substituted negative five directly into this and solved for y."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "We got it right. Now I said there was two ways to tackle it. I kind of just did it, I guess you could say the naive way. I just substituted negative five directly into this and solved for y. And then I substituted y equals positive eight directly into this and then solved for x. Another way that I could have done it that actually probably would have been, or for sure would have been the easier way to do it, is ahead of time to try to simplify this expression. So what I could have done right from the get-go is said, hey, let's put all my x's on one side and all my y's on the other side."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "I just substituted negative five directly into this and solved for y. And then I substituted y equals positive eight directly into this and then solved for x. Another way that I could have done it that actually probably would have been, or for sure would have been the easier way to do it, is ahead of time to try to simplify this expression. So what I could have done right from the get-go is said, hey, let's put all my x's on one side and all my y's on the other side. So this is negative three x plus seven y is equal to five x plus two y. And let's say I want to get all my y's on the left and all my x's on the right. So I don't want this negative three x on the left, so I'd want to add three x."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "So what I could have done right from the get-go is said, hey, let's put all my x's on one side and all my y's on the other side. So this is negative three x plus seven y is equal to five x plus two y. And let's say I want to get all my y's on the left and all my x's on the right. So I don't want this negative three x on the left, so I'd want to add three x. Adding three x would cancel this out, but I can't just do it on the left-hand side, I'd have to do it on the right-hand side as well. And then if I want to get rid of this two y on the right, I could subtract two y from the right, but of course I'd also want to do it from the left. And then what am I left with?"}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "So I don't want this negative three x on the left, so I'd want to add three x. Adding three x would cancel this out, but I can't just do it on the left-hand side, I'd have to do it on the right-hand side as well. And then if I want to get rid of this two y on the right, I could subtract two y from the right, but of course I'd also want to do it from the left. And then what am I left with? So negative three x plus three x is zero, seven y minus two y is five y. And then I have five x plus three x is eight x. Two y minus two y is zero."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "And then what am I left with? So negative three x plus three x is zero, seven y minus two y is five y. And then I have five x plus three x is eight x. Two y minus two y is zero. And then if I wanted to, I could solve for y, I could divide both sides by five and I'd get y is equal to 8 5ths x. So this right over here represents the same exact equation as this over here, it's just written in a different way. All of the xy pairs that satisfy this would satisfy this and vice versa."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "Two y minus two y is zero. And then if I wanted to, I could solve for y, I could divide both sides by five and I'd get y is equal to 8 5ths x. So this right over here represents the same exact equation as this over here, it's just written in a different way. All of the xy pairs that satisfy this would satisfy this and vice versa. And this is much easier, because if x is now negative five, if x is negative five, y would be 8 5ths times negative five. Well that's going to be negative eight. And when y is equal to eight, well you actually could even do this up here."}, {"video_title": "Graphing solutions to two-variable linear equations example 2   Algebra I   Khan Academy.mp3", "Sentence": "All of the xy pairs that satisfy this would satisfy this and vice versa. And this is much easier, because if x is now negative five, if x is negative five, y would be 8 5ths times negative five. Well that's going to be negative eight. And when y is equal to eight, well you actually could even do this up here. You could say five times eight is equal to eight x. And then you could see, well five times eight is the same thing as eight times five. So x would be equal to five."}, {"video_title": "Patterns in raising 1 and -1 to different powers   Pre-Algebra   Khan Academy.mp3", "Sentence": "Let's think about exponents with ones and zeros. So let's take the number one, and let's raise it to the eighth power. So we've already seen that there's two ways of thinking about this. You could literally view this as taking eight ones and then multiplying them together. So let's do that. So you have one, two, three, four, five, six, seven, eight ones. And then you're going to multiply them together."}, {"video_title": "Patterns in raising 1 and -1 to different powers   Pre-Algebra   Khan Academy.mp3", "Sentence": "You could literally view this as taking eight ones and then multiplying them together. So let's do that. So you have one, two, three, four, five, six, seven, eight ones. And then you're going to multiply them together. And if you were to do that, you would get, well, 1 times 1 is 1 times 1. It doesn't matter how many times you multiply 1 by 1. You are going to just get 1."}, {"video_title": "Patterns in raising 1 and -1 to different powers   Pre-Algebra   Khan Academy.mp3", "Sentence": "And then you're going to multiply them together. And if you were to do that, you would get, well, 1 times 1 is 1 times 1. It doesn't matter how many times you multiply 1 by 1. You are going to just get 1. You are just going to get 1. And you could imagine, I did it eight times. I multiplied eight ones."}, {"video_title": "Patterns in raising 1 and -1 to different powers   Pre-Algebra   Khan Academy.mp3", "Sentence": "You are going to just get 1. You are just going to get 1. And you could imagine, I did it eight times. I multiplied eight ones. But even if this was 80, or if this was 800, or if this was 8 million, if I just multiplied 1, if I had 8 million ones and I multiplied them all together, it would still be equal to 1. So 1 to any power is just going to be equal to 1. You would say, hey, what about 1 to the 0th power?"}, {"video_title": "Patterns in raising 1 and -1 to different powers   Pre-Algebra   Khan Academy.mp3", "Sentence": "I multiplied eight ones. But even if this was 80, or if this was 800, or if this was 8 million, if I just multiplied 1, if I had 8 million ones and I multiplied them all together, it would still be equal to 1. So 1 to any power is just going to be equal to 1. You would say, hey, what about 1 to the 0th power? 1 to the 0th power. Well, we've already said anything to the 0th power, except for 0. It's actually up for debate."}, {"video_title": "Patterns in raising 1 and -1 to different powers   Pre-Algebra   Khan Academy.mp3", "Sentence": "You would say, hey, what about 1 to the 0th power? 1 to the 0th power. Well, we've already said anything to the 0th power, except for 0. It's actually up for debate. But anything to the 0th power is going to be equal to 1. And just as a little bit of intuition here, you could literally view this as our other definition of exponentiation, which is you start with a 1. And this number says how many times you're going to multiply that 1 times this number."}, {"video_title": "Patterns in raising 1 and -1 to different powers   Pre-Algebra   Khan Academy.mp3", "Sentence": "It's actually up for debate. But anything to the 0th power is going to be equal to 1. And just as a little bit of intuition here, you could literally view this as our other definition of exponentiation, which is you start with a 1. And this number says how many times you're going to multiply that 1 times this number. So 1 times 1, 0 times, is just going to be 1. And that was a little bit clearer when we did it like this, where we said 2 to the, let's say, 4th power is equal to, this was the other definition of exponentiation we had, which is you start with a 1. And then you multiply it by 2 four times."}, {"video_title": "Patterns in raising 1 and -1 to different powers   Pre-Algebra   Khan Academy.mp3", "Sentence": "And this number says how many times you're going to multiply that 1 times this number. So 1 times 1, 0 times, is just going to be 1. And that was a little bit clearer when we did it like this, where we said 2 to the, let's say, 4th power is equal to, this was the other definition of exponentiation we had, which is you start with a 1. And then you multiply it by 2 four times. So times 2, times 2, times 2, times 2, which is equal to, let's see, this is equal to 16. So here, if you start with a 1, and then you multiply it by 1, 0 times, you're still going to have that 1 right over there. That's why anything that's not 0 to the 1 power is going to be equal to 1."}, {"video_title": "Patterns in raising 1 and -1 to different powers   Pre-Algebra   Khan Academy.mp3", "Sentence": "And then you multiply it by 2 four times. So times 2, times 2, times 2, times 2, which is equal to, let's see, this is equal to 16. So here, if you start with a 1, and then you multiply it by 1, 0 times, you're still going to have that 1 right over there. That's why anything that's not 0 to the 1 power is going to be equal to 1. Now let's try some other interesting scenarios. Let's try some negative numbers. So let's take negative 1."}, {"video_title": "Patterns in raising 1 and -1 to different powers   Pre-Algebra   Khan Academy.mp3", "Sentence": "That's why anything that's not 0 to the 1 power is going to be equal to 1. Now let's try some other interesting scenarios. Let's try some negative numbers. So let's take negative 1. And let's first raise it to the 0 power. So once again, this is just going based on this definition. This is starting with a 1 and then multiplying it by this number 0 times."}, {"video_title": "Patterns in raising 1 and -1 to different powers   Pre-Algebra   Khan Academy.mp3", "Sentence": "So let's take negative 1. And let's first raise it to the 0 power. So once again, this is just going based on this definition. This is starting with a 1 and then multiplying it by this number 0 times. Well, that means we're just not going to multiply it by this number. So you're just going to get a 1. Let's try negative 1 to the 1st power."}, {"video_title": "Patterns in raising 1 and -1 to different powers   Pre-Algebra   Khan Academy.mp3", "Sentence": "This is starting with a 1 and then multiplying it by this number 0 times. Well, that means we're just not going to multiply it by this number. So you're just going to get a 1. Let's try negative 1 to the 1st power. Well, anything to the 1st power, you could view this. And I like going with this definition as opposed to this one right over here. If we were to make them consistent, if you were to make this definition consistent with this, you would say, hey, let's start with a 1."}, {"video_title": "Patterns in raising 1 and -1 to different powers   Pre-Algebra   Khan Academy.mp3", "Sentence": "Let's try negative 1 to the 1st power. Well, anything to the 1st power, you could view this. And I like going with this definition as opposed to this one right over here. If we were to make them consistent, if you were to make this definition consistent with this, you would say, hey, let's start with a 1. And then multiply it by 1 8 times. And you're still going to get a 1 right over here. But let's do this with negative 1."}, {"video_title": "Patterns in raising 1 and -1 to different powers   Pre-Algebra   Khan Academy.mp3", "Sentence": "If we were to make them consistent, if you were to make this definition consistent with this, you would say, hey, let's start with a 1. And then multiply it by 1 8 times. And you're still going to get a 1 right over here. But let's do this with negative 1. So we're going to start with a 1. And then we're going to multiply it by negative 1 1 times. Times negative 1."}, {"video_title": "Patterns in raising 1 and -1 to different powers   Pre-Algebra   Khan Academy.mp3", "Sentence": "But let's do this with negative 1. So we're going to start with a 1. And then we're going to multiply it by negative 1 1 times. Times negative 1. And this is, of course, going to be equal to negative 1. Now let's take negative 1. And let's take it to the 2nd power."}, {"video_title": "Patterns in raising 1 and -1 to different powers   Pre-Algebra   Khan Academy.mp3", "Sentence": "Times negative 1. And this is, of course, going to be equal to negative 1. Now let's take negative 1. And let's take it to the 2nd power. We often say that we are squaring it when we take something to the 2nd power. So negative 1 to the 2nd power. Well, we could start with a 1."}, {"video_title": "Patterns in raising 1 and -1 to different powers   Pre-Algebra   Khan Academy.mp3", "Sentence": "And let's take it to the 2nd power. We often say that we are squaring it when we take something to the 2nd power. So negative 1 to the 2nd power. Well, we could start with a 1. And then multiply it by negative 1 2 times. Multiply it by negative 1 twice. And what's this going to be equal to?"}, {"video_title": "Patterns in raising 1 and -1 to different powers   Pre-Algebra   Khan Academy.mp3", "Sentence": "Well, we could start with a 1. And then multiply it by negative 1 2 times. Multiply it by negative 1 twice. And what's this going to be equal to? And once again, by our old definition, you could also just say, hey, ignoring this 1, because that's not going to change the value. We took 2 negative 1's and we're multiplying them. Well, negative 1 times negative 1 is 1."}, {"video_title": "Patterns in raising 1 and -1 to different powers   Pre-Algebra   Khan Academy.mp3", "Sentence": "And what's this going to be equal to? And once again, by our old definition, you could also just say, hey, ignoring this 1, because that's not going to change the value. We took 2 negative 1's and we're multiplying them. Well, negative 1 times negative 1 is 1. And I think you see a pattern forming. Let's take negative 1 to the 3rd power. What's this going to be equal to?"}, {"video_title": "Patterns in raising 1 and -1 to different powers   Pre-Algebra   Khan Academy.mp3", "Sentence": "Well, negative 1 times negative 1 is 1. And I think you see a pattern forming. Let's take negative 1 to the 3rd power. What's this going to be equal to? Well, by this definition, you start with a 1 and then you multiply it by negative 1 3 times. So negative 1 times negative 1 times negative 1. Or you could just think of it as you're taking 3 negative 1's and you're multiplying it, because this 1 doesn't change the value."}, {"video_title": "Patterns in raising 1 and -1 to different powers   Pre-Algebra   Khan Academy.mp3", "Sentence": "What's this going to be equal to? Well, by this definition, you start with a 1 and then you multiply it by negative 1 3 times. So negative 1 times negative 1 times negative 1. Or you could just think of it as you're taking 3 negative 1's and you're multiplying it, because this 1 doesn't change the value. And this is going to be equal to negative 1 times negative 1 is positive 1 times negative 1 is negative 1. So you see the pattern. Negative 1 to the 0th power is 1, negative 1 to the 1st power is negative 1."}, {"video_title": "Patterns in raising 1 and -1 to different powers   Pre-Algebra   Khan Academy.mp3", "Sentence": "Or you could just think of it as you're taking 3 negative 1's and you're multiplying it, because this 1 doesn't change the value. And this is going to be equal to negative 1 times negative 1 is positive 1 times negative 1 is negative 1. So you see the pattern. Negative 1 to the 0th power is 1, negative 1 to the 1st power is negative 1. Then you multiply it by negative 1 again to get positive 1. Then you multiply it by negative 1 again to get negative 1. And the pattern you might be seeing is is if you take negative 1 to an odd power, you're going to get negative 1."}, {"video_title": "Patterns in raising 1 and -1 to different powers   Pre-Algebra   Khan Academy.mp3", "Sentence": "Negative 1 to the 0th power is 1, negative 1 to the 1st power is negative 1. Then you multiply it by negative 1 again to get positive 1. Then you multiply it by negative 1 again to get negative 1. And the pattern you might be seeing is is if you take negative 1 to an odd power, you're going to get negative 1. And if you take it to an even power, you're going to get 1. because a negative times a negative is going to be the positive. And you're going to have an even number of negatives, so you're always going to have negative times negatives. So this right over here, this is even."}, {"video_title": "Patterns in raising 1 and -1 to different powers   Pre-Algebra   Khan Academy.mp3", "Sentence": "And the pattern you might be seeing is is if you take negative 1 to an odd power, you're going to get negative 1. And if you take it to an even power, you're going to get 1. because a negative times a negative is going to be the positive. And you're going to have an even number of negatives, so you're always going to have negative times negatives. So this right over here, this is even. Even is going to be positive 1. And you could see that if you went to negative 1 to the fourth power. Negative 1 to the fourth power."}, {"video_title": "Patterns in raising 1 and -1 to different powers   Pre-Algebra   Khan Academy.mp3", "Sentence": "So this right over here, this is even. Even is going to be positive 1. And you could see that if you went to negative 1 to the fourth power. Negative 1 to the fourth power. Well, you could start with a 1 and then multiply it by negative 1 four times. So negative 1 times negative 1 times negative 1 times negative 1, which is just going to be equal to positive 1. So if someone were to ask you, we already established that if someone were to take 1 to the 1 millionth power, this is just going to be equal to 1."}, {"video_title": "Patterns in raising 1 and -1 to different powers   Pre-Algebra   Khan Academy.mp3", "Sentence": "Negative 1 to the fourth power. Well, you could start with a 1 and then multiply it by negative 1 four times. So negative 1 times negative 1 times negative 1 times negative 1, which is just going to be equal to positive 1. So if someone were to ask you, we already established that if someone were to take 1 to the 1 millionth power, this is just going to be equal to 1. If someone told you, let's take negative 1 and raise it to the 1 millionth power, well, 1 million is an even number. So this is still going to be equal to positive 1. But if you took negative 1 to the 999,999th power, this is an odd number."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "And so he storms out of the room, and then a few seconds later, he storms back in. He says, my fault, my apologies. I realize now, I realize now what the mistake was. There was a slight, I guess, typing error or writing error. The first week when they went to the market and bought two pounds of apples and one pounds of bananas, it wasn't a $3 cost. It was a $5 cost. It was a $5 cost."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "There was a slight, I guess, typing error or writing error. The first week when they went to the market and bought two pounds of apples and one pounds of bananas, it wasn't a $3 cost. It was a $5 cost. It was a $5 cost. Now surely, considering how smart you and this bird seem to be, you surely could figure out what is the per pound cost of apples and what is the per pound cost of bananas. So you think for a little bit. Is there now going to be a solution?"}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "It was a $5 cost. Now surely, considering how smart you and this bird seem to be, you surely could figure out what is the per pound cost of apples and what is the per pound cost of bananas. So you think for a little bit. Is there now going to be a solution? So let's break it down using the exact same variables. You say, well, if a is the cost of apples per pound and b is the cost of bananas, this first constraint tells us that two pounds of apples, so two pounds of apples are going to cost 2a, because it's $a per pound. And one pound of bananas is going to cost b dollars, because it's one pound times b dollars per pound, is now going to cost $5."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "Is there now going to be a solution? So let's break it down using the exact same variables. You say, well, if a is the cost of apples per pound and b is the cost of bananas, this first constraint tells us that two pounds of apples, so two pounds of apples are going to cost 2a, because it's $a per pound. And one pound of bananas is going to cost b dollars, because it's one pound times b dollars per pound, is now going to cost $5. This is the corrected. This is the corrected number. And we saw from the last scenario, this information hasn't changed."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "And one pound of bananas is going to cost b dollars, because it's one pound times b dollars per pound, is now going to cost $5. This is the corrected. This is the corrected number. And we saw from the last scenario, this information hasn't changed. Six pounds of apples is going to cost 6a, six pounds times $a per pound. And three pounds of bananas is going to cost 3b, three pounds times b dollars per pound. The total cost of the apples and bananas in this trip we are given is $15."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "And we saw from the last scenario, this information hasn't changed. Six pounds of apples is going to cost 6a, six pounds times $a per pound. And three pounds of bananas is going to cost 3b, three pounds times b dollars per pound. The total cost of the apples and bananas in this trip we are given is $15. So once again, you say, well, let me try to solve this maybe through elimination. And once again, you say, well, let me cancel out the a's. I have 2a here."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "The total cost of the apples and bananas in this trip we are given is $15. So once again, you say, well, let me try to solve this maybe through elimination. And once again, you say, well, let me cancel out the a's. I have 2a here. I have 6a here. If I multiply the 2a here by negative 3, then this will become a negative 6a, and it might be able to cancel out with all of this business. So you do that."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "I have 2a here. I have 6a here. If I multiply the 2a here by negative 3, then this will become a negative 6a, and it might be able to cancel out with all of this business. So you do that. You multiply this entire equation. You can't just multiply one term. You have to multiply the entire equation times negative 3 if you want the equation to still hold."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "So you do that. You multiply this entire equation. You can't just multiply one term. You have to multiply the entire equation times negative 3 if you want the equation to still hold. And so we're multiplying by negative 3. So 2a times negative 3 is negative 6a. b times negative 3 is negative 3b."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "You have to multiply the entire equation times negative 3 if you want the equation to still hold. And so we're multiplying by negative 3. So 2a times negative 3 is negative 6a. b times negative 3 is negative 3b. And then 5 times negative 3 is negative 15. And now something fishy starts to look like it's about to happen. Because when you add the left-hand side of this blue equation or this purplish equation to the green one, you get 0."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "b times negative 3 is negative 3b. And then 5 times negative 3 is negative 15. And now something fishy starts to look like it's about to happen. Because when you add the left-hand side of this blue equation or this purplish equation to the green one, you get 0. All of these things right over here just cancel out. And on the right-hand side, 15 minus 15, that is also equal to 0. And you get 0 equals 0, which seems a little bit better than the last time you worked through it."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "Because when you add the left-hand side of this blue equation or this purplish equation to the green one, you get 0. All of these things right over here just cancel out. And on the right-hand side, 15 minus 15, that is also equal to 0. And you get 0 equals 0, which seems a little bit better than the last time you worked through it. Last time we got 0 equals 6. But 0 equals 0 doesn't really tell you anything about the x's and y's. This is true."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "And you get 0 equals 0, which seems a little bit better than the last time you worked through it. Last time we got 0 equals 6. But 0 equals 0 doesn't really tell you anything about the x's and y's. This is true. This is absolutely true that 0 does definitely equal 0. But it doesn't tell you any information about x and y. And so then the bird whispers in the king's ear."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "This is true. This is absolutely true that 0 does definitely equal 0. But it doesn't tell you any information about x and y. And so then the bird whispers in the king's ear. And then the king says, well, the bird says you should graph it to figure out what's actually going on. And so you've learned that listening to the bird actually makes a lot of sense. So you try to graph these two constraints."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "And so then the bird whispers in the king's ear. And then the king says, well, the bird says you should graph it to figure out what's actually going on. And so you've learned that listening to the bird actually makes a lot of sense. So you try to graph these two constraints. So let's do it the same way. We'll have a b-axis. That's our b-axis."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "So you try to graph these two constraints. So let's do it the same way. We'll have a b-axis. That's our b-axis. And we will have our a-axis. Let me mark off some markers here. 1, 2, 3, 4, 5, and 1, 2, 3, 4, 5."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "That's our b-axis. And we will have our a-axis. Let me mark off some markers here. 1, 2, 3, 4, 5, and 1, 2, 3, 4, 5. So this first equation right over here, if we subtract 2a from both sides, I'm just going to put it into slope-intercept form, you get b is equal to negative 2a plus 5. All I did is subtract 2a from both sides. And if we were to graph that, our b-intercept, when a is equal to 0, b is equal to 5."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "1, 2, 3, 4, 5, and 1, 2, 3, 4, 5. So this first equation right over here, if we subtract 2a from both sides, I'm just going to put it into slope-intercept form, you get b is equal to negative 2a plus 5. All I did is subtract 2a from both sides. And if we were to graph that, our b-intercept, when a is equal to 0, b is equal to 5. So that's right over here. And our slope is negative 2. Every time you add 1 to a, so if a goes from 0 to 1, b is going to go down by 2."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "And if we were to graph that, our b-intercept, when a is equal to 0, b is equal to 5. So that's right over here. And our slope is negative 2. Every time you add 1 to a, so if a goes from 0 to 1, b is going to go down by 2. So go down by 2, go down by 2. So this first white equation looks like this if we graph the solution set. These are all of the prices for bananas and apples that meet this constraint."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "Every time you add 1 to a, so if a goes from 0 to 1, b is going to go down by 2. So go down by 2, go down by 2. So this first white equation looks like this if we graph the solution set. These are all of the prices for bananas and apples that meet this constraint. Now let's graph the second equation. If we subtract 6a from both sides, we get 3b is equal to negative 6a plus 15. And now we can divide both sides by 3."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "These are all of the prices for bananas and apples that meet this constraint. Now let's graph the second equation. If we subtract 6a from both sides, we get 3b is equal to negative 6a plus 15. And now we can divide both sides by 3. Divide everything by 3. We are left with b is equal to negative 2a plus 5. Well, this is interesting."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "And now we can divide both sides by 3. Divide everything by 3. We are left with b is equal to negative 2a plus 5. Well, this is interesting. This looks very similar, or it looks exactly the same. Our b-intercept is 5, and our slope is negative 2a. So this is essentially the same line."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "Well, this is interesting. This looks very similar, or it looks exactly the same. Our b-intercept is 5, and our slope is negative 2a. So this is essentially the same line. So these are essentially the same constraints. And so you start to look at it a little bit confused. And you say, OK, I see why we got 0 equals 0."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "So this is essentially the same line. So these are essentially the same constraints. And so you start to look at it a little bit confused. And you say, OK, I see why we got 0 equals 0. There's actually an infinite number of solutions. You pick any x, and then the corresponding y for each of these could be a solution for either of these things. So there's an infinite number of solutions."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "And you say, OK, I see why we got 0 equals 0. There's actually an infinite number of solutions. You pick any x, and then the corresponding y for each of these could be a solution for either of these things. So there's an infinite number of solutions. But you start to wonder, why is this happening? And so the bird whispers again into the king's ear. And the king says, well, the bird says this is because in both trips to the market, the same ratio of apples and bananas was bought."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "So there's an infinite number of solutions. But you start to wonder, why is this happening? And so the bird whispers again into the king's ear. And the king says, well, the bird says this is because in both trips to the market, the same ratio of apples and bananas was bought. In the green trip versus the white trip, bought three times as many apples, bought three times as many bananas, and you had three times the cost. So in any situation, for any prices, per pound prices of apples and bananas, if you buy exactly three times the number of apples, three times the number of bananas, and have three times the cost, that could be true for any prices. And so this is actually consistent."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "And the king says, well, the bird says this is because in both trips to the market, the same ratio of apples and bananas was bought. In the green trip versus the white trip, bought three times as many apples, bought three times as many bananas, and you had three times the cost. So in any situation, for any prices, per pound prices of apples and bananas, if you buy exactly three times the number of apples, three times the number of bananas, and have three times the cost, that could be true for any prices. And so this is actually consistent. We can't say that our begla is lying to us. But it's not giving us enough information. This is what we call a consistent system."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "And so this is actually consistent. We can't say that our begla is lying to us. But it's not giving us enough information. This is what we call a consistent system. It's consistent information here. So let me write this down. This is consistent."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "This is what we call a consistent system. It's consistent information here. So let me write this down. This is consistent. It is consistent. 0 equals 0. There's no shadiness going on here."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "This is consistent. It is consistent. 0 equals 0. There's no shadiness going on here. But it's not enough information. This system of equations is dependent. It is dependent."}, {"video_title": "Infinite solutions to systems   Systems of equations and inequalities   Algebra II   Khan Academy.mp3", "Sentence": "There's no shadiness going on here. But it's not enough information. This system of equations is dependent. It is dependent. And you have an infinite number of solutions. Any point on this line represents a solution. So you tell our begla, well, if you really want us to figure this out, you need to give us more information."}, {"video_title": "Examples of linear and exponential relationships.mp3", "Sentence": "And what I would like to do in this video is figure out whether each of these relationships, whether they are either linear relationships, exponential relationships, or neither. And like always, pause this video and see if you can figure it out yourself. So let's look at this first relationship right over here. And the key way to tell whether we're dealing with a linear or exponential or neither relationship is think about, okay, for a given change in x, and you see each time here, we are increasing x by the same amount. So we're increasing x by three. So given that we are increasing x by a constant amount, by three each time, does y increase by a constant amount, in which case we would be dealing with a linear relationship? Or is there a constant ratio between successive terms when you increase x by a constant amount, in which case we would be dealing with an exponential relationship?"}, {"video_title": "Examples of linear and exponential relationships.mp3", "Sentence": "And the key way to tell whether we're dealing with a linear or exponential or neither relationship is think about, okay, for a given change in x, and you see each time here, we are increasing x by the same amount. So we're increasing x by three. So given that we are increasing x by a constant amount, by three each time, does y increase by a constant amount, in which case we would be dealing with a linear relationship? Or is there a constant ratio between successive terms when you increase x by a constant amount, in which case we would be dealing with an exponential relationship? So let's see. Here we're going from negative two to five. So we are adding seven."}, {"video_title": "Examples of linear and exponential relationships.mp3", "Sentence": "Or is there a constant ratio between successive terms when you increase x by a constant amount, in which case we would be dealing with an exponential relationship? So let's see. Here we're going from negative two to five. So we are adding seven. When x increases by three, y increases by seven. When x is increasing by three, y increases by seven again. When x increases by three, y increases by seven again."}, {"video_title": "Examples of linear and exponential relationships.mp3", "Sentence": "So we are adding seven. When x increases by three, y increases by seven. When x is increasing by three, y increases by seven again. When x increases by three, y increases by seven again. So here it is clearly a linear relationship. Linear relationship. In fact, you can even, relationship."}, {"video_title": "Examples of linear and exponential relationships.mp3", "Sentence": "When x increases by three, y increases by seven again. So here it is clearly a linear relationship. Linear relationship. In fact, you can even, relationship. You could even plot this on a line if you assume that these are samples on a line. You could think even about the slope of that line. For a change in x, for a given change in x, the change in y is always constant."}, {"video_title": "Examples of linear and exponential relationships.mp3", "Sentence": "In fact, you can even, relationship. You could even plot this on a line if you assume that these are samples on a line. You could think even about the slope of that line. For a change in x, for a given change in x, the change in y is always constant. When our change in x is three, our change in y is always seven. So this is clearly a linear relationship. Now let's look at this one."}, {"video_title": "Examples of linear and exponential relationships.mp3", "Sentence": "For a change in x, for a given change in x, the change in y is always constant. When our change in x is three, our change in y is always seven. So this is clearly a linear relationship. Now let's look at this one. Let's see. Looks like our x's are changing by one each time. So plus one."}, {"video_title": "Examples of linear and exponential relationships.mp3", "Sentence": "Now let's look at this one. Let's see. Looks like our x's are changing by one each time. So plus one. Now what are y's changing by? Here it changes by two. Then it changes by six."}, {"video_title": "Examples of linear and exponential relationships.mp3", "Sentence": "So plus one. Now what are y's changing by? Here it changes by two. Then it changes by six. All right, it's clearly not linear. Then it changes by 18. Clearly not a linear relationship."}, {"video_title": "Examples of linear and exponential relationships.mp3", "Sentence": "Then it changes by six. All right, it's clearly not linear. Then it changes by 18. Clearly not a linear relationship. If this was linear, this would be the same amount, same delta, same change in y for every time because we have the same change in x. So let's test to see if it's exponential. If it's an exponential, for each of these constant changes in x's, when we increase x by one every time, our ratio of successive y should be the same."}, {"video_title": "Examples of linear and exponential relationships.mp3", "Sentence": "Clearly not a linear relationship. If this was linear, this would be the same amount, same delta, same change in y for every time because we have the same change in x. So let's test to see if it's exponential. If it's an exponential, for each of these constant changes in x's, when we increase x by one every time, our ratio of successive y should be the same. Or another way to think about it is what are we multiplying y by? So to go from one to three, you multiply by three. To go from three to nine, you multiply by three."}, {"video_title": "Examples of linear and exponential relationships.mp3", "Sentence": "If it's an exponential, for each of these constant changes in x's, when we increase x by one every time, our ratio of successive y should be the same. Or another way to think about it is what are we multiplying y by? So to go from one to three, you multiply by three. To go from three to nine, you multiply by three. To go from nine to 27, you multiply by three. So in a situation where every time you increase x by a fixed amount, in this case one, and the corresponding y's get multiplied by some fixed amount, then you are dealing with an exponential relationship. Exponential."}, {"video_title": "Comparing linear rates example.mp3", "Sentence": "Then, 42 minutes later, a dragon also flew east over the castle. The dragon flew 225 kilometers per hour. Assume both the griffin and the dragon continue flying east at the same speeds. How many minutes will the dragon have flown since passing the castle when it catches up to the griffin? And they also ask us, how many kilometers east of the castle will they be at that time? So pause this video and see if you can figure this out before we do this together. All right, so the question is, how many minutes will the dragon have flown since passing the castle when it catches up to the griffin?"}, {"video_title": "Comparing linear rates example.mp3", "Sentence": "How many minutes will the dragon have flown since passing the castle when it catches up to the griffin? And they also ask us, how many kilometers east of the castle will they be at that time? So pause this video and see if you can figure this out before we do this together. All right, so the question is, how many minutes will the dragon have flown since passing the castle when it catches up to the griffin? So let's set that variable to be equal to t, the number of minutes that the dragon has flown. Dragon flown since castle, since castle, and catches up. Catches up."}, {"video_title": "Comparing linear rates example.mp3", "Sentence": "All right, so the question is, how many minutes will the dragon have flown since passing the castle when it catches up to the griffin? So let's set that variable to be equal to t, the number of minutes that the dragon has flown. Dragon flown since castle, since castle, and catches up. Catches up. So let's think about the distance that the dragon would have flown in that t minutes. Well, the dragon's flying at 225 kilometers per hour, so the distance is going to be the rate, 225 kilometers per hour times the time, so times t minutes, but we have to be careful. This is in minutes while the rate is given in kilometers per hour."}, {"video_title": "Comparing linear rates example.mp3", "Sentence": "Catches up. So let's think about the distance that the dragon would have flown in that t minutes. Well, the dragon's flying at 225 kilometers per hour, so the distance is going to be the rate, 225 kilometers per hour times the time, so times t minutes, but we have to be careful. This is in minutes while the rate is given in kilometers per hour. So we have to make sure that our units work out. And so for every one hour, we have 60 minutes. And we can see here that the units indeed do work out."}, {"video_title": "Comparing linear rates example.mp3", "Sentence": "This is in minutes while the rate is given in kilometers per hour. So we have to make sure that our units work out. And so for every one hour, we have 60 minutes. And we can see here that the units indeed do work out. This hour cancels with that hour in the numerator and the denominator, and this minute cancels out with this minute. And so the distance that the dragon would have flown after t minutes is going to be 225t over 60 kilometers. So let me write it this way."}, {"video_title": "Comparing linear rates example.mp3", "Sentence": "And we can see here that the units indeed do work out. This hour cancels with that hour in the numerator and the denominator, and this minute cancels out with this minute. And so the distance that the dragon would have flown after t minutes is going to be 225t over 60 kilometers. So let me write it this way. 225 over 60t kilometers. Now we could try to simplify this, but I'll leave it like this for now. Maybe I'll simplify it a little bit later."}, {"video_title": "Comparing linear rates example.mp3", "Sentence": "So let me write it this way. 225 over 60t kilometers. Now we could try to simplify this, but I'll leave it like this for now. Maybe I'll simplify it a little bit later. Now let's think about how far the griffin would have flown. So they tell us that the griffin is flying at 50 kilometers per hour. So 50 kilometers per hour."}, {"video_title": "Comparing linear rates example.mp3", "Sentence": "Maybe I'll simplify it a little bit later. Now let's think about how far the griffin would have flown. So they tell us that the griffin is flying at 50 kilometers per hour. So 50 kilometers per hour. And how long would the griffin have flown by that point? Well, the griffin passed the castle 42 minutes before the dragon passed it. So if t is how many minutes that the dragon has been flying east of the castle, well, then the griffin is going to be t plus 42 minutes."}, {"video_title": "Comparing linear rates example.mp3", "Sentence": "So 50 kilometers per hour. And how long would the griffin have flown by that point? Well, the griffin passed the castle 42 minutes before the dragon passed it. So if t is how many minutes that the dragon has been flying east of the castle, well, then the griffin is going to be t plus 42 minutes. So t plus 42 minutes is how long that the griffin has been traveling east of the castle. And then once again, we have to make sure that our units work out. So we're gonna say one hour for every 60 minutes."}, {"video_title": "Comparing linear rates example.mp3", "Sentence": "So if t is how many minutes that the dragon has been flying east of the castle, well, then the griffin is going to be t plus 42 minutes. So t plus 42 minutes is how long that the griffin has been traveling east of the castle. And then once again, we have to make sure that our units work out. So we're gonna say one hour for every 60 minutes. The minutes cancel out, the hours cancel out. And so we are going to be left with 50 over 60, or I could write 5 6ths times t plus 42 kilometers. Or if we wanna simplify this even more, this is going to be 5 6ths t plus, let's see, 5 6ths of 42."}, {"video_title": "Comparing linear rates example.mp3", "Sentence": "So we're gonna say one hour for every 60 minutes. The minutes cancel out, the hours cancel out. And so we are going to be left with 50 over 60, or I could write 5 6ths times t plus 42 kilometers. Or if we wanna simplify this even more, this is going to be 5 6ths t plus, let's see, 5 6ths of 42. 42 divided by six is seven times five is 35, plus 35 kilometers. So we know that they would have flown the exact same distance, because we're talking about when the dragon catches up with the griffin. So these two things need to be equal to each other, and then we can just solve for t. So let's do that."}, {"video_title": "Comparing linear rates example.mp3", "Sentence": "Or if we wanna simplify this even more, this is going to be 5 6ths t plus, let's see, 5 6ths of 42. 42 divided by six is seven times five is 35, plus 35 kilometers. So we know that they would have flown the exact same distance, because we're talking about when the dragon catches up with the griffin. So these two things need to be equal to each other, and then we can just solve for t. So let's do that. We get 225 over 60 t, and we know that both sides are in kilometers, so just for the sake of simplicity, I won't write the units here. So this is going to be equal to 5 6ths t plus 35. And now let us solve for t. We can subtract 5 6ths t from both sides, or actually, since I already have 60 as a denominator, I could subtract 50 over 60 t from both sides, which is the same thing as 5 6ths t. So I'm going to have 225 over 60 minus 50 over 60, and then all of that times t is equal to 35."}, {"video_title": "Comparing linear rates example.mp3", "Sentence": "So these two things need to be equal to each other, and then we can just solve for t. So let's do that. We get 225 over 60 t, and we know that both sides are in kilometers, so just for the sake of simplicity, I won't write the units here. So this is going to be equal to 5 6ths t plus 35. And now let us solve for t. We can subtract 5 6ths t from both sides, or actually, since I already have 60 as a denominator, I could subtract 50 over 60 t from both sides, which is the same thing as 5 6ths t. So I'm going to have 225 over 60 minus 50 over 60, and then all of that times t is equal to 35. And so let me get myself a little bit more real estate. So this is going to be simplified as 175 over 60 t is equal to 35, or that if I just multiply both sides by 60 over 175, I'll get that t is equal to 35 times 60 over 175. And you might recognize that 35 is the same thing as five times seven, and 175 is the same thing as 25 times seven."}, {"video_title": "Comparing linear rates example.mp3", "Sentence": "And now let us solve for t. We can subtract 5 6ths t from both sides, or actually, since I already have 60 as a denominator, I could subtract 50 over 60 t from both sides, which is the same thing as 5 6ths t. So I'm going to have 225 over 60 minus 50 over 60, and then all of that times t is equal to 35. And so let me get myself a little bit more real estate. So this is going to be simplified as 175 over 60 t is equal to 35, or that if I just multiply both sides by 60 over 175, I'll get that t is equal to 35 times 60 over 175. And you might recognize that 35 is the same thing as five times seven, and 175 is the same thing as 25 times seven. So these sevens cancel out, and then if we divide both this and this by five, this becomes a one, this becomes a five, and then 60 divided by five is equal to 12. And so remember, t was in minutes, so the answer to the first part of the question is 12 minutes. So let's go back up to what they were asking us."}, {"video_title": "Comparing linear rates example.mp3", "Sentence": "And you might recognize that 35 is the same thing as five times seven, and 175 is the same thing as 25 times seven. So these sevens cancel out, and then if we divide both this and this by five, this becomes a one, this becomes a five, and then 60 divided by five is equal to 12. And so remember, t was in minutes, so the answer to the first part of the question is 12 minutes. So let's go back up to what they were asking us. How many minutes will the dragon have flown since passing the castle when it catches up to the griffin? Well, we define that as t, and then we got 12 minutes. Now, the next part of the question is how many kilometers east of the castle will they be at that time?"}, {"video_title": "Comparing linear rates example.mp3", "Sentence": "So let's go back up to what they were asking us. How many minutes will the dragon have flown since passing the castle when it catches up to the griffin? Well, we define that as t, and then we got 12 minutes. Now, the next part of the question is how many kilometers east of the castle will they be at that time? So to figure out how many kilometers east of the castle, we just have to calculate this expression or this expression when t is equal to 12. So I'm going to use this first one. So you're going to have 225 over 60 times 12 is going to give us, let's see, 60 and 12 are both divisible by 12, so you get that one over five."}, {"video_title": "Comparing linear rates example.mp3", "Sentence": "Now, the next part of the question is how many kilometers east of the castle will they be at that time? So to figure out how many kilometers east of the castle, we just have to calculate this expression or this expression when t is equal to 12. So I'm going to use this first one. So you're going to have 225 over 60 times 12 is going to give us, let's see, 60 and 12 are both divisible by 12, so you get that one over five. And if you divide 225 by five, that is going to give us 45, and the units all work out to kilometers. So we answered the first two parts of the question. And the second part of this question, they tell us that Latanya and Jair both wrote correct inequalities for the times in minutes when the dragon is farther east of the castle than the griffin is."}, {"video_title": "Comparing linear rates example.mp3", "Sentence": "So you're going to have 225 over 60 times 12 is going to give us, let's see, 60 and 12 are both divisible by 12, so you get that one over five. And if you divide 225 by five, that is going to give us 45, and the units all work out to kilometers. So we answered the first two parts of the question. And the second part of this question, they tell us that Latanya and Jair both wrote correct inequalities for the times in minutes when the dragon is farther east of the castle than the griffin is. Latanya wrote t is greater than 12, and Jair wrote t is greater than 54. How did Latanya and Jair define their variables? Well, Latanya defined her variables the exact same way that I define mine, because we got t is equal to 12 when the dragon passes up the griffin."}, {"video_title": "Comparing linear rates example.mp3", "Sentence": "And the second part of this question, they tell us that Latanya and Jair both wrote correct inequalities for the times in minutes when the dragon is farther east of the castle than the griffin is. Latanya wrote t is greater than 12, and Jair wrote t is greater than 54. How did Latanya and Jair define their variables? Well, Latanya defined her variables the exact same way that I define mine, because we got t is equal to 12 when the dragon passes up the griffin. So for t is greater than 12, the dragon is farther east of the castle than the griffin. So we did the exact same thing as Latanya. But what Jair did, if he got t is greater than 54, is he must have defined t as being equal to the number of minutes since the griffin, the slower, the first but slower creature, passed the castle."}, {"video_title": "Comparing linear rates example.mp3", "Sentence": "Well, Latanya defined her variables the exact same way that I define mine, because we got t is equal to 12 when the dragon passes up the griffin. So for t is greater than 12, the dragon is farther east of the castle than the griffin. So we did the exact same thing as Latanya. But what Jair did, if he got t is greater than 54, is he must have defined t as being equal to the number of minutes since the griffin, the slower, the first but slower creature, passed the castle. And so he would have gotten t is greater than 54. And then if you wanted to know how many minutes since the dragon passed the castle, because the dragon got there 42 minutes later, he would have subtracted 42 from that. And that's how you connect these two numbers over here."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Let's do a few more problems that bring together the concepts that we learned in the last two videos. So let's say we have the inequality 4x plus 3 is less than negative 1. So let's find all of the x's that satisfy this. So the first thing I'd like to do is get rid of this 3. So let's subtract 3 from both sides of this equation. So the left-hand side is just going to end up being 4x. These 3's cancel out."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So the first thing I'd like to do is get rid of this 3. So let's subtract 3 from both sides of this equation. So the left-hand side is just going to end up being 4x. These 3's cancel out. That just ends up with a 0. No reason to change the inequality just yet. We're just adding and subtracting from both sides."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "These 3's cancel out. That just ends up with a 0. No reason to change the inequality just yet. We're just adding and subtracting from both sides. In this case, subtracting. That doesn't change the inequality as long as we're subtracting the same value. We have negative 1 minus 3."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "We're just adding and subtracting from both sides. In this case, subtracting. That doesn't change the inequality as long as we're subtracting the same value. We have negative 1 minus 3. That is negative 4. Negative 1 minus 3 is negative 4. And then we'll want to divide both sides of this equation by 4."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "We have negative 1 minus 3. That is negative 4. Negative 1 minus 3 is negative 4. And then we'll want to divide both sides of this equation by 4. Let's divide both sides of this equation by 4. Once again, when you multiply or divide both sides of an inequality by a positive number, it doesn't change the inequality. So the left-hand side is just x. x is less than negative 4 divided by 4 is negative 1. x is less than negative 1."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "And then we'll want to divide both sides of this equation by 4. Let's divide both sides of this equation by 4. Once again, when you multiply or divide both sides of an inequality by a positive number, it doesn't change the inequality. So the left-hand side is just x. x is less than negative 4 divided by 4 is negative 1. x is less than negative 1. Or we could write this in interval notation. All of the x's from negative infinity to negative 1, but not including negative 1. So we put a parenthesis right there."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So the left-hand side is just x. x is less than negative 4 divided by 4 is negative 1. x is less than negative 1. Or we could write this in interval notation. All of the x's from negative infinity to negative 1, but not including negative 1. So we put a parenthesis right there. Let's do a slightly harder one. Let's say we have 5x is greater than 8x plus 27. Let's get all our x's on the left-hand side."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So we put a parenthesis right there. Let's do a slightly harder one. Let's say we have 5x is greater than 8x plus 27. Let's get all our x's on the left-hand side. The best way to do that is subtract 8x from both sides. You subtract 8x from both sides. The left-hand side becomes 5x minus 8x."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Let's get all our x's on the left-hand side. The best way to do that is subtract 8x from both sides. You subtract 8x from both sides. The left-hand side becomes 5x minus 8x. That's negative 3x. We still have a greater than sign. We're just adding or subtracting the same quantity to both sides."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "The left-hand side becomes 5x minus 8x. That's negative 3x. We still have a greater than sign. We're just adding or subtracting the same quantity to both sides. These 8x's cancel out and you're just left with a 27. So you have negative 3x is greater than 27. Now, to just turn this into an x, we want to divide both sides by negative 3."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "We're just adding or subtracting the same quantity to both sides. These 8x's cancel out and you're just left with a 27. So you have negative 3x is greater than 27. Now, to just turn this into an x, we want to divide both sides by negative 3. But remember, when you multiply or divide both sides of an inequality by a negative number, you swap the inequality. So if we divide both sides of this by negative 3, we have to swap this inequality. It will go from being a greater than sign to a less than sign."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Now, to just turn this into an x, we want to divide both sides by negative 3. But remember, when you multiply or divide both sides of an inequality by a negative number, you swap the inequality. So if we divide both sides of this by negative 3, we have to swap this inequality. It will go from being a greater than sign to a less than sign. And this is a bit of a way that I remember greater than. The left-hand side just looks bigger. This is greater than."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "It will go from being a greater than sign to a less than sign. And this is a bit of a way that I remember greater than. The left-hand side just looks bigger. This is greater than. If you just imagine this height, that height is greater than that height right there, which is just a point. I don't know if that confuses you or not. This is less than."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "This is greater than. If you just imagine this height, that height is greater than that height right there, which is just a point. I don't know if that confuses you or not. This is less than. This little point is less than the distance of that big opening. That's how I remember it. But anyway, 3x over negative 3."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "This is less than. This little point is less than the distance of that big opening. That's how I remember it. But anyway, 3x over negative 3. So anyway, now that we divided both sides by a negative number, by negative 3, we swapped the inequality from greater than to less than. In the left-hand side, the negative 3's cancel out. You get x is less than 27 over negative 3, which is negative 9."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "But anyway, 3x over negative 3. So anyway, now that we divided both sides by a negative number, by negative 3, we swapped the inequality from greater than to less than. In the left-hand side, the negative 3's cancel out. You get x is less than 27 over negative 3, which is negative 9. Or in interval notation, it would be everything from negative infinity to negative 9, not including negative 9. If you wanted to do it as a number line, it would look like this. This would be negative 9."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "You get x is less than 27 over negative 3, which is negative 9. Or in interval notation, it would be everything from negative infinity to negative 9, not including negative 9. If you wanted to do it as a number line, it would look like this. This would be negative 9. Maybe this would be negative 8. Maybe this would be negative 10. You would start at negative 9, not include it because we don't have an equal sign here."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "This would be negative 9. Maybe this would be negative 8. Maybe this would be negative 10. You would start at negative 9, not include it because we don't have an equal sign here. You go all the way, everything less than that, all the way down as we see to negative infinity. Let's do a nice, hairy problem. Let's say we have 8x minus 5 times 4x plus 1 is greater than or equal to negative 1 plus 2 times 4x minus 3."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "You would start at negative 9, not include it because we don't have an equal sign here. You go all the way, everything less than that, all the way down as we see to negative infinity. Let's do a nice, hairy problem. Let's say we have 8x minus 5 times 4x plus 1 is greater than or equal to negative 1 plus 2 times 4x minus 3. This might seem very daunting, but if we just simplify it step by step, you'll see it's no harder than any of the other problems we've tackled. Let's just simplify this. You get 8x minus, let's distribute this negative 5."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Let's say we have 8x minus 5 times 4x plus 1 is greater than or equal to negative 1 plus 2 times 4x minus 3. This might seem very daunting, but if we just simplify it step by step, you'll see it's no harder than any of the other problems we've tackled. Let's just simplify this. You get 8x minus, let's distribute this negative 5. Let me say 8x and then distribute the negative 5. Negative 5 times 4x is negative 20x. Negative 5, when I say negative 5, I'm talking about this whole thing."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "You get 8x minus, let's distribute this negative 5. Let me say 8x and then distribute the negative 5. Negative 5 times 4x is negative 20x. Negative 5, when I say negative 5, I'm talking about this whole thing. Negative 5 times 1 is negative 5. Then that's going to be greater than or equal to negative 1 plus 2 times 4x is 8x. 2 times negative 3 is negative 6."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Negative 5, when I say negative 5, I'm talking about this whole thing. Negative 5 times 1 is negative 5. Then that's going to be greater than or equal to negative 1 plus 2 times 4x is 8x. 2 times negative 3 is negative 6. Now we can merge these two terms. 8x minus 20x is negative 12x minus 5 is greater than or equal to, we can merge these constant terms, negative 1 minus 6, that's negative 7. Then we have this plus 8x left over."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "2 times negative 3 is negative 6. Now we can merge these two terms. 8x minus 20x is negative 12x minus 5 is greater than or equal to, we can merge these constant terms, negative 1 minus 6, that's negative 7. Then we have this plus 8x left over. I like to get all my x terms on the left-hand side, so let's subtract 8x from both sides of this equation. Let's subtract 8x from both sides of this equation. From both sides, I'm subtracting 8x."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Then we have this plus 8x left over. I like to get all my x terms on the left-hand side, so let's subtract 8x from both sides of this equation. Let's subtract 8x from both sides of this equation. From both sides, I'm subtracting 8x. This left-hand side, negative 12 minus 8, that's negative 20. Negative 20x minus 5. Once again, no reason to change the inequality just yet."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "From both sides, I'm subtracting 8x. This left-hand side, negative 12 minus 8, that's negative 20. Negative 20x minus 5. Once again, no reason to change the inequality just yet. All we're doing is simplifying the sides or adding and subtracting from them. The right-hand side becomes, this thing cancels out, 8x minus 8x, that's 0. You're just left with a negative 7."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Once again, no reason to change the inequality just yet. All we're doing is simplifying the sides or adding and subtracting from them. The right-hand side becomes, this thing cancels out, 8x minus 8x, that's 0. You're just left with a negative 7. Now I want to get rid of this negative 5, so let's add 5 to both sides of this equation. The left-hand side, you're just left with a negative 20x. The 5 and these 5s cancel out."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "You're just left with a negative 7. Now I want to get rid of this negative 5, so let's add 5 to both sides of this equation. The left-hand side, you're just left with a negative 20x. The 5 and these 5s cancel out. No reason to change the inequality just yet. Negative 7 plus 5, that's negative 2. Now we're at an interesting point."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "The 5 and these 5s cancel out. No reason to change the inequality just yet. Negative 7 plus 5, that's negative 2. Now we're at an interesting point. We have negative 20x is greater than or equal to negative 2. If this was an equation or really any type of even inequality, we want to divide both sides by negative 20, but we have to remember, when you multiply or divide both sides of an inequality by a negative number, you have to swap the inequality, so let's remember that. If we divide this side by negative 20 and we divide this side by negative 20, all I did is took both of these sides, divided by negative 20."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Now we're at an interesting point. We have negative 20x is greater than or equal to negative 2. If this was an equation or really any type of even inequality, we want to divide both sides by negative 20, but we have to remember, when you multiply or divide both sides of an inequality by a negative number, you have to swap the inequality, so let's remember that. If we divide this side by negative 20 and we divide this side by negative 20, all I did is took both of these sides, divided by negative 20. We have to swap the inequality. The greater than or equal to has to become a less than or equal sign. Of course, these cancel out."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "If we divide this side by negative 20 and we divide this side by negative 20, all I did is took both of these sides, divided by negative 20. We have to swap the inequality. The greater than or equal to has to become a less than or equal sign. Of course, these cancel out. You get x is less than or equal to, the negatives cancel out. 2 over 20 is 1 over 10. If we were writing it in interval notation, the upper bound would be 1 over 10."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Of course, these cancel out. You get x is less than or equal to, the negatives cancel out. 2 over 20 is 1 over 10. If we were writing it in interval notation, the upper bound would be 1 over 10. Notice we're including it because we have an equal sign, less than or equal, so we're including 1 over 10, and we're going to go all the way down to negative infinity, everything less than or equal to 1 over 10. This is just another way of writing that. Just for fun, let's draw the number line right here."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "If we were writing it in interval notation, the upper bound would be 1 over 10. Notice we're including it because we have an equal sign, less than or equal, so we're including 1 over 10, and we're going to go all the way down to negative infinity, everything less than or equal to 1 over 10. This is just another way of writing that. Just for fun, let's draw the number line right here. This is maybe 0. That is 1. 1 over 10 might be over here."}, {"video_title": "Multi-step inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Just for fun, let's draw the number line right here. This is maybe 0. That is 1. 1 over 10 might be over here. Everything less than or equal to 1 over 10. We're going to include the 1 tenth and everything less than that is included in the solutions. You could try out any value less than 1 over 10 and verify that it will satisfy this inequality."}, {"video_title": "Multiplying and dividing with inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "And I also want to introduce you to some other types of notations for describing the solution set of an inequality. So let's do a couple of examples. So let's say I had negative 0.5 is x is less than or equal to 7.5. Now if this was an equality, your natural impulse is to say, hey, let's divide both sides by the coefficient on the x term, and that is a completely legitimate thing to do. Divide both sides by negative 0.5. The important thing you need to realize, though, when you do it with an inequality, is that when you multiply or divide both sides of the equation by a negative number, you swap the inequality. Think of it this way."}, {"video_title": "Multiplying and dividing with inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Now if this was an equality, your natural impulse is to say, hey, let's divide both sides by the coefficient on the x term, and that is a completely legitimate thing to do. Divide both sides by negative 0.5. The important thing you need to realize, though, when you do it with an inequality, is that when you multiply or divide both sides of the equation by a negative number, you swap the inequality. Think of it this way. I'll do a simple example here. If I were to tell you that 1 is less than 2, I think you would agree with that. 1 is definitely less than 2."}, {"video_title": "Multiplying and dividing with inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Think of it this way. I'll do a simple example here. If I were to tell you that 1 is less than 2, I think you would agree with that. 1 is definitely less than 2. Now what happens if I multiply both sides of this by negative 1? Negative 1 versus negative 2. Well, all of a sudden, negative 2 is more negative than negative 1."}, {"video_title": "Multiplying and dividing with inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "1 is definitely less than 2. Now what happens if I multiply both sides of this by negative 1? Negative 1 versus negative 2. Well, all of a sudden, negative 2 is more negative than negative 1. So here, negative 2 is actually less than negative 1. Now this isn't a proof, but I think it'll give you comfort on why you're swapping the sign. If something is larger, when you take the negative of both of it, it'll be more negative, or vice versa."}, {"video_title": "Multiplying and dividing with inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Well, all of a sudden, negative 2 is more negative than negative 1. So here, negative 2 is actually less than negative 1. Now this isn't a proof, but I think it'll give you comfort on why you're swapping the sign. If something is larger, when you take the negative of both of it, it'll be more negative, or vice versa. So that's why if we're going to multiply both sides of this equation or divide both sides of the equation by a negative number, we need to swap the sign. So let's multiply both sides of this equation. Dividing by 0.5 is the same thing as multiplying by 2."}, {"video_title": "Multiplying and dividing with inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "If something is larger, when you take the negative of both of it, it'll be more negative, or vice versa. So that's why if we're going to multiply both sides of this equation or divide both sides of the equation by a negative number, we need to swap the sign. So let's multiply both sides of this equation. Dividing by 0.5 is the same thing as multiplying by 2. Our whole goal here is to have a 1 coefficient there. So let's multiply both sides of this equation by negative 2. So we have negative 2 times negative 0.5."}, {"video_title": "Multiplying and dividing with inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Dividing by 0.5 is the same thing as multiplying by 2. Our whole goal here is to have a 1 coefficient there. So let's multiply both sides of this equation by negative 2. So we have negative 2 times negative 0.5. You might say, hey, how did Sal get this 2 here? My brain is just thinking, what can I multiply negative 0.5 by to get 1? And negative 0.5 is the same thing as negative 1 half."}, {"video_title": "Multiplying and dividing with inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So we have negative 2 times negative 0.5. You might say, hey, how did Sal get this 2 here? My brain is just thinking, what can I multiply negative 0.5 by to get 1? And negative 0.5 is the same thing as negative 1 half. The inverse of that is negative 2. So I'm multiplying negative 2 times both sides of this equation, and I have the 7.5 on the other side. I'm going to multiply that by negative 2 as well."}, {"video_title": "Multiplying and dividing with inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "And negative 0.5 is the same thing as negative 1 half. The inverse of that is negative 2. So I'm multiplying negative 2 times both sides of this equation, and I have the 7.5 on the other side. I'm going to multiply that by negative 2 as well. And remember, when you multiply or divide both sides of an inequality by a negative, you swap the inequality. You had less than or equal, now it will be greater than or equal. So the left-hand side, negative 2 times negative 0.5 is just 1, you get x is greater than or equal to 7.5 times negative 2, that's negative 15, which is our solution set."}, {"video_title": "Multiplying and dividing with inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "I'm going to multiply that by negative 2 as well. And remember, when you multiply or divide both sides of an inequality by a negative, you swap the inequality. You had less than or equal, now it will be greater than or equal. So the left-hand side, negative 2 times negative 0.5 is just 1, you get x is greater than or equal to 7.5 times negative 2, that's negative 15, which is our solution set. All x's larger than negative 15 will satisfy this equation. I challenge you to try it. For example, 0 will work."}, {"video_title": "Multiplying and dividing with inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So the left-hand side, negative 2 times negative 0.5 is just 1, you get x is greater than or equal to 7.5 times negative 2, that's negative 15, which is our solution set. All x's larger than negative 15 will satisfy this equation. I challenge you to try it. For example, 0 will work. 0 is greater than negative 15. But try something like negative 16. Negative 16 will not work."}, {"video_title": "Multiplying and dividing with inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "For example, 0 will work. 0 is greater than negative 15. But try something like negative 16. Negative 16 will not work. Negative 16 times 0.5 is 8, which is not less than 7.5. So the solution set is all of the x's, let me draw a number line here, greater than negative 15. So that is negative 15 there, maybe that's negative 16, that's negative 14."}, {"video_title": "Multiplying and dividing with inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Negative 16 will not work. Negative 16 times 0.5 is 8, which is not less than 7.5. So the solution set is all of the x's, let me draw a number line here, greater than negative 15. So that is negative 15 there, maybe that's negative 16, that's negative 14. Greater than or equal to negative 15 is the solution. Now, you might also see solution sets to inequalities written in interval notation. An interval notation, it just takes a little getting used to."}, {"video_title": "Multiplying and dividing with inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So that is negative 15 there, maybe that's negative 16, that's negative 14. Greater than or equal to negative 15 is the solution. Now, you might also see solution sets to inequalities written in interval notation. An interval notation, it just takes a little getting used to. We want to include negative 15, so our lower bound to our interval is negative 15. And putting this bracket here means that we're going to include negative 15. The set includes the bottom boundary, it includes negative 15."}, {"video_title": "Multiplying and dividing with inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "An interval notation, it just takes a little getting used to. We want to include negative 15, so our lower bound to our interval is negative 15. And putting this bracket here means that we're going to include negative 15. The set includes the bottom boundary, it includes negative 15. And we're going to go all the way to infinity. And we put a curly, a parentheses here. Parentheses normally means that you're not including the upper bound."}, {"video_title": "Multiplying and dividing with inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "The set includes the bottom boundary, it includes negative 15. And we're going to go all the way to infinity. And we put a curly, a parentheses here. Parentheses normally means that you're not including the upper bound. You also do it for infinity, because infinity really isn't a normal number, so to speak. You can't just say, oh, I'm at infinity. You're never at infinity."}, {"video_title": "Multiplying and dividing with inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Parentheses normally means that you're not including the upper bound. You also do it for infinity, because infinity really isn't a normal number, so to speak. You can't just say, oh, I'm at infinity. You're never at infinity. So that's why you put that parentheses. But the parentheses tends to mean that you don't include that boundary, but you also use it with infinity. So this and this are the exact same thing."}, {"video_title": "Multiplying and dividing with inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "You're never at infinity. So that's why you put that parentheses. But the parentheses tends to mean that you don't include that boundary, but you also use it with infinity. So this and this are the exact same thing. Sometimes you might also see set notations, where the solution of that, they might say x is a real number such that, that little line, that vertical line thing, just means such that x is greater than or equal to negative 15. So this is the set, these curly brackets mean the set of all real numbers, or the set of all numbers where x is a real number such that x is greater than or equal to negative 15. All of this, this, and this are all equivalent."}, {"video_title": "Multiplying and dividing with inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So this and this are the exact same thing. Sometimes you might also see set notations, where the solution of that, they might say x is a real number such that, that little line, that vertical line thing, just means such that x is greater than or equal to negative 15. So this is the set, these curly brackets mean the set of all real numbers, or the set of all numbers where x is a real number such that x is greater than or equal to negative 15. All of this, this, and this are all equivalent. Let's keep that in mind and do a couple of more examples. So let's say we had 75x is greater than or equal to 125. So here we can just divide both sides by 75."}, {"video_title": "Multiplying and dividing with inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "All of this, this, and this are all equivalent. Let's keep that in mind and do a couple of more examples. So let's say we had 75x is greater than or equal to 125. So here we can just divide both sides by 75. And since 75 is a positive number, you don't have to change the inequality. So you get x is greater than or equal to 125 over 75. And if you divide the numerator and denominator by 25, this is 5 over 3."}, {"video_title": "Multiplying and dividing with inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So here we can just divide both sides by 75. And since 75 is a positive number, you don't have to change the inequality. So you get x is greater than or equal to 125 over 75. And if you divide the numerator and denominator by 25, this is 5 over 3. So x is greater than or equal to 5 thirds. Or we could write the solution set being from including 5 thirds to infinity. And once again, if you were to graph it on a number line, 5 thirds is what?"}, {"video_title": "Multiplying and dividing with inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "And if you divide the numerator and denominator by 25, this is 5 over 3. So x is greater than or equal to 5 thirds. Or we could write the solution set being from including 5 thirds to infinity. And once again, if you were to graph it on a number line, 5 thirds is what? That's 1 and 2 thirds. So you have 0, 1, 2, and 1 and 2 thirds will be right around there. We're going to include it."}, {"video_title": "Multiplying and dividing with inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "And once again, if you were to graph it on a number line, 5 thirds is what? That's 1 and 2 thirds. So you have 0, 1, 2, and 1 and 2 thirds will be right around there. We're going to include it. That right there is 5 thirds. And everything greater than or equal to that will be included in our solution set. Let's do another one."}, {"video_title": "Multiplying and dividing with inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "We're going to include it. That right there is 5 thirds. And everything greater than or equal to that will be included in our solution set. Let's do another one. Let's say we have x over negative 3 is greater than negative 10 over 9. So we want to just isolate the x on the left-hand side. So let's multiply both sides by negative 3."}, {"video_title": "Multiplying and dividing with inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Let's do another one. Let's say we have x over negative 3 is greater than negative 10 over 9. So we want to just isolate the x on the left-hand side. So let's multiply both sides by negative 3. This is essentially the coefficient you could imagine is negative 1 over 3. So we want to multiply by the inverse, which would be negative 3. So if you multiply both sides by negative 3, you get negative 3 times, this you could rewrite it as negative 1 third x."}, {"video_title": "Multiplying and dividing with inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So let's multiply both sides by negative 3. This is essentially the coefficient you could imagine is negative 1 over 3. So we want to multiply by the inverse, which would be negative 3. So if you multiply both sides by negative 3, you get negative 3 times, this you could rewrite it as negative 1 third x. And on this side, you have negative 10 over 9 times negative 3. And the inequality will switch because we are multiplying or dividing by a negative number. So the inequality will switch."}, {"video_title": "Multiplying and dividing with inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So if you multiply both sides by negative 3, you get negative 3 times, this you could rewrite it as negative 1 third x. And on this side, you have negative 10 over 9 times negative 3. And the inequality will switch because we are multiplying or dividing by a negative number. So the inequality will switch. It will go from greater than to less than. So the left-hand side of the equation just becomes an x. That was the whole point."}, {"video_title": "Multiplying and dividing with inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So the inequality will switch. It will go from greater than to less than. So the left-hand side of the equation just becomes an x. That was the whole point. That cancels out with that. The negatives cancel out. x is less than."}, {"video_title": "Multiplying and dividing with inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "That was the whole point. That cancels out with that. The negatives cancel out. x is less than. And then you have a negative times a negative. That will make it a positive. Then if you divide the numerator and the denominator by 3, you get a 1 and a 3."}, {"video_title": "Multiplying and dividing with inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "x is less than. And then you have a negative times a negative. That will make it a positive. Then if you divide the numerator and the denominator by 3, you get a 1 and a 3. So x is less than 10 over 3. So if we were to write this in interval notation, the solution set, the upper bound will be 10 over 3. And it won't include 10 over 3."}, {"video_title": "Multiplying and dividing with inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Then if you divide the numerator and the denominator by 3, you get a 1 and a 3. So x is less than 10 over 3. So if we were to write this in interval notation, the solution set, the upper bound will be 10 over 3. And it won't include 10 over 3. This isn't less than or equal to. So we're going to put a parentheses here. Notice here it included 5 thirds."}, {"video_title": "Multiplying and dividing with inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "And it won't include 10 over 3. This isn't less than or equal to. So we're going to put a parentheses here. Notice here it included 5 thirds. We put a bracket. Here we're not including 10 thirds. We put a parentheses."}, {"video_title": "Multiplying and dividing with inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Notice here it included 5 thirds. We put a bracket. Here we're not including 10 thirds. We put a parentheses. And it'll go from 10 over 3 all the way down to negative infinity. And everything less than 10 over 3 is in our solution set. Let's draw that."}, {"video_title": "Multiplying and dividing with inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "We put a parentheses. And it'll go from 10 over 3 all the way down to negative infinity. And everything less than 10 over 3 is in our solution set. Let's draw that. Let's draw the solution set. So 10 over 3. So we might have 0, 1, 2, 3, 4."}, {"video_title": "Multiplying and dividing with inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Let's draw that. Let's draw the solution set. So 10 over 3. So we might have 0, 1, 2, 3, 4. 10 over 3 is 3 and 1 third. So it might sit, let me do it in a different color. It might be over here."}, {"video_title": "Multiplying and dividing with inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So we might have 0, 1, 2, 3, 4. 10 over 3 is 3 and 1 third. So it might sit, let me do it in a different color. It might be over here. We're not going to include that. It's less than 10 over 3. 10 over 3 is not in the solution set."}, {"video_title": "Multiplying and dividing with inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "It might be over here. We're not going to include that. It's less than 10 over 3. 10 over 3 is not in the solution set. That is 10 over 3 right there. And everything less than that, but not including 10 over 3, is in our solution set. Let's do one more."}, {"video_title": "Multiplying and dividing with inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "10 over 3 is not in the solution set. That is 10 over 3 right there. And everything less than that, but not including 10 over 3, is in our solution set. Let's do one more. Let's do one more. Say we have x over negative 15 is less than 8. So once again, let's multiply both sides of this equation by negative 15."}, {"video_title": "Multiplying and dividing with inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Let's do one more. Let's do one more. Say we have x over negative 15 is less than 8. So once again, let's multiply both sides of this equation by negative 15. So negative 15 times x over negative 15. Then you have an 8 times a negative 15. And when you multiply both sides of an inequality by a negative number, or divide both sides by a negative number, you swap the inequality."}, {"video_title": "Multiplying and dividing with inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "So once again, let's multiply both sides of this equation by negative 15. So negative 15 times x over negative 15. Then you have an 8 times a negative 15. And when you multiply both sides of an inequality by a negative number, or divide both sides by a negative number, you swap the inequality. It's less than, you change it to greater than. And now the left-hand side just becomes an x, because these guys cancel out. x is greater than 8 times 15 is 80 plus 40 is 120."}, {"video_title": "Multiplying and dividing with inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "And when you multiply both sides of an inequality by a negative number, or divide both sides by a negative number, you swap the inequality. It's less than, you change it to greater than. And now the left-hand side just becomes an x, because these guys cancel out. x is greater than 8 times 15 is 80 plus 40 is 120. So negative 120. Is that right? 80 plus 40."}, {"video_title": "Multiplying and dividing with inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "x is greater than 8 times 15 is 80 plus 40 is 120. So negative 120. Is that right? 80 plus 40. Negative 120. Or we could write the solution set as starting at negative 120. But we're not including negative 120."}, {"video_title": "Multiplying and dividing with inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "80 plus 40. Negative 120. Or we could write the solution set as starting at negative 120. But we're not including negative 120. We don't have an equal sign here. Going all the way up to infinity. And if we were to graph it, we draw the number line here."}, {"video_title": "Multiplying and dividing with inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "But we're not including negative 120. We don't have an equal sign here. Going all the way up to infinity. And if we were to graph it, we draw the number line here. I'll do a real quick one. Let's say that that is negative 120. Maybe 0 sitting up here."}, {"video_title": "Multiplying and dividing with inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "And if we were to graph it, we draw the number line here. I'll do a real quick one. Let's say that that is negative 120. Maybe 0 sitting up here. This would be negative 121. This would be negative 119. We are not going to include negative 120, because we don't have an equal sign there."}, {"video_title": "Multiplying and dividing with inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "Maybe 0 sitting up here. This would be negative 121. This would be negative 119. We are not going to include negative 120, because we don't have an equal sign there. It's going to be everything greater than negative 120. All of these things that I'm shading in green would satisfy the inequality. And you could even try it out."}, {"video_title": "Multiplying and dividing with inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "We are not going to include negative 120, because we don't have an equal sign there. It's going to be everything greater than negative 120. All of these things that I'm shading in green would satisfy the inequality. And you could even try it out. Does 0 work? 0 over 15? Yeah, that's 0."}, {"video_title": "Multiplying and dividing with inequalities   Linear inequalities   Algebra I   Khan Academy.mp3", "Sentence": "And you could even try it out. Does 0 work? 0 over 15? Yeah, that's 0. That's definitely less than 8. I mean, that doesn't prove it to you, but you could try any of these numbers, and they should work. Anyway, hopefully you found that helpful."}, {"video_title": "Percent from fraction models.mp3", "Sentence": "So why don't you pause this video and see if you can figure that out. So let's see, the whole is divided into one, two, three, four, five, six, seven, eight, nine, 10 equal sections, of which one, two, three, four, five, six, seven are actually filled in, that's the shaded area. So one way to think about it is, 7 tenths are shaded in. But how do we express this fraction as a percent? They're asking for a percent. Well remember, per cent, it literally means per hundred. Cent, same root as the word hundred, you see it in cents or century."}, {"video_title": "Percent from fraction models.mp3", "Sentence": "But how do we express this fraction as a percent? They're asking for a percent. Well remember, per cent, it literally means per hundred. Cent, same root as the word hundred, you see it in cents or century. And so, can we write this as per hundred instead of per 10? Well, seven per 10 is the same thing as 70 per hundred, or 70% And how did I go from 7 tenths to 70 over 100? Well, I just multiply both the numerator and the denominator by 10."}, {"video_title": "Percent from fraction models.mp3", "Sentence": "Cent, same root as the word hundred, you see it in cents or century. And so, can we write this as per hundred instead of per 10? Well, seven per 10 is the same thing as 70 per hundred, or 70% And how did I go from 7 tenths to 70 over 100? Well, I just multiply both the numerator and the denominator by 10. And once you do more and more percents, you'll get a hang of it, you say, oh, 7 tenths, that's the same thing as 70 per 100, which is 70%. Let's do another example. Here we're told 100% is shown on the following tape diagram."}, {"video_title": "Percent from fraction models.mp3", "Sentence": "Well, I just multiply both the numerator and the denominator by 10. And once you do more and more percents, you'll get a hang of it, you say, oh, 7 tenths, that's the same thing as 70 per 100, which is 70%. Let's do another example. Here we're told 100% is shown on the following tape diagram. So just this amount right over here is 100%. And then they ask us, what percent is represented by the entire tape diagram? So by this entire thing right over here."}, {"video_title": "Percent from fraction models.mp3", "Sentence": "Here we're told 100% is shown on the following tape diagram. So just this amount right over here is 100%. And then they ask us, what percent is represented by the entire tape diagram? So by this entire thing right over here. Pause this video and see if you can answer that. Well, one way to think about 100%, 100% is equivalent to a whole. And now we have three times as much of that for the entire tape diagram."}, {"video_title": "Percent from fraction models.mp3", "Sentence": "So by this entire thing right over here. Pause this video and see if you can answer that. Well, one way to think about 100%, 100% is equivalent to a whole. And now we have three times as much of that for the entire tape diagram. So you could view this as three wholes, or you could say that's 100%, we have another 100% right over here, and then we have another 100% right over here. So the whole tape diagram, that would be 300%. Let's do another example."}, {"video_title": "Percent from fraction models.mp3", "Sentence": "And now we have three times as much of that for the entire tape diagram. So you could view this as three wholes, or you could say that's 100%, we have another 100% right over here, and then we have another 100% right over here. So the whole tape diagram, that would be 300%. Let's do another example. This is strangely fun. And I'll see, it says, the large rectangle below represents one whole. So that's this whole thing is one whole."}, {"video_title": "Percent from fraction models.mp3", "Sentence": "Let's do another example. This is strangely fun. And I'll see, it says, the large rectangle below represents one whole. So that's this whole thing is one whole. What percentage is represented by the shaded area? So pause the video and see if you can figure that out again. So let's just express it as a fraction first."}, {"video_title": "Percent from fraction models.mp3", "Sentence": "So that's this whole thing is one whole. What percentage is represented by the shaded area? So pause the video and see if you can figure that out again. So let's just express it as a fraction first. So we have a total of one, two, three, four, five, six, seven, eight, nine, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20 squares. So out of those 20 squares, we see that six of them are actually shaded in. So 6 20ths, can we write that as per 100?"}, {"video_title": "Percent from fraction models.mp3", "Sentence": "So let's just express it as a fraction first. So we have a total of one, two, three, four, five, six, seven, eight, nine, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20 squares. So out of those 20 squares, we see that six of them are actually shaded in. So 6 20ths, can we write that as per 100? Well, let's see. If I were to go from 20 to 100, I multiply by five. And so if I multiply the numerator by five, I'll get the same value."}, {"video_title": "Percent from fraction models.mp3", "Sentence": "So 6 20ths, can we write that as per 100? Well, let's see. If I were to go from 20 to 100, I multiply by five. And so if I multiply the numerator by five, I'll get the same value. Six times five is 30. So six per 20 is the same thing as 30 per 100, which is the same thing as 30%, which literally means per 100. So this is 30%."}, {"video_title": "Percent from fraction models.mp3", "Sentence": "And so if I multiply the numerator by five, I'll get the same value. Six times five is 30. So six per 20 is the same thing as 30 per 100, which is the same thing as 30%, which literally means per 100. So this is 30%. Let's do one last example. Here we are told, which large rectangle below represents one whole? So this is a whole, and then this whole thing right over here is another whole."}, {"video_title": "Percent from fraction models.mp3", "Sentence": "So this is 30%. Let's do one last example. Here we are told, which large rectangle below represents one whole? So this is a whole, and then this whole thing right over here is another whole. What percentage is represented by the shaded area? Again, pause the video. See if you can answer that."}, {"video_title": "Percent from fraction models.mp3", "Sentence": "So this is a whole, and then this whole thing right over here is another whole. What percentage is represented by the shaded area? Again, pause the video. See if you can answer that. So this one, we have shaded in a whole, so that is 100%. And then over here, we have shaded in one, two, three, four fifths of the whole. So four fifths, if I were to express it as per 100, what would it be?"}, {"video_title": "Percent from fraction models.mp3", "Sentence": "See if you can answer that. So this one, we have shaded in a whole, so that is 100%. And then over here, we have shaded in one, two, three, four fifths of the whole. So four fifths, if I were to express it as per 100, what would it be? Five times 20 is 100, so four times 20 is 80. So four fifths, or 80 hundredths, is filled out here, or you could say 80 per 100, which is the same thing as 80%. So this right over here is 80%."}, {"video_title": "Percent from fraction models.mp3", "Sentence": "So four fifths, if I were to express it as per 100, what would it be? Five times 20 is 100, so four times 20 is 80. So four fifths, or 80 hundredths, is filled out here, or you could say 80 per 100, which is the same thing as 80%. So this right over here is 80%. So what percent is represented by the shaded area? Well, we have 100%, and then we have 80%, so we have 180%. It's more than a whole."}, {"video_title": "Proportion word problem (example 2)   7th grade   Khan Academy.mp3", "Sentence": "We're told that Mika can eat 21 hot dogs in 66 minutes. She wants to know how many minutes, M, would it take her to eat 35 hot dogs if she can keep up the same pace. So a big clue is the same pace. I have to remove a hair from my tongue, alright. A big clue is the same pace. That means that the hot dogs, hot dogs per minute, is going to be constant, is always going to be the same. Always the same."}, {"video_title": "Proportion word problem (example 2)   7th grade   Khan Academy.mp3", "Sentence": "I have to remove a hair from my tongue, alright. A big clue is the same pace. That means that the hot dogs, hot dogs per minute, is going to be constant, is always going to be the same. Always the same. Because this is essentially the pace. Her hot dogs per minute are going to stay the same. It's going to stay at the same pace."}, {"video_title": "Proportion word problem (example 2)   7th grade   Khan Academy.mp3", "Sentence": "Always the same. Because this is essentially the pace. Her hot dogs per minute are going to stay the same. It's going to stay at the same pace. So it tells us that she can eat 21 hot dogs in 66 minutes. So her hot dogs per minute, at least up here, is 21 hot dogs in 66 minutes. So 21 hot dogs in 66 minutes."}, {"video_title": "Proportion word problem (example 2)   7th grade   Khan Academy.mp3", "Sentence": "It's going to stay at the same pace. So it tells us that she can eat 21 hot dogs in 66 minutes. So her hot dogs per minute, at least up here, is 21 hot dogs in 66 minutes. So 21 hot dogs in 66 minutes. Well if her pace is always going to be the same, well it's going to take her, this ratio over here, is going to be the ratio between 35 hot dogs and however long it takes her to eat 35 hot dogs. So once again, hot dogs per minute has to be a constant because it's going to be the same pace. Hot dogs per minute."}, {"video_title": "Proportion word problem (example 2)   7th grade   Khan Academy.mp3", "Sentence": "So 21 hot dogs in 66 minutes. Well if her pace is always going to be the same, well it's going to take her, this ratio over here, is going to be the ratio between 35 hot dogs and however long it takes her to eat 35 hot dogs. So once again, hot dogs per minute has to be a constant because it's going to be the same pace. Hot dogs per minute. If 21 hot dogs take 66 minutes, 35 hot dogs take M minutes, these two ratios are going to be the same. We're dealing with a proportional relationship that's going to be happening at the same rate. And then we're left with a situation where we just have to solve for M. And there's a bunch of different ways that you could tackle this."}, {"video_title": "Proportion word problem (example 2)   7th grade   Khan Academy.mp3", "Sentence": "Hot dogs per minute. If 21 hot dogs take 66 minutes, 35 hot dogs take M minutes, these two ratios are going to be the same. We're dealing with a proportional relationship that's going to be happening at the same rate. And then we're left with a situation where we just have to solve for M. And there's a bunch of different ways that you could tackle this. The easiest way that I can think of doing it is, I don't like this M sitting here in the denominator, so let's multiply both sides by M. Let me do that in a different color. So I multiply that side by M, and this side by M. And so what do we get? On the left-hand side, we have 21 over 66 M, 21 over 66 times M, times M, is equal to, well, you divide by M and multiply by M, those are going to cancel out and you're just going to have 35."}, {"video_title": "Proportion word problem (example 2)   7th grade   Khan Academy.mp3", "Sentence": "And then we're left with a situation where we just have to solve for M. And there's a bunch of different ways that you could tackle this. The easiest way that I can think of doing it is, I don't like this M sitting here in the denominator, so let's multiply both sides by M. Let me do that in a different color. So I multiply that side by M, and this side by M. And so what do we get? On the left-hand side, we have 21 over 66 M, 21 over 66 times M, times M, is equal to, well, you divide by M and multiply by M, those are going to cancel out and you're just going to have 35. So now you just have to solve for M, and the best way I can think of doing that is multiply both sides times the reciprocal, both sides times the reciprocal of the coefficient on the M. So let's multiply both sides by, let's multiply both sides by 66 over 21. Once again, I've just swapped the numerator and the denominator here to get the reciprocal, but I can't just do it to one side of an equation, I have to do it to both sides, otherwise it's not going to be equal anymore. So times 66 over 21."}, {"video_title": "Proportion word problem (example 2)   7th grade   Khan Academy.mp3", "Sentence": "On the left-hand side, we have 21 over 66 M, 21 over 66 times M, times M, is equal to, well, you divide by M and multiply by M, those are going to cancel out and you're just going to have 35. So now you just have to solve for M, and the best way I can think of doing that is multiply both sides times the reciprocal, both sides times the reciprocal of the coefficient on the M. So let's multiply both sides by, let's multiply both sides by 66 over 21. Once again, I've just swapped the numerator and the denominator here to get the reciprocal, but I can't just do it to one side of an equation, I have to do it to both sides, otherwise it's not going to be equal anymore. So times 66 over 21. This is just going to be one, you multiply something times its reciprocal, you're just going to end up with one. So you're going to be left with M, is equal to, now 35 times 66 divided by 21. Well, 35 is the same thing as, 35 is five times seven, and 21 is three times seven."}, {"video_title": "Proportion word problem (example 2)   7th grade   Khan Academy.mp3", "Sentence": "So times 66 over 21. This is just going to be one, you multiply something times its reciprocal, you're just going to end up with one. So you're going to be left with M, is equal to, now 35 times 66 divided by 21. Well, 35 is the same thing as, 35 is five times seven, and 21 is three times seven. So you're multiplying by seven up here, and here you have a seven in the denominator, you're dividing by seven, so they're going to cancel out. So this is going to simplify to five times 66 over three, and then we could simplify it even more, because 66 is the same thing as three times 22. Three times 22."}, {"video_title": "Proportion word problem (example 2)   7th grade   Khan Academy.mp3", "Sentence": "Well, 35 is the same thing as, 35 is five times seven, and 21 is three times seven. So you're multiplying by seven up here, and here you have a seven in the denominator, you're dividing by seven, so they're going to cancel out. So this is going to simplify to five times 66 over three, and then we could simplify it even more, because 66 is the same thing as three times 22. Three times 22. And so you have a three in the numerator, you're multiplying by three, and a three in the denominator, dividing by three, three divided by three is one. So you're left with five times 22, which is 110. So it would take her M minutes to eat 35 hot dogs at the same pace."}, {"video_title": "Proportion word problem (example 2)   7th grade   Khan Academy.mp3", "Sentence": "Three times 22. And so you have a three in the numerator, you're multiplying by three, and a three in the denominator, dividing by three, three divided by three is one. So you're left with five times 22, which is 110. So it would take her M minutes to eat 35 hot dogs at the same pace. Now, when some of you might have tackled it, you might have had a different equation set up here. Instead of thinking of hot dogs per minute, you might have thought about minutes per hot dog. And so in that situation, if you thought in terms of minutes per hot dog, you might have said, okay, look, it took Micah 66 minutes to eat 21 hot dogs, and it's going to take her M minutes to eat 35 hot dogs, and if it's the same pace, then these two rates are going to be equal."}, {"video_title": "Proportion word problem (example 2)   7th grade   Khan Academy.mp3", "Sentence": "So it would take her M minutes to eat 35 hot dogs at the same pace. Now, when some of you might have tackled it, you might have had a different equation set up here. Instead of thinking of hot dogs per minute, you might have thought about minutes per hot dog. And so in that situation, if you thought in terms of minutes per hot dog, you might have said, okay, look, it took Micah 66 minutes to eat 21 hot dogs, and it's going to take her M minutes to eat 35 hot dogs, and if it's the same pace, then these two rates are going to be equal. They have to be the same pace. And so then you can solve for M, and actually this one's easier to solve for M. You just multiply both sides by 35. Multiply both sides by 35, and you're left with, on the right-hand side, you're left with just an M, and on the left-hand side, same idea."}, {"video_title": "Proportion word problem (example 2)   7th grade   Khan Academy.mp3", "Sentence": "And so in that situation, if you thought in terms of minutes per hot dog, you might have said, okay, look, it took Micah 66 minutes to eat 21 hot dogs, and it's going to take her M minutes to eat 35 hot dogs, and if it's the same pace, then these two rates are going to be equal. They have to be the same pace. And so then you can solve for M, and actually this one's easier to solve for M. You just multiply both sides by 35. Multiply both sides by 35, and you're left with, on the right-hand side, you're left with just an M, and on the left-hand side, same idea. You're taking 35. You have 35 times 66, 21st, which we already figured out is 110. So 110 is equal to M. So once again, multiple ways to tackle it, but it's important that we got the same answer."}, {"video_title": "Multiplying three numbers in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "All this meant when I wrote these parentheses times next to each other, I'm just going to multiply this expression times this expression times this expression. And since everything is involved in multiplication, it actually doesn't matter what order I multiply in. And so with that in mind, I can swap the order here. This is going to be the same thing as 1.45. That's that right there times 9.2 times 9.2 times 3.01 times 3.01 times 10 to the 8th. Let me do that in that purple color. Times 10 to the 8th, so times 10 to the 8th times 10 to the negative 12th power, 10 to the negative 12th power times 10 to the negative 5th power times 10 to the negative 5th power."}, {"video_title": "Multiplying three numbers in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "This is going to be the same thing as 1.45. That's that right there times 9.2 times 9.2 times 3.01 times 3.01 times 10 to the 8th. Let me do that in that purple color. Times 10 to the 8th, so times 10 to the 8th times 10 to the negative 12th power, 10 to the negative 12th power times 10 to the negative 5th power times 10 to the negative 5th power. And this is useful because now I have all of my powers of 10 right over here. I can put parentheses around that. And I have all of my non-powers of 10 right over there."}, {"video_title": "Multiplying three numbers in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "Times 10 to the 8th, so times 10 to the 8th times 10 to the negative 12th power, 10 to the negative 12th power times 10 to the negative 5th power times 10 to the negative 5th power. And this is useful because now I have all of my powers of 10 right over here. I can put parentheses around that. And I have all of my non-powers of 10 right over there. And so I can simplify it if I have the same base 10 or right over here, so I can add the exponents. This is going to be 10 to the eight minus 12 minus five power, minus five power. And then all of this on the left-hand side, let me get a calculator out."}, {"video_title": "Multiplying three numbers in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "And I have all of my non-powers of 10 right over there. And so I can simplify it if I have the same base 10 or right over here, so I can add the exponents. This is going to be 10 to the eight minus 12 minus five power, minus five power. And then all of this on the left-hand side, let me get a calculator out. I have 1.45, you could do it by hand, but this is a little bit faster and less likely to make a careless mistake, times 9.2 times 3.01, which is equal to 40.1534. So this is equal to 40.1534. And of course, this is going to be multiplied times 10 to this thing."}, {"video_title": "Multiplying three numbers in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "And then all of this on the left-hand side, let me get a calculator out. I have 1.45, you could do it by hand, but this is a little bit faster and less likely to make a careless mistake, times 9.2 times 3.01, which is equal to 40.1534. So this is equal to 40.1534. And of course, this is going to be multiplied times 10 to this thing. And so if you simplify this exponent, you get 40.1534 times 10 to the eight minus 12 is negative four minus five is negative nine, 10 to the negative nine power. Now, you might be tempted to say that this is already in scientific notation because I have some number here times some power of 10, but this is not quite official scientific notation. And that's because in order for it to be in scientific notation, this number right over here has to be greater than or equal to one and less than 10."}, {"video_title": "Multiplying three numbers in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "And of course, this is going to be multiplied times 10 to this thing. And so if you simplify this exponent, you get 40.1534 times 10 to the eight minus 12 is negative four minus five is negative nine, 10 to the negative nine power. Now, you might be tempted to say that this is already in scientific notation because I have some number here times some power of 10, but this is not quite official scientific notation. And that's because in order for it to be in scientific notation, this number right over here has to be greater than or equal to one and less than 10. And this is obviously not less than 10. Essentially for it to be in scientific notation, you want a non-zero digit right over here, and then you want your decimal and then the rest of everything else. So here, and you want a non-zero single digit over here."}, {"video_title": "Multiplying three numbers in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "And that's because in order for it to be in scientific notation, this number right over here has to be greater than or equal to one and less than 10. And this is obviously not less than 10. Essentially for it to be in scientific notation, you want a non-zero digit right over here, and then you want your decimal and then the rest of everything else. So here, and you want a non-zero single digit over here. Here we obviously have, here we have two digits. This is larger than 10, or this is greater than or equal to 10. You want this thing to be less than 10 and greater than or equal to one."}, {"video_title": "Multiplying three numbers in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "So here, and you want a non-zero single digit over here. Here we obviously have, here we have two digits. This is larger than 10, or this is greater than or equal to 10. You want this thing to be less than 10 and greater than or equal to one. So the best way to do that is to write this thing right over here in scientific notation. This is the same thing as 4.01534 times 10. And one way to think about it is to go from 40 to four, we had to move this decimal over to the left."}, {"video_title": "Multiplying three numbers in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "You want this thing to be less than 10 and greater than or equal to one. So the best way to do that is to write this thing right over here in scientific notation. This is the same thing as 4.01534 times 10. And one way to think about it is to go from 40 to four, we had to move this decimal over to the left. Moving a decimal over to the left to go from 40 to four, you're dividing by 10. So you have to multiply by 10 so it all equals out. Divide by 10 and then multiply by 10."}, {"video_title": "Multiplying three numbers in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "And one way to think about it is to go from 40 to four, we had to move this decimal over to the left. Moving a decimal over to the left to go from 40 to four, you're dividing by 10. So you have to multiply by 10 so it all equals out. Divide by 10 and then multiply by 10. Or another way to write it, or another way to think about it is 4.0 and all this stuff times 10 is going to be 40.1534. And so you're going to have four, all of this times 10 to the first power, that's the same thing as 10, times this thing, times 10 to the negative ninth power. And then once again, powers of 10, so it's 10 to the first times 10 to the negative nine is going to be 10 to the negative eighth power."}, {"video_title": "Multiplying three numbers in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "Divide by 10 and then multiply by 10. Or another way to write it, or another way to think about it is 4.0 and all this stuff times 10 is going to be 40.1534. And so you're going to have four, all of this times 10 to the first power, that's the same thing as 10, times this thing, times 10 to the negative ninth power. And then once again, powers of 10, so it's 10 to the first times 10 to the negative nine is going to be 10 to the negative eighth power. 10 to the negative eighth power and we still have this 4.01534 times 10 to the negative eight and now we have written it in scientific, now we have written it in scientific notation. Now they wanted us to express it in both decimal and scientific notation. And when they're asking us to write it in decimal notation, they essentially want us to multiply this out, expand this out."}, {"video_title": "Multiplying three numbers in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "And then once again, powers of 10, so it's 10 to the first times 10 to the negative nine is going to be 10 to the negative eighth power. 10 to the negative eighth power and we still have this 4.01534 times 10 to the negative eight and now we have written it in scientific, now we have written it in scientific notation. Now they wanted us to express it in both decimal and scientific notation. And when they're asking us to write it in decimal notation, they essentially want us to multiply this out, expand this out. And so the way to think about it, write these digits out. So I have 4.01534. And if I'm just looking at this number, I start with the decimal right over here."}, {"video_title": "Multiplying three numbers in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "And when they're asking us to write it in decimal notation, they essentially want us to multiply this out, expand this out. And so the way to think about it, write these digits out. So I have 4.01534. And if I'm just looking at this number, I start with the decimal right over here. Now every time I divide by 10, or if I multiply by 10 to the negative one, I'm moving this over to the left one spot. So 10 to the negative one, if I multiply by 10 to the negative one, that's the same thing as dividing by 10. And so I'm moving the decimal over to the left one."}, {"video_title": "Multiplying three numbers in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "And if I'm just looking at this number, I start with the decimal right over here. Now every time I divide by 10, or if I multiply by 10 to the negative one, I'm moving this over to the left one spot. So 10 to the negative one, if I multiply by 10 to the negative one, that's the same thing as dividing by 10. And so I'm moving the decimal over to the left one. Here, I'm multiplying by 10 to the negative eight. Or you could say I'm dividing by 10 to the eighth power. So I'm going to want to move the decimal to the left eight times."}, {"video_title": "Multiplying three numbers in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "And so I'm moving the decimal over to the left one. Here, I'm multiplying by 10 to the negative eight. Or you could say I'm dividing by 10 to the eighth power. So I'm going to want to move the decimal to the left eight times. So move a decimal to left eight times. And one way to remember it, look, this is a very, very, very, very small number. If I multiply this, I should get a smaller number."}, {"video_title": "Multiplying three numbers in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "So I'm going to want to move the decimal to the left eight times. So move a decimal to left eight times. And one way to remember it, look, this is a very, very, very, very small number. If I multiply this, I should get a smaller number. So I should be moving the decimal to the left. If this was a positive eight, then this would be a very large number. And so if I multiply by a large power of 10, I'm going to be moving the decimal to the right."}, {"video_title": "Multiplying three numbers in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "If I multiply this, I should get a smaller number. So I should be moving the decimal to the left. If this was a positive eight, then this would be a very large number. And so if I multiply by a large power of 10, I'm going to be moving the decimal to the right. So this whole thing should evaluate to being smaller than 4.01534. So I move the decimal eight times to the left, eight times to the left. I move it one time to the left to get it right over here."}, {"video_title": "Multiplying three numbers in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "And so if I multiply by a large power of 10, I'm going to be moving the decimal to the right. So this whole thing should evaluate to being smaller than 4.01534. So I move the decimal eight times to the left, eight times to the left. I move it one time to the left to get it right over here. And then the next seven times, I'm just going to add zeros. So one, two, three, four, five, six, seven zeros. And I'll put a zero in front of the decimal just to clarify it."}, {"video_title": "Multiplying three numbers in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "I move it one time to the left to get it right over here. And then the next seven times, I'm just going to add zeros. So one, two, three, four, five, six, seven zeros. And I'll put a zero in front of the decimal just to clarify it. So now I notice if you include this digit right over here, I have a total of eight digits. I have an eight, I have eight, sorry, I have seven zeros and this digit gives us eight. So again, one, two, three, four, five, six, seven, eight."}, {"video_title": "Multiplying three numbers in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "And I'll put a zero in front of the decimal just to clarify it. So now I notice if you include this digit right over here, I have a total of eight digits. I have an eight, I have eight, sorry, I have seven zeros and this digit gives us eight. So again, one, two, three, four, five, six, seven, eight. The best way to think about it is I started with the decimal right here. I moved once, twice, three, four, five, six, seven, eight times. That's what multiplying times 10 to the negative eight did for us."}, {"video_title": "Multiplying three numbers in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "So again, one, two, three, four, five, six, seven, eight. The best way to think about it is I started with the decimal right here. I moved once, twice, three, four, five, six, seven, eight times. That's what multiplying times 10 to the negative eight did for us. And I get this number right over here. And when you see a number like this, you start to appreciate why we write things in scientific notation. This is much easier to, it takes less space to write."}, {"video_title": "Multiplying three numbers in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "That's what multiplying times 10 to the negative eight did for us. And I get this number right over here. And when you see a number like this, you start to appreciate why we write things in scientific notation. This is much easier to, it takes less space to write. And you immediately know roughly how big this number is. This is much harder to write. You might even forget a zero when you write it or you might add a zero."}, {"video_title": "Multiplying three numbers in scientific notation (example)   Pre-Algebra   Khan Academy.mp3", "Sentence": "This is much easier to, it takes less space to write. And you immediately know roughly how big this number is. This is much harder to write. You might even forget a zero when you write it or you might add a zero. And now the person has to sit and count the zeros to figure out essentially how large or get a rough sense of how large this thing is. It's one, two, three, four, five, six, seven zeros and you have this digit right here. That's what gets us to that eight."}, {"video_title": "Multiplying binomials intro   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "And we want to write that product in standard quadratic form, which is just a fancy way of saying a form where you have some coefficient on the second degree term, ax squared, plus some coefficient b on the first degree term, plus the constant term. So this right over here would be standard quadratic form. So that's the form that we want to express this product in. And I encourage you to pause the video and try to work through it on your own. All right, now let's work through this. And the key when we're multiplying two binomials like this, or actually when we're multiplying any polynomials, is just to remember the distributive property that we all by this point know quite well. So what we could view this as is we can distribute this x minus four, this entire expression, over the x and the seven."}, {"video_title": "Multiplying binomials intro   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "And I encourage you to pause the video and try to work through it on your own. All right, now let's work through this. And the key when we're multiplying two binomials like this, or actually when we're multiplying any polynomials, is just to remember the distributive property that we all by this point know quite well. So what we could view this as is we can distribute this x minus four, this entire expression, over the x and the seven. So we could say that this is the same thing as x minus four times x, plus x minus four times seven. So let's write that. So x minus four times x, or we could write this as x times x minus four."}, {"video_title": "Multiplying binomials intro   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "So what we could view this as is we can distribute this x minus four, this entire expression, over the x and the seven. So we could say that this is the same thing as x minus four times x, plus x minus four times seven. So let's write that. So x minus four times x, or we could write this as x times x minus four. That's distributing the, or multiplying the x minus four times x. That's right there. Plus seven times x minus four."}, {"video_title": "Multiplying binomials intro   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "So x minus four times x, or we could write this as x times x minus four. That's distributing the, or multiplying the x minus four times x. That's right there. Plus seven times x minus four. Times x minus four. Notice, all we did is distribute the x minus four. We took this whole thing and we multiplied it by each term over here."}, {"video_title": "Multiplying binomials intro   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "Plus seven times x minus four. Times x minus four. Notice, all we did is distribute the x minus four. We took this whole thing and we multiplied it by each term over here. We multiplied x by x minus four, and we multiplied seven by x minus four. Now we see that we have these, I guess you'd call them two separate terms, and to simplify each of them, or to multiply them out, we just have to distribute in this first, we have to distribute this blue x, and over here we have to distribute this blue seven. So let's do that."}, {"video_title": "Multiplying binomials intro   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "We took this whole thing and we multiplied it by each term over here. We multiplied x by x minus four, and we multiplied seven by x minus four. Now we see that we have these, I guess you'd call them two separate terms, and to simplify each of them, or to multiply them out, we just have to distribute in this first, we have to distribute this blue x, and over here we have to distribute this blue seven. So let's do that. So here we could say x times x is going to be x squared. X times, we have a negative here, so we could say negative four is going to be negative four x. And just like that, we get x squared minus four x."}, {"video_title": "Multiplying binomials intro   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "So let's do that. So here we could say x times x is going to be x squared. X times, we have a negative here, so we could say negative four is going to be negative four x. And just like that, we get x squared minus four x. And then over here, we have seven times x, so that's going to be plus seven x. And then we have seven times the negative four, which is negative 28. And we are almost done."}, {"video_title": "Multiplying binomials intro   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "And just like that, we get x squared minus four x. And then over here, we have seven times x, so that's going to be plus seven x. And then we have seven times the negative four, which is negative 28. And we are almost done. We can simplify it a little bit more. We have two first degree terms here. If I have negative four x's, and to that I add seven x's, what is that going to be?"}, {"video_title": "Multiplying binomials intro   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "And we are almost done. We can simplify it a little bit more. We have two first degree terms here. If I have negative four x's, and to that I add seven x's, what is that going to be? Well, those two terms together, these two terms together are going to be negative four plus seven x's. Negative four plus seven. Negative four plus seven x's."}, {"video_title": "Multiplying binomials intro   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "If I have negative four x's, and to that I add seven x's, what is that going to be? Well, those two terms together, these two terms together are going to be negative four plus seven x's. Negative four plus seven. Negative four plus seven x's. So all I'm doing here, I'm making it very clear that I'm adding these two coefficients, and then we have all the other terms. We have the x squared, x squared plus this, and then we have the minus 28. And we're at the home stretch."}, {"video_title": "Multiplying binomials intro   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "Negative four plus seven x's. So all I'm doing here, I'm making it very clear that I'm adding these two coefficients, and then we have all the other terms. We have the x squared, x squared plus this, and then we have the minus 28. And we're at the home stretch. This would simplify to x squared. Now negative four plus seven is three, so this is going to be plus three x. That's what these two middle terms simplify to, to three x, and then we have minus 28."}, {"video_title": "Multiplying binomials intro   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "And we're at the home stretch. This would simplify to x squared. Now negative four plus seven is three, so this is going to be plus three x. That's what these two middle terms simplify to, to three x, and then we have minus 28. Minus 28. And just like that, we are done. And a fun thing to think about, and notice it's in the same form."}, {"video_title": "Multiplying binomials intro   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "That's what these two middle terms simplify to, to three x, and then we have minus 28. Minus 28. And just like that, we are done. And a fun thing to think about, and notice it's in the same form. If we were to compare, a is one, b is three, and c is negative 28. But it's interesting here to look at the pattern. When we multiply these two binomials, especially these two binomials, where the coefficient on the x term was a one."}, {"video_title": "Multiplying binomials intro   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "And a fun thing to think about, and notice it's in the same form. If we were to compare, a is one, b is three, and c is negative 28. But it's interesting here to look at the pattern. When we multiply these two binomials, especially these two binomials, where the coefficient on the x term was a one. Notice, we have x times x, that's what actually forms the x squared term over here. We have negative four, let me do this in a new color. We have negative four times, that's not a new color."}, {"video_title": "Multiplying binomials intro   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "When we multiply these two binomials, especially these two binomials, where the coefficient on the x term was a one. Notice, we have x times x, that's what actually forms the x squared term over here. We have negative four, let me do this in a new color. We have negative four times, that's not a new color. We have negative four times seven, which is going to be negative 28. And then how did we get this middle term? How did we get this three x?"}, {"video_title": "Multiplying binomials intro   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "We have negative four times, that's not a new color. We have negative four times seven, which is going to be negative 28. And then how did we get this middle term? How did we get this three x? Well, you had the negative four x plus the seven x, or you had the negative four plus the seven times x. You had the negative four plus the seven times x. So hopefully you see a little bit of a pattern here."}, {"video_title": "Multiplying binomials intro   Mathematics II   High School Math   Khan Academy.mp3", "Sentence": "How did we get this three x? Well, you had the negative four x plus the seven x, or you had the negative four plus the seven times x. You had the negative four plus the seven times x. So hopefully you see a little bit of a pattern here. If you're multiplying two binomials where the coefficients on the x term are both one, it's going to be x squared, and then the last term, the constant term, is going to be the product of these two constants, negative four and seven. And then the first degree term, right over here, its coefficient is going to be the sum of these two constants, negative four and seven. Now this might, you could view this pattern, if you practice it, as just something that'll help you multiply binomials a little bit faster."}, {"video_title": "Percent word problem example 1   Ratios, rates, and percentages   6th grade   Khan Academy.mp3", "Sentence": "Let's say I go to the fruit store today and they have a sale on guavas. Everything is 30% off. This is for guavas. And it's only today. So I say, you know what, let me go buy a bunch of guavas. So I go and I buy 6 guavas. So I buy 6 guavas."}, {"video_title": "Percent word problem example 1   Ratios, rates, and percentages   6th grade   Khan Academy.mp3", "Sentence": "And it's only today. So I say, you know what, let me go buy a bunch of guavas. So I go and I buy 6 guavas. So I buy 6 guavas. It ends up when I go to the register, and we're assuming no tax for it's a grocery and I live in a state where they don't tax groceries. So for the 6 guavas, they charge me, I get the 30% off, they charge me $12.60. So this is the 30% off sale price on 6 guavas."}, {"video_title": "Percent word problem example 1   Ratios, rates, and percentages   6th grade   Khan Academy.mp3", "Sentence": "So I buy 6 guavas. It ends up when I go to the register, and we're assuming no tax for it's a grocery and I live in a state where they don't tax groceries. So for the 6 guavas, they charge me, I get the 30% off, they charge me $12.60. So this is the 30% off sale price on 6 guavas. I go home and then my wife tells me, you know, Sal, can you go get 2 more guavas tomorrow? I say sure. So the next day I go and I want to buy 2 more guavas."}, {"video_title": "Percent word problem example 1   Ratios, rates, and percentages   6th grade   Khan Academy.mp3", "Sentence": "So this is the 30% off sale price on 6 guavas. I go home and then my wife tells me, you know, Sal, can you go get 2 more guavas tomorrow? I say sure. So the next day I go and I want to buy 2 more guavas. So 2 guavas. But now the sale is off. There's no more 30%."}, {"video_title": "Percent word problem example 1   Ratios, rates, and percentages   6th grade   Khan Academy.mp3", "Sentence": "So the next day I go and I want to buy 2 more guavas. So 2 guavas. But now the sale is off. There's no more 30%. That was only that first day that I bought the 6. So how much are those 2 guavas going to cost me? How much are those 2 guavas going to cost at full price?"}, {"video_title": "Percent word problem example 1   Ratios, rates, and percentages   6th grade   Khan Academy.mp3", "Sentence": "There's no more 30%. That was only that first day that I bought the 6. So how much are those 2 guavas going to cost me? How much are those 2 guavas going to cost at full price? At full price. So a good place to start is to think about how much would those 6 guavas cost us at full price? This is the sale price right here."}, {"video_title": "Percent word problem example 1   Ratios, rates, and percentages   6th grade   Khan Academy.mp3", "Sentence": "How much are those 2 guavas going to cost at full price? At full price. So a good place to start is to think about how much would those 6 guavas cost us at full price? This is the sale price right here. How much would those have cost me at full price? So let's do a little bit of algebra here. Let me pick a suitable color for the algebra."}, {"video_title": "Percent word problem example 1   Ratios, rates, and percentages   6th grade   Khan Academy.mp3", "Sentence": "This is the sale price right here. How much would those have cost me at full price? So let's do a little bit of algebra here. Let me pick a suitable color for the algebra. So let's say that x is equal to the cost of 6 guavas at full price. So essentially if we take 30% off of this we should get $12.60. So let's do that."}, {"video_title": "Percent word problem example 1   Ratios, rates, and percentages   6th grade   Khan Academy.mp3", "Sentence": "Let me pick a suitable color for the algebra. So let's say that x is equal to the cost of 6 guavas at full price. So essentially if we take 30% off of this we should get $12.60. So let's do that. So if we have the full price of 6 guavas, we're going to take 30% off of that so that's the same thing as.30 or I could just write.3. I could ignore that 0 if I like. Actually let me write it like this."}, {"video_title": "Percent word problem example 1   Ratios, rates, and percentages   6th grade   Khan Academy.mp3", "Sentence": "So let's do that. So if we have the full price of 6 guavas, we're going to take 30% off of that so that's the same thing as.30 or I could just write.3. I could ignore that 0 if I like. Actually let me write it like this. My wife is always bugging me to write 0s before decimals. So that's the full price of 6 guavas minus 0.30 times the full price of guavas. So I'm literally just taking 30% off of the full price off of the full price."}, {"video_title": "Percent word problem example 1   Ratios, rates, and percentages   6th grade   Khan Academy.mp3", "Sentence": "Actually let me write it like this. My wife is always bugging me to write 0s before decimals. So that's the full price of 6 guavas minus 0.30 times the full price of guavas. So I'm literally just taking 30% off of the full price off of the full price. This is how we figure out the sale price. This is going to be equal to that $12.60 right there. That's going to be equal to $12.60."}, {"video_title": "Percent word problem example 1   Ratios, rates, and percentages   6th grade   Khan Academy.mp3", "Sentence": "So I'm literally just taking 30% off of the full price off of the full price. This is how we figure out the sale price. This is going to be equal to that $12.60 right there. That's going to be equal to $12.60. I just took 30% off of the full price. And now we just do algebra. We can imagine there's a 1 in front."}, {"video_title": "Percent word problem example 1   Ratios, rates, and percentages   6th grade   Khan Academy.mp3", "Sentence": "That's going to be equal to $12.60. I just took 30% off of the full price. And now we just do algebra. We can imagine there's a 1 in front. You know x is the same thing as 1x. So 1x minus 0.3x is going to be equal to.7x. So we get.7x or you could say.70 if you like."}, {"video_title": "Percent word problem example 1   Ratios, rates, and percentages   6th grade   Khan Academy.mp3", "Sentence": "We can imagine there's a 1 in front. You know x is the same thing as 1x. So 1x minus 0.3x is going to be equal to.7x. So we get.7x or you could say.70 if you like. Same number. 0.7x is equal to $12.60. Once you get used to these problems, you might just skip straight to this step right here."}, {"video_title": "Percent word problem example 1   Ratios, rates, and percentages   6th grade   Khan Academy.mp3", "Sentence": "So we get.7x or you could say.70 if you like. Same number. 0.7x is equal to $12.60. Once you get used to these problems, you might just skip straight to this step right here. You say, hey, 70% of the full price is equal to my sale price. I took 30% off. This is 70% of the full price."}, {"video_title": "Percent word problem example 1   Ratios, rates, and percentages   6th grade   Khan Academy.mp3", "Sentence": "Once you get used to these problems, you might just skip straight to this step right here. You say, hey, 70% of the full price is equal to my sale price. I took 30% off. This is 70% of the full price. You might just skip to this step once you get used to these problems a little bit. Let's solve for x. Divide both sides by.7. x is equal to $12.60 divided by 0.7."}, {"video_title": "Percent word problem example 1   Ratios, rates, and percentages   6th grade   Khan Academy.mp3", "Sentence": "This is 70% of the full price. You might just skip to this step once you get used to these problems a little bit. Let's solve for x. Divide both sides by.7. x is equal to $12.60 divided by 0.7. We could use a calculator, but it's always good to get a little bit of practice dividing decimals. 0.7 goes into $12.60. Let's multiply both of these numbers by 10, which is what we do when we move both of their decimals one to the right."}, {"video_title": "Percent word problem example 1   Ratios, rates, and percentages   6th grade   Khan Academy.mp3", "Sentence": "x is equal to $12.60 divided by 0.7. We could use a calculator, but it's always good to get a little bit of practice dividing decimals. 0.7 goes into $12.60. Let's multiply both of these numbers by 10, which is what we do when we move both of their decimals one to the right. So the.7 becomes a 7. Ignore that right there. The $12.60 becomes $126."}, {"video_title": "Percent word problem example 1   Ratios, rates, and percentages   6th grade   Khan Academy.mp3", "Sentence": "Let's multiply both of these numbers by 10, which is what we do when we move both of their decimals one to the right. So the.7 becomes a 7. Ignore that right there. The $12.60 becomes $126. Put the decimal right there. We're ready to just do straight up long division. This is now a 7, not a.7."}, {"video_title": "Percent word problem example 1   Ratios, rates, and percentages   6th grade   Khan Academy.mp3", "Sentence": "The $12.60 becomes $126. Put the decimal right there. We're ready to just do straight up long division. This is now a 7, not a.7. So 7 goes into 12 one time. 1 times 7 is 7. 12 minus 7 is 5."}, {"video_title": "Percent word problem example 1   Ratios, rates, and percentages   6th grade   Khan Academy.mp3", "Sentence": "This is now a 7, not a.7. So 7 goes into 12 one time. 1 times 7 is 7. 12 minus 7 is 5. Bring down the 6. 7 goes into 56 eight times. 8 times 7 is 56."}, {"video_title": "Percent word problem example 1   Ratios, rates, and percentages   6th grade   Khan Academy.mp3", "Sentence": "12 minus 7 is 5. Bring down the 6. 7 goes into 56 eight times. 8 times 7 is 56. Then we have no remainder. It's 18, and there's nothing behind the decimal point. In our case, it's $18."}, {"video_title": "Percent word problem example 1   Ratios, rates, and percentages   6th grade   Khan Academy.mp3", "Sentence": "8 times 7 is 56. Then we have no remainder. It's 18, and there's nothing behind the decimal point. In our case, it's $18. x is equal to $18. Remember what x was. x was the full price of 6 guavas."}, {"video_title": "Percent word problem example 1   Ratios, rates, and percentages   6th grade   Khan Academy.mp3", "Sentence": "In our case, it's $18. x is equal to $18. Remember what x was. x was the full price of 6 guavas. x was the full price of 6 guavas. x is the full price of 6 guavas. Now the question is, how much will 2 guavas cost me the full price?"}, {"video_title": "Percent word problem example 1   Ratios, rates, and percentages   6th grade   Khan Academy.mp3", "Sentence": "x was the full price of 6 guavas. x was the full price of 6 guavas. x is the full price of 6 guavas. Now the question is, how much will 2 guavas cost me the full price? Well, this is full price of 6. So you immediately could figure out what's the full price of 1 guava. You divide 18 by 6."}, {"video_title": "Percent word problem example 1   Ratios, rates, and percentages   6th grade   Khan Academy.mp3", "Sentence": "Now the question is, how much will 2 guavas cost me the full price? Well, this is full price of 6. So you immediately could figure out what's the full price of 1 guava. You divide 18 by 6. 18 divided by 6 is $3. That's $3 per guava at full price. And they're asking us, we want 2 guavas."}, {"video_title": "Percent word problem example 1   Ratios, rates, and percentages   6th grade   Khan Academy.mp3", "Sentence": "You divide 18 by 6. 18 divided by 6 is $3. That's $3 per guava at full price. And they're asking us, we want 2 guavas. So 2 guavas is going to be 2 times $3. So this is going to be $6. Another way you could have done it, you could have just said, hey, 6 at full price are going to cost me $18."}, {"video_title": "Percent word problem example 1   Ratios, rates, and percentages   6th grade   Khan Academy.mp3", "Sentence": "And they're asking us, we want 2 guavas. So 2 guavas is going to be 2 times $3. So this is going to be $6. Another way you could have done it, you could have just said, hey, 6 at full price are going to cost me $18. 2 is 1 third of 6. So 1 third of $18 is $6. So just to give a quick review of what we did, we said the sale price on 6 guavas, $12.60."}, {"video_title": "Percent word problem example 1   Ratios, rates, and percentages   6th grade   Khan Academy.mp3", "Sentence": "Another way you could have done it, you could have just said, hey, 6 at full price are going to cost me $18. 2 is 1 third of 6. So 1 third of $18 is $6. So just to give a quick review of what we did, we said the sale price on 6 guavas, $12.60. That's 30% off the full price. Or you could say this is 70% of the full price. And so you could say 30% x is the full price of 6 guavas."}, {"video_title": "Percent word problem example 1   Ratios, rates, and percentages   6th grade   Khan Academy.mp3", "Sentence": "So just to give a quick review of what we did, we said the sale price on 6 guavas, $12.60. That's 30% off the full price. Or you could say this is 70% of the full price. And so you could say 30% x is the full price of 6 guavas. You could say the full price of 6 guavas minus 30% of the full price of 6 guavas is equal to $12.60. And that's equivalent to saying 70% of the full price is $12.60. Then we just solve this algebraically."}, {"video_title": "Percent word problem example 1   Ratios, rates, and percentages   6th grade   Khan Academy.mp3", "Sentence": "And so you could say 30% x is the full price of 6 guavas. You could say the full price of 6 guavas minus 30% of the full price of 6 guavas is equal to $12.60. And that's equivalent to saying 70% of the full price is $12.60. Then we just solve this algebraically. Divide both sides by.7. And then we got x, the full price of 6 guavas is $18. Or that's $3 per guava or $6 for 2."}, {"video_title": "Graph from slope-intercept equation example   Algebra I   Khan Academy.mp3", "Sentence": "Now, whenever you see an equation in this form, this is called slope intercept form. And the general way of writing it is y is equal to mx plus b. Where m is the slope, and here in this case m is equal to 1 3rd, so let me write that down. And b is the y intercept. So in this case b is equal to negative 2. And you know that b is the y intercept because we know that the y intercept occurs when x is equal to 0. So if x is equal to 0 in either of these situations, this term just becomes 0 and y will be equal to b."}, {"video_title": "Graph from slope-intercept equation example   Algebra I   Khan Academy.mp3", "Sentence": "And b is the y intercept. So in this case b is equal to negative 2. And you know that b is the y intercept because we know that the y intercept occurs when x is equal to 0. So if x is equal to 0 in either of these situations, this term just becomes 0 and y will be equal to b. So that's what we mean by b is the y intercept. So whenever you look at an equation in this form, it's actually fairly straightforward to graph this line. b is the y intercept, in this case it is negative 2."}, {"video_title": "Graph from slope-intercept equation example   Algebra I   Khan Academy.mp3", "Sentence": "So if x is equal to 0 in either of these situations, this term just becomes 0 and y will be equal to b. So that's what we mean by b is the y intercept. So whenever you look at an equation in this form, it's actually fairly straightforward to graph this line. b is the y intercept, in this case it is negative 2. So that means that this line must intersect the y axis at y is equal to negative 2. So it's this point right here. Negative 1, negative 2."}, {"video_title": "Graph from slope-intercept equation example   Algebra I   Khan Academy.mp3", "Sentence": "b is the y intercept, in this case it is negative 2. So that means that this line must intersect the y axis at y is equal to negative 2. So it's this point right here. Negative 1, negative 2. This is the point 0, negative 2. If you don't believe me, there's nothing magical about this. Try evaluating or try solving for y when x is equal to 0."}, {"video_title": "Graph from slope-intercept equation example   Algebra I   Khan Academy.mp3", "Sentence": "Negative 1, negative 2. This is the point 0, negative 2. If you don't believe me, there's nothing magical about this. Try evaluating or try solving for y when x is equal to 0. When x is equal to 0, this term cancels out and you're just left with y is equal to negative 2. So that's the y intercept right there. Now, this 1 3rd tells us the slope of the line."}, {"video_title": "Graph from slope-intercept equation example   Algebra I   Khan Academy.mp3", "Sentence": "Try evaluating or try solving for y when x is equal to 0. When x is equal to 0, this term cancels out and you're just left with y is equal to negative 2. So that's the y intercept right there. Now, this 1 3rd tells us the slope of the line. How much do we change in y for any change in x? So this tells us that 1 3rd, so that right there is the slope. So it tells us that 1 3rd is equal to the change in y over the change in x."}, {"video_title": "Graph from slope-intercept equation example   Algebra I   Khan Academy.mp3", "Sentence": "Now, this 1 3rd tells us the slope of the line. How much do we change in y for any change in x? So this tells us that 1 3rd, so that right there is the slope. So it tells us that 1 3rd is equal to the change in y over the change in x. Or another way to think about it, if x changes by 3, then y will change by 1. So let me graph that. So we know that this point is on the graph."}, {"video_title": "Graph from slope-intercept equation example   Algebra I   Khan Academy.mp3", "Sentence": "So it tells us that 1 3rd is equal to the change in y over the change in x. Or another way to think about it, if x changes by 3, then y will change by 1. So let me graph that. So we know that this point is on the graph. That's the y intercept. The slope tells us that if x changes by 3, so let me go 3 to the right, 1, 2, 3, that y will change by 1. So this must also be a point on the graph."}, {"video_title": "Graph from slope-intercept equation example   Algebra I   Khan Academy.mp3", "Sentence": "So we know that this point is on the graph. That's the y intercept. The slope tells us that if x changes by 3, so let me go 3 to the right, 1, 2, 3, that y will change by 1. So this must also be a point on the graph. And we could keep doing that. If x changes by 3, y changes by 1. If x goes down by 3, y will go down by 1."}, {"video_title": "Graph from slope-intercept equation example   Algebra I   Khan Academy.mp3", "Sentence": "So this must also be a point on the graph. And we could keep doing that. If x changes by 3, y changes by 1. If x goes down by 3, y will go down by 1. If x goes down by 6, y will go down by 2. It's that same ratio. So 1, 2, 3, 4, 5, 6, 1, 2."}, {"video_title": "Graph from slope-intercept equation example   Algebra I   Khan Academy.mp3", "Sentence": "If x goes down by 3, y will go down by 1. If x goes down by 6, y will go down by 2. It's that same ratio. So 1, 2, 3, 4, 5, 6, 1, 2. And you can see all of these points are on the line. And the line is the graph of this equation up here. So let me graph it."}, {"video_title": "Slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "There's a lot of different ways that you can represent a linear equation. So for example, if you had the linear equation y is equal to two x plus three, that's one way to represent it, but I could represent this in an infinite number of ways. I could, let's see, I could subtract two x from both sides. I could write this as negative two x plus y is equal to three. I could manipulate it in ways where I get it to, and I'm not gonna do it right now, but this is another way of writing that same thing. Y minus five is equal to two times x minus one. You could actually simplify this and you could get either this equation here or that equation up on top."}, {"video_title": "Slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "I could write this as negative two x plus y is equal to three. I could manipulate it in ways where I get it to, and I'm not gonna do it right now, but this is another way of writing that same thing. Y minus five is equal to two times x minus one. You could actually simplify this and you could get either this equation here or that equation up on top. These are all equivalent. You can get from one to the other with logical algebraic operations. So there's an infinite number of ways to represent a given linear equation, but what I wanna focus on in this video is this representation in particular because this one is a very useful representation of a linear equation, and we'll see in future videos this one and this one can also be useful depending on what you are looking for, but we're gonna focus on this one."}, {"video_title": "Slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "You could actually simplify this and you could get either this equation here or that equation up on top. These are all equivalent. You can get from one to the other with logical algebraic operations. So there's an infinite number of ways to represent a given linear equation, but what I wanna focus on in this video is this representation in particular because this one is a very useful representation of a linear equation, and we'll see in future videos this one and this one can also be useful depending on what you are looking for, but we're gonna focus on this one. And this one right over here, it's often called slope-intercept form. Slope-intercept form. And hopefully in a few minutes it will be obvious why it is called slope-intercept form."}, {"video_title": "Slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "So there's an infinite number of ways to represent a given linear equation, but what I wanna focus on in this video is this representation in particular because this one is a very useful representation of a linear equation, and we'll see in future videos this one and this one can also be useful depending on what you are looking for, but we're gonna focus on this one. And this one right over here, it's often called slope-intercept form. Slope-intercept form. And hopefully in a few minutes it will be obvious why it is called slope-intercept form. And before I explain that to you, let's just try to graph this thing. I'm gonna try to graph it. I'm just gonna plot some points here."}, {"video_title": "Slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "And hopefully in a few minutes it will be obvious why it is called slope-intercept form. And before I explain that to you, let's just try to graph this thing. I'm gonna try to graph it. I'm just gonna plot some points here. So x comma y, and I'm gonna pick some x values where it's easy to calculate the y values. So maybe the easiest is if x is equal to zero. If x is equal to zero, then two times zero is zero."}, {"video_title": "Slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "I'm just gonna plot some points here. So x comma y, and I'm gonna pick some x values where it's easy to calculate the y values. So maybe the easiest is if x is equal to zero. If x is equal to zero, then two times zero is zero. That term goes away, and you're only left with this term right over here, y is equal to three. Y is equal to three. And so if we were to plot this, actually let me start plotting it."}, {"video_title": "Slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "If x is equal to zero, then two times zero is zero. That term goes away, and you're only left with this term right over here, y is equal to three. Y is equal to three. And so if we were to plot this, actually let me start plotting it. So that is my y-axis. And let me do the x-axis. So that can be my x, oh that's not as straight as I would like it."}, {"video_title": "Slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "And so if we were to plot this, actually let me start plotting it. So that is my y-axis. And let me do the x-axis. So that can be my x, oh that's not as straight as I would like it. So that looks pretty good. All right, that is my x-axis. And let me mark off some hash marks here."}, {"video_title": "Slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "So that can be my x, oh that's not as straight as I would like it. So that looks pretty good. All right, that is my x-axis. And let me mark off some hash marks here. So this is x equals one, x equals two, x equals three, this is y equals, let me do this, y equals one, y equals two, y equals three, and obviously I can keep going, I can keep going. This would be y is equal to negative one. This would be x is equal to negative one, negative two, negative three, so on and so forth."}, {"video_title": "Slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "And let me mark off some hash marks here. So this is x equals one, x equals two, x equals three, this is y equals, let me do this, y equals one, y equals two, y equals three, and obviously I can keep going, I can keep going. This would be y is equal to negative one. This would be x is equal to negative one, negative two, negative three, so on and so forth. So this point right over here, zero comma three, this is x is zero, y is three. Well, the point that represents when x is equal to zero and y equals three, this is, we're right on the y-axis. If there of a line going through it and this line contains this point, this is going to be the y-intercept."}, {"video_title": "Slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "This would be x is equal to negative one, negative two, negative three, so on and so forth. So this point right over here, zero comma three, this is x is zero, y is three. Well, the point that represents when x is equal to zero and y equals three, this is, we're right on the y-axis. If there of a line going through it and this line contains this point, this is going to be the y-intercept. So one way to think about it, the reason why this is called slope-intercept form, is it's very easy to calculate the y-intercept. The y-intercept here is going to happen when it's written in this form, it's going to happen when x is equal to zero and y is equal to three. It's going to be this point right over here."}, {"video_title": "Slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "If there of a line going through it and this line contains this point, this is going to be the y-intercept. So one way to think about it, the reason why this is called slope-intercept form, is it's very easy to calculate the y-intercept. The y-intercept here is going to happen when it's written in this form, it's going to happen when x is equal to zero and y is equal to three. It's going to be this point right over here. So it's very easy to figure out the intercept, the y-intercept from this form. Now you might be saying, oh, well it's a slope-intercept form, it must also be easy to figure out the slope from this form. And if you made that conclusion, you would be correct, and we're about to see that in a few seconds."}, {"video_title": "Slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "It's going to be this point right over here. So it's very easy to figure out the intercept, the y-intercept from this form. Now you might be saying, oh, well it's a slope-intercept form, it must also be easy to figure out the slope from this form. And if you made that conclusion, you would be correct, and we're about to see that in a few seconds. So let's plot some more points here, and I'm just going to keep increasing x by one. So if you increase x by one, so we could write that our delta x, our change in x, delta Greek letter, this triangle's Greek letter delta represents change in, change in x here is one. We just increased x by one."}, {"video_title": "Slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "And if you made that conclusion, you would be correct, and we're about to see that in a few seconds. So let's plot some more points here, and I'm just going to keep increasing x by one. So if you increase x by one, so we could write that our delta x, our change in x, delta Greek letter, this triangle's Greek letter delta represents change in, change in x here is one. We just increased x by one. What's going to be our corresponding change in y? What's going to be our change in y? So let's see, when x is equal to one, you have two times one plus three is going to be five."}, {"video_title": "Slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "We just increased x by one. What's going to be our corresponding change in y? What's going to be our change in y? So let's see, when x is equal to one, you have two times one plus three is going to be five. So our change in y is going to be two. Let's do that again. Let's increase our x by one, change in x is equal to one."}, {"video_title": "Slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "So let's see, when x is equal to one, you have two times one plus three is going to be five. So our change in y is going to be two. Let's do that again. Let's increase our x by one, change in x is equal to one. So then if we go from, if we're going to increase by one, we're going to go from x equals one to x equals two, what's our corresponding change in y? Well, when x is equal to two, two times two is four plus three is seven. Well, our change in y, our change in y is equal to two."}, {"video_title": "Slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "Let's increase our x by one, change in x is equal to one. So then if we go from, if we're going to increase by one, we're going to go from x equals one to x equals two, what's our corresponding change in y? Well, when x is equal to two, two times two is four plus three is seven. Well, our change in y, our change in y is equal to two. We went from five, when x went from one to two, y went from five to seven. So for every one that we increase x, y is increasing by two. So for this linear equation, our change in y over change in x is always going to be, our change in y is two when our change in x is one, or it's equal to two."}, {"video_title": "Slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "Well, our change in y, our change in y is equal to two. We went from five, when x went from one to two, y went from five to seven. So for every one that we increase x, y is increasing by two. So for this linear equation, our change in y over change in x is always going to be, our change in y is two when our change in x is one, or it's equal to two. Or we could say that our slope is equal to two. And let's just graph this to make sure that we understand this. So when x equals one, y is equal to five."}, {"video_title": "Slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "So for this linear equation, our change in y over change in x is always going to be, our change in y is two when our change in x is one, or it's equal to two. Or we could say that our slope is equal to two. And let's just graph this to make sure that we understand this. So when x equals one, y is equal to five. And actually we're going to have to graph five up here. So when x is equal to one, y is equal to, and actually this is a little bit higher. Let me clean this up a little bit."}, {"video_title": "Slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "So when x equals one, y is equal to five. And actually we're going to have to graph five up here. So when x is equal to one, y is equal to, and actually this is a little bit higher. Let me clean this up a little bit. So this one, let me erase that a little bit. Just like that. So that's y is equal to four, and this is y is equal to five."}, {"video_title": "Slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "Let me clean this up a little bit. So this one, let me erase that a little bit. Just like that. So that's y is equal to four, and this is y is equal to five. So when x is one, y is equal to five. So it's that point right over there. So our line is going to look, you only need two points to define a line."}, {"video_title": "Slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "So that's y is equal to four, and this is y is equal to five. So when x is one, y is equal to five. So it's that point right over there. So our line is going to look, you only need two points to define a line. Our line is going to look like, let me do this in this color right over here. Our line is going to look like, is going to look, is going to look something like, is going to look, let me see if I can, I didn't draw it completely at scale, but it's going to look something like this. This is the line, this is the line, y is equal to two x plus three."}, {"video_title": "Slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "So our line is going to look, you only need two points to define a line. Our line is going to look like, let me do this in this color right over here. Our line is going to look like, is going to look, is going to look something like, is going to look, let me see if I can, I didn't draw it completely at scale, but it's going to look something like this. This is the line, this is the line, y is equal to two x plus three. And we already figured out that its slope is equal to two. Our change, when our change in x is one, when our change in x is one, our change in y is two. If our change in x was negative one, if our change in x was negative one, our change in y is negative two."}, {"video_title": "Slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "This is the line, this is the line, y is equal to two x plus three. And we already figured out that its slope is equal to two. Our change, when our change in x is one, when our change in x is one, our change in y is two. If our change in x was negative one, if our change in x was negative one, our change in y is negative two. And you could see that. If from zero we went down one, if we went to negative one, then what's our y going to be? Two times negative one is negative two, plus three is one."}, {"video_title": "Slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "If our change in x was negative one, if our change in x was negative one, our change in y is negative two. And you could see that. If from zero we went down one, if we went to negative one, then what's our y going to be? Two times negative one is negative two, plus three is one. So we see that the point one, or the point negative one comma one is on the line as well. So the slope here, our change in y or change in x, if we're going from, between any two points on this line, is always going to be two. But where did you see two in this original equation?"}, {"video_title": "Slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "Two times negative one is negative two, plus three is one. So we see that the point one, or the point negative one comma one is on the line as well. So the slope here, our change in y or change in x, if we're going from, between any two points on this line, is always going to be two. But where did you see two in this original equation? Well, you see the two right over here. And when you write something in slope intercept form, where you explicitly solve for y, y is equal to some constant times x to the first power, plus some other constant, the second one is going to be your intercept, your y, or it's going to be a way to figure out the y intercept. The intercept itself is this point, the point at which the line intersects the y axis."}, {"video_title": "Slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "But where did you see two in this original equation? Well, you see the two right over here. And when you write something in slope intercept form, where you explicitly solve for y, y is equal to some constant times x to the first power, plus some other constant, the second one is going to be your intercept, your y, or it's going to be a way to figure out the y intercept. The intercept itself is this point, the point at which the line intersects the y axis. And then this two is going to represent your slope. And that makes sense, because every time you increase x by one, you're going to multiply that by two, so you're going to increase y by two. So this is just a kind of a, get your feet wet with the idea of slope intercept form, but you'll see, at least for me, this is the easiest form for me to think about what the graph of something looks like."}, {"video_title": "Slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "The intercept itself is this point, the point at which the line intersects the y axis. And then this two is going to represent your slope. And that makes sense, because every time you increase x by one, you're going to multiply that by two, so you're going to increase y by two. So this is just a kind of a, get your feet wet with the idea of slope intercept form, but you'll see, at least for me, this is the easiest form for me to think about what the graph of something looks like. Because if you were given another linear equation, let's say y is equal to negative x, negative x plus two. Well, immediately you say, okay, look, my y intercept is going to be the point zero comma two, so I'm going to intersect the y axis right at that point. And then I have a slope of, the coefficient here is really just negative one."}, {"video_title": "Slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "So this is just a kind of a, get your feet wet with the idea of slope intercept form, but you'll see, at least for me, this is the easiest form for me to think about what the graph of something looks like. Because if you were given another linear equation, let's say y is equal to negative x, negative x plus two. Well, immediately you say, okay, look, my y intercept is going to be the point zero comma two, so I'm going to intersect the y axis right at that point. And then I have a slope of, the coefficient here is really just negative one. So I have a slope of negative one. So as we increase x by one, we're going to decrease y by one. Increase x by one, you're going to decrease y by one."}, {"video_title": "Slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "And then I have a slope of, the coefficient here is really just negative one. So I have a slope of negative one. So as we increase x by one, we're going to decrease y by one. Increase x by one, you're going to decrease y by one. If you increase x by two, you're going to decrease y by two. And so our line is going to look something like this. Let me see if I can draw it relatively neatly."}, {"video_title": "Slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "Increase x by one, you're going to decrease y by one. If you increase x by two, you're going to decrease y by two. And so our line is going to look something like this. Let me see if I can draw it relatively neatly. It's going to look something, it's, let me, I can do it a little bit better than that. It's because my graph paper is hand-drawn. It's not ideal."}, {"video_title": "Slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "Let me see if I can draw it relatively neatly. It's going to look something, it's, let me, I can do it a little bit better than that. It's because my graph paper is hand-drawn. It's not ideal. But I think you get, you get the point. It's going to look something like that. So from slope-intercept form, very easy to figure out what the y intercept is, and very easy to figure out the slope."}, {"video_title": "Slope-intercept form   Algebra I   Khan Academy.mp3", "Sentence": "It's not ideal. But I think you get, you get the point. It's going to look something like that. So from slope-intercept form, very easy to figure out what the y intercept is, and very easy to figure out the slope. The slope here, slope here is negative one. That's this negative one right over here. And the y intercept, y intercept is the point zero comma two."}, {"video_title": "Example of solving a one-step equation   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "We have a plus 5 is equal to 54. Now all this is saying is that we have some number, some variable a, and if I add 5 to it, I will get 54. And you might be able to do this in your head, but we're going to do it a little bit more systematically, because that'll be helpful for you when we do more complicated problems. So in general, whenever you have an equation like this, we want to have the variable. We want this a all by itself on one side of the equation. We want to isolate it. It's already on the left-hand side, so let's try to get rid of everything else on the left-hand side."}, {"video_title": "Example of solving a one-step equation   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So in general, whenever you have an equation like this, we want to have the variable. We want this a all by itself on one side of the equation. We want to isolate it. It's already on the left-hand side, so let's try to get rid of everything else on the left-hand side. Well, the only other thing on the left-hand side is this positive 5. Well, the best way to get rid of a plus 5 or of a positive 5 is to subtract 5. So let's subtract 5."}, {"video_title": "Example of solving a one-step equation   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "It's already on the left-hand side, so let's try to get rid of everything else on the left-hand side. Well, the only other thing on the left-hand side is this positive 5. Well, the best way to get rid of a plus 5 or of a positive 5 is to subtract 5. So let's subtract 5. But remember, this says a plus 5 is equal to 54. If we want the equality to still hold, anything we do to the left-hand side of this equation, we have to do to the right side of the equation. So we also have to subtract 54 from the right."}, {"video_title": "Example of solving a one-step equation   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So let's subtract 5. But remember, this says a plus 5 is equal to 54. If we want the equality to still hold, anything we do to the left-hand side of this equation, we have to do to the right side of the equation. So we also have to subtract 54 from the right. So we have a plus 5 minus 5. Well, that's just going to be a plus 0, because you add 5 and you subtract 5, they cancel out. So a plus 0 is just a."}, {"video_title": "Example of solving a one-step equation   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So we also have to subtract 54 from the right. So we have a plus 5 minus 5. Well, that's just going to be a plus 0, because you add 5 and you subtract 5, they cancel out. So a plus 0 is just a. And then 54 minus 5, that is 49. And we're done. We have solved for a. a is equal to 49."}, {"video_title": "Example of solving a one-step equation   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So a plus 0 is just a. And then 54 minus 5, that is 49. And we're done. We have solved for a. a is equal to 49. And now we can check it. And we can check it by just substituting 49 back for a in our original equation. So instead of writing a plus 5 is equal to 54, let's see if 49 plus 5 is equal to 54."}, {"video_title": "Example of solving a one-step equation   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "We have solved for a. a is equal to 49. And now we can check it. And we can check it by just substituting 49 back for a in our original equation. So instead of writing a plus 5 is equal to 54, let's see if 49 plus 5 is equal to 54. So we're just substituting it back in. 49 plus 5 is equal to 54. We're trying to check this."}, {"video_title": "Example of solving a one-step equation   Linear equations   Algebra I   Khan Academy.mp3", "Sentence": "So instead of writing a plus 5 is equal to 54, let's see if 49 plus 5 is equal to 54. So we're just substituting it back in. 49 plus 5 is equal to 54. We're trying to check this. 49 plus 5 is 54. And that, indeed, is equal to 54. So it all checks out."}, {"video_title": "How to write algebraic expressions with parentheses   Algebra I   Khan Academy.mp3", "Sentence": "So let's do it step by step. First we're going to have this expression, negative 5 plus something. So it's going to be negative 5 plus the quantity of 4 times x. The quantity of 4 times x, well that's just going to be 4x. So it's going to be negative 5 plus 4x. So that's this expression up here. Now take the product of negative 8."}, {"video_title": "How to write algebraic expressions with parentheses   Algebra I   Khan Academy.mp3", "Sentence": "The quantity of 4 times x, well that's just going to be 4x. So it's going to be negative 5 plus 4x. So that's this expression up here. Now take the product of negative 8. So we're going to take negative 8 and we're going to multiply the product of negative 8 and that expression. So we're going to take negative 8 and multiply it. So that expression is this thing right over here."}, {"video_title": "How to write algebraic expressions with parentheses   Algebra I   Khan Academy.mp3", "Sentence": "Now take the product of negative 8. So we're going to take negative 8 and we're going to multiply the product of negative 8 and that expression. So we're going to take negative 8 and multiply it. So that expression is this thing right over here. So the product, if we say the product of negative 8 and that expression, it's going to be negative 8 times that expression. That expression is negative 5 plus 4x. So that's negative 8, that's that expression."}, {"video_title": "How to write algebraic expressions with parentheses   Algebra I   Khan Academy.mp3", "Sentence": "So that expression is this thing right over here. So the product, if we say the product of negative 8 and that expression, it's going to be negative 8 times that expression. That expression is negative 5 plus 4x. So that's negative 8, that's that expression. The product of the two, so we could put a multiplication sign there or we could just leave that out. In implicit it would mean multiplication. Take the product of negative 8 and that expression and then add 6."}, {"video_title": "How to write algebraic expressions with parentheses   Algebra I   Khan Academy.mp3", "Sentence": "So that's negative 8, that's that expression. The product of the two, so we could put a multiplication sign there or we could just leave that out. In implicit it would mean multiplication. Take the product of negative 8 and that expression and then add 6. So that's and then add 6. So that would be then adding 6 right over here. So we could write it as negative 8, open parentheses, negative 5 plus 4x and then add 6."}, {"video_title": "How to write algebraic expressions with parentheses   Algebra I   Khan Academy.mp3", "Sentence": "Take the product of negative 8 and that expression and then add 6. So that's and then add 6. So that would be then adding 6 right over here. So we could write it as negative 8, open parentheses, negative 5 plus 4x and then add 6. Let's do one more. First consider the expression, the sum of 7 and, so that's going to be 7 plus something, and the product of negative 2 and x. The product of negative 2 and x is negative 2x."}, {"video_title": "How to write algebraic expressions with parentheses   Algebra I   Khan Academy.mp3", "Sentence": "So we could write it as negative 8, open parentheses, negative 5 plus 4x and then add 6. Let's do one more. First consider the expression, the sum of 7 and, so that's going to be 7 plus something, and the product of negative 2 and x. The product of negative 2 and x is negative 2x. Negative 2x, so 7 plus negative 2x. We could write that as 7 minus 2x. So this is equal to 7 minus 2x."}, {"video_title": "How to write algebraic expressions with parentheses   Algebra I   Khan Academy.mp3", "Sentence": "The product of negative 2 and x is negative 2x. Negative 2x, so 7 plus negative 2x. We could write that as 7 minus 2x. So this is equal to 7 minus 2x. These are the same expression. What expression would be 4 plus, so now we're saying 4 plus, 4 plus the quantity of 2 times that expression. So it's going to be 4 plus some quantity, 4 plus the quantity, I'll put that in parentheses, the quantity of 2 times, do this in magenta, of 2 times that expression or in yellow."}, {"video_title": "How to write algebraic expressions with parentheses   Algebra I   Khan Academy.mp3", "Sentence": "So this is equal to 7 minus 2x. These are the same expression. What expression would be 4 plus, so now we're saying 4 plus, 4 plus the quantity of 2 times that expression. So it's going to be 4 plus some quantity, 4 plus the quantity, I'll put that in parentheses, the quantity of 2 times, do this in magenta, of 2 times that expression or in yellow. 2 times that expression, that, let me do this in blue, that expression is this thing right over here. So the 4 plus the quantity of 2 times that expression, 2 times 7 minus 2x. And we are done."}, {"video_title": "Multiplying & dividing powers (integer exponents)   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "And I encourage you to pause the video and think about it on your own. Well, there's a couple of ways to do this. One, you say, oh look, I'm multiplying two things that have the same base, so this is going to be that base, four, and then I add the exponents. Four to the negative three plus five power, which is equal to four to the second power. And that's just a straightforward exponent property, but you can also think about why does that actually make sense? Four to the negative three power, that is one over four to the third power. Or you could view that as one over four times four times four and then four to the fifth, that's five fours being multiplied together, so it's times four times four times four times four times four."}, {"video_title": "Multiplying & dividing powers (integer exponents)   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "Four to the negative three plus five power, which is equal to four to the second power. And that's just a straightforward exponent property, but you can also think about why does that actually make sense? Four to the negative three power, that is one over four to the third power. Or you could view that as one over four times four times four and then four to the fifth, that's five fours being multiplied together, so it's times four times four times four times four times four. And so notice, when you multiply this out, you're gonna have five fours in the numerator and three fours in the denominator. And so three of these in the denominator are gonna cancel out with three of these in the numerator. And so you're gonna be left with five minus three, or negative three plus five fours."}, {"video_title": "Multiplying & dividing powers (integer exponents)   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "Or you could view that as one over four times four times four and then four to the fifth, that's five fours being multiplied together, so it's times four times four times four times four times four. And so notice, when you multiply this out, you're gonna have five fours in the numerator and three fours in the denominator. And so three of these in the denominator are gonna cancel out with three of these in the numerator. And so you're gonna be left with five minus three, or negative three plus five fours. So this four times four is the same thing as four squared. Now let's do one with variables. So let's say that you have a to the negative four power times a to the, let's say a squared."}, {"video_title": "Multiplying & dividing powers (integer exponents)   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "And so you're gonna be left with five minus three, or negative three plus five fours. So this four times four is the same thing as four squared. Now let's do one with variables. So let's say that you have a to the negative four power times a to the, let's say a squared. What is that going to be? Well, once again, you have the same base. In this case, it's a."}, {"video_title": "Multiplying & dividing powers (integer exponents)   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "So let's say that you have a to the negative four power times a to the, let's say a squared. What is that going to be? Well, once again, you have the same base. In this case, it's a. And so, and since I'm multiplying them, you can just add the exponents. So it's gonna be a to the negative four plus two power, which is equal to a to the negative two power. And once again, it should make sense."}, {"video_title": "Multiplying & dividing powers (integer exponents)   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "In this case, it's a. And so, and since I'm multiplying them, you can just add the exponents. So it's gonna be a to the negative four plus two power, which is equal to a to the negative two power. And once again, it should make sense. This right over here, that is one over a times a times a times a. And then this is times a times a. So that cancels with that, that cancels with that, and you're still left with one over a times a, which is the same thing as a to the negative two power."}, {"video_title": "Multiplying & dividing powers (integer exponents)   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "And once again, it should make sense. This right over here, that is one over a times a times a times a. And then this is times a times a. So that cancels with that, that cancels with that, and you're still left with one over a times a, which is the same thing as a to the negative two power. Now let's do it with some quotients. So what if I were to ask you, what is 12 to the negative seven divided by 12 to the negative five power? Well, when you're dividing, you subtract exponents if you have the same base."}, {"video_title": "Multiplying & dividing powers (integer exponents)   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "So that cancels with that, that cancels with that, and you're still left with one over a times a, which is the same thing as a to the negative two power. Now let's do it with some quotients. So what if I were to ask you, what is 12 to the negative seven divided by 12 to the negative five power? Well, when you're dividing, you subtract exponents if you have the same base. So this is going to be equal to 12 to the negative seven minus negative five power. You're subtracting the bottom exponent. And so this is going to be equal to 12 to the, well, subtracting a negative is the same thing as adding the positive, a 12 to the negative two power."}, {"video_title": "Multiplying & dividing powers (integer exponents)   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "Well, when you're dividing, you subtract exponents if you have the same base. So this is going to be equal to 12 to the negative seven minus negative five power. You're subtracting the bottom exponent. And so this is going to be equal to 12 to the, well, subtracting a negative is the same thing as adding the positive, a 12 to the negative two power. And once again, we just have to think about why does this actually make sense? Well, you can actually rewrite this. 12 to the negative seven divided by 12 to the negative five, that's the same thing as 12 to the negative seven times 12 to the fifth power."}, {"video_title": "Multiplying & dividing powers (integer exponents)   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "And so this is going to be equal to 12 to the, well, subtracting a negative is the same thing as adding the positive, a 12 to the negative two power. And once again, we just have to think about why does this actually make sense? Well, you can actually rewrite this. 12 to the negative seven divided by 12 to the negative five, that's the same thing as 12 to the negative seven times 12 to the fifth power. If we take the reciprocal of, if we take the reciprocal of this right over here, you would make the exponent positive. And then you get exactly what we were doing in those previous examples with products. And so let's just do one more with variables for good measure."}, {"video_title": "Multiplying & dividing powers (integer exponents)   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "12 to the negative seven divided by 12 to the negative five, that's the same thing as 12 to the negative seven times 12 to the fifth power. If we take the reciprocal of, if we take the reciprocal of this right over here, you would make the exponent positive. And then you get exactly what we were doing in those previous examples with products. And so let's just do one more with variables for good measure. Let's say I have x to the negative 20th power divided by x to the fifth power. Well, once again, we have the same base and we're taking a quotient. So this is going to be x to the negative 20 minus five because we have this one right over here in the denominator."}, {"video_title": "Multiplying & dividing powers (integer exponents)   Mathematics I   High School Math   Khan Academy.mp3", "Sentence": "And so let's just do one more with variables for good measure. Let's say I have x to the negative 20th power divided by x to the fifth power. Well, once again, we have the same base and we're taking a quotient. So this is going to be x to the negative 20 minus five because we have this one right over here in the denominator. So this is going to be equal to x to the negative 25th power. And once again, you could view our original expression as x to the negative 20th. And having an x to the fifth in the denominator, dividing by x to the fifth, is the same thing as multiplying by x to the negative five."}, {"video_title": "Linear equation word problems.mp3", "Sentence": "He set the temperature as high as it could go. Q represents the temperature in Quinn's home in degrees Celsius after T minutes. They say Q is equal to 15 plus 0.4 T. What was the temperature when Quinn returned from vacation? So pause this video and see if you can work this out on your own. All right, so they wanna know the temperature, and you might get a little confused, say hey, maybe T is for temperature. No, T is time in minutes. Temperature is Q. Q represents the temperature."}, {"video_title": "Linear equation word problems.mp3", "Sentence": "So pause this video and see if you can work this out on your own. All right, so they wanna know the temperature, and you might get a little confused, say hey, maybe T is for temperature. No, T is time in minutes. Temperature is Q. Q represents the temperature. So they really wanna know is what was Q when Quinn returned from vacation? Well, right when Quinn returned from vacation, that is when T is equal to zero. So this is equivalent to saying what is Q, our temperature, when zero minutes have elapsed?"}, {"video_title": "Linear equation word problems.mp3", "Sentence": "Temperature is Q. Q represents the temperature. So they really wanna know is what was Q when Quinn returned from vacation? Well, right when Quinn returned from vacation, that is when T is equal to zero. So this is equivalent to saying what is Q, our temperature, when zero minutes have elapsed? Well, if you go back to this original equation, we see that Q is equal to 15 plus 0.4 times the amount of elapsed time in minutes, so that's times zero. So that's just going to be 15 degrees Celsius. If you're familiar with slope-intercept form, you could think of it as our temperature is equal to 0.4 times the elapsed time plus 15."}, {"video_title": "Linear equation word problems.mp3", "Sentence": "So this is equivalent to saying what is Q, our temperature, when zero minutes have elapsed? Well, if you go back to this original equation, we see that Q is equal to 15 plus 0.4 times the amount of elapsed time in minutes, so that's times zero. So that's just going to be 15 degrees Celsius. If you're familiar with slope-intercept form, you could think of it as our temperature is equal to 0.4 times the elapsed time plus 15. So T equals zero, you're left with just this term, which in many cases we view as our y-intercept. What is going on right when we're just getting started, right when our horizontal variable is equal to zero, and our horizontal variable in this situation is elapsed time. How much does the temperature increase every minute?"}, {"video_title": "Linear equation word problems.mp3", "Sentence": "If you're familiar with slope-intercept form, you could think of it as our temperature is equal to 0.4 times the elapsed time plus 15. So T equals zero, you're left with just this term, which in many cases we view as our y-intercept. What is going on right when we're just getting started, right when our horizontal variable is equal to zero, and our horizontal variable in this situation is elapsed time. How much does the temperature increase every minute? There's a couple of ways you could think about this. One, if you recognize this as slope-intercept form, you could see that 0.4 is the slope. So that says for every one minute change in time, you're going to have an increase in temperature by 0.4 degrees Celsius."}, {"video_title": "Linear equation word problems.mp3", "Sentence": "How much does the temperature increase every minute? There's a couple of ways you could think about this. One, if you recognize this as slope-intercept form, you could see that 0.4 is the slope. So that says for every one minute change in time, you're going to have an increase in temperature by 0.4 degrees Celsius. So you could do it that way. You could try out some values. You could say, all right, let me think about what Q is going to be based on T. So time T equals zero, right when he got home, we already figured out that the temperature is 15 degrees Celsius."}, {"video_title": "Linear equation word problems.mp3", "Sentence": "So that says for every one minute change in time, you're going to have an increase in temperature by 0.4 degrees Celsius. So you could do it that way. You could try out some values. You could say, all right, let me think about what Q is going to be based on T. So time T equals zero, right when he got home, we already figured out that the temperature is 15 degrees Celsius. At T equals one, what happens? Well, it's going to be 15 plus 0.4 times one. Well, that's just going to be 15.4."}, {"video_title": "Linear equation word problems.mp3", "Sentence": "You could say, all right, let me think about what Q is going to be based on T. So time T equals zero, right when he got home, we already figured out that the temperature is 15 degrees Celsius. At T equals one, what happens? Well, it's going to be 15 plus 0.4 times one. Well, that's just going to be 15.4. Notice, when we increased our time by one, our temperature increased by 0.4 degrees Celsius, by the slope. And it would happen again. If we increased time by another minute, if we go from one to two, we would get to 15.8."}, {"video_title": "Linear equation word problems.mp3", "Sentence": "Well, that's just going to be 15.4. Notice, when we increased our time by one, our temperature increased by 0.4 degrees Celsius, by the slope. And it would happen again. If we increased time by another minute, if we go from one to two, we would get to 15.8. We would increase temperature by another 0.4. How much will the temperature increase if Quinn leaves the heat on for 20 minutes? Pause the video and see if you can answer that."}, {"video_title": "Linear equation word problems.mp3", "Sentence": "If we increased time by another minute, if we go from one to two, we would get to 15.8. We would increase temperature by another 0.4. How much will the temperature increase if Quinn leaves the heat on for 20 minutes? Pause the video and see if you can answer that. All right, now we have to be careful here. They're not asking us what is the temperature after 20 minutes. They're saying how much will the temperature increase if he leaves the heat on for 20 minutes."}, {"video_title": "Linear equation word problems.mp3", "Sentence": "Pause the video and see if you can answer that. All right, now we have to be careful here. They're not asking us what is the temperature after 20 minutes. They're saying how much will the temperature increase if he leaves the heat on for 20 minutes. If we just want to know what is the temperature after 20 minutes, we would just say, okay, what is Q when T is equal to 20? So it'd be 15 plus 0.4 times 20. 0.4 times 20 is eight."}, {"video_title": "Linear equation word problems.mp3", "Sentence": "They're saying how much will the temperature increase if he leaves the heat on for 20 minutes. If we just want to know what is the temperature after 20 minutes, we would just say, okay, what is Q when T is equal to 20? So it'd be 15 plus 0.4 times 20. 0.4 times 20 is eight. Eight plus 15 is 23. So it's 23 degrees Celsius after 20 minutes. But that's not what they're asking us."}, {"video_title": "Linear equation word problems.mp3", "Sentence": "0.4 times 20 is eight. Eight plus 15 is 23. So it's 23 degrees Celsius after 20 minutes. But that's not what they're asking us. They're asking how much will the temperature increase? Well, what did we start from? We started from 15 degrees Celsius, and now after 20 minutes, we have gone to 23 degrees Celsius so we have increased by eight degrees Celsius."}, {"video_title": "Sum and product of rational numbers   Rational and irrational numbers   Algebra I   Khan Academy.mp3", "Sentence": "What I want to do in this video is think about whether the product or sums of rational numbers are definitely going to be rational. So let's just first think about the product of rational numbers. So if I have one rational number, and actually let me instead of writing out the word rational, let me just represent it as a ratio of two integers. So I have one rational number right over there. I can represent it as a over b. And I'm going to multiply it times another rational number. And I can represent that as the ratio of two integers, m and n. And so what is this product going to be?"}, {"video_title": "Sum and product of rational numbers   Rational and irrational numbers   Algebra I   Khan Academy.mp3", "Sentence": "So I have one rational number right over there. I can represent it as a over b. And I'm going to multiply it times another rational number. And I can represent that as the ratio of two integers, m and n. And so what is this product going to be? Well, the numerator, I'm going to have am. I'm going to have a times m. And the denominator, I'm going to have b times n. b times n. Well, a is an integer, m is an integer, so you have an integer in the numerator. And b is an integer, and n is an integer, so you have an integer in the denominator."}, {"video_title": "Sum and product of rational numbers   Rational and irrational numbers   Algebra I   Khan Academy.mp3", "Sentence": "And I can represent that as the ratio of two integers, m and n. And so what is this product going to be? Well, the numerator, I'm going to have am. I'm going to have a times m. And the denominator, I'm going to have b times n. b times n. Well, a is an integer, m is an integer, so you have an integer in the numerator. And b is an integer, and n is an integer, so you have an integer in the denominator. So now the product is the ratio of two integers right over here. So the product is also rational. So this thing is also rational."}, {"video_title": "Sum and product of rational numbers   Rational and irrational numbers   Algebra I   Khan Academy.mp3", "Sentence": "And b is an integer, and n is an integer, so you have an integer in the denominator. So now the product is the ratio of two integers right over here. So the product is also rational. So this thing is also rational. So if you give me the product of any two rational numbers, you're going to end up with a rational number. Let's see if the same thing is true for the sum of two rational numbers. So let's say my first rational number is a over b, and that my second can be represented as a over b."}, {"video_title": "Sum and product of rational numbers   Rational and irrational numbers   Algebra I   Khan Academy.mp3", "Sentence": "So this thing is also rational. So if you give me the product of any two rational numbers, you're going to end up with a rational number. Let's see if the same thing is true for the sum of two rational numbers. So let's say my first rational number is a over b, and that my second can be represented as a over b. And my second rational number can be represented as m over n. Well, how would I add these two? Well, I can find a common denominator, and the easiest one is b times n. So let me multiply this fraction, let me multiply this one times n in the numerator and n in the denominator. And let me multiply this one times b in the numerator and b in the denominator."}, {"video_title": "Sum and product of rational numbers   Rational and irrational numbers   Algebra I   Khan Academy.mp3", "Sentence": "So let's say my first rational number is a over b, and that my second can be represented as a over b. And my second rational number can be represented as m over n. Well, how would I add these two? Well, I can find a common denominator, and the easiest one is b times n. So let me multiply this fraction, let me multiply this one times n in the numerator and n in the denominator. And let me multiply this one times b in the numerator and b in the denominator. Now we've written them so they have a common denominator of bn. And so this is going to be equal to an plus bm, all of that over b times n. So b times n we've just already talked about. This is definitely going to be an integer right over here."}, {"video_title": "Sum and product of rational numbers   Rational and irrational numbers   Algebra I   Khan Academy.mp3", "Sentence": "And let me multiply this one times b in the numerator and b in the denominator. Now we've written them so they have a common denominator of bn. And so this is going to be equal to an plus bm, all of that over b times n. So b times n we've just already talked about. This is definitely going to be an integer right over here. And then what do we have up here? Well, we have a times n, which is an integer. b times m is another integer."}, {"video_title": "Sum and product of rational numbers   Rational and irrational numbers   Algebra I   Khan Academy.mp3", "Sentence": "This is definitely going to be an integer right over here. And then what do we have up here? Well, we have a times n, which is an integer. b times m is another integer. The sum of two integers is going to be an integer. So you have an integer over an integer. You have the ratio of two integers."}, {"video_title": "Sum and product of rational numbers   Rational and irrational numbers   Algebra I   Khan Academy.mp3", "Sentence": "b times m is another integer. The sum of two integers is going to be an integer. So you have an integer over an integer. You have the ratio of two integers. So the sum of two rational numbers is going to give you another. So this one right over here was rational, and this one right over here is rational. So you take the product of two rational numbers, you get a rational number."}, {"video_title": "Graphing piecewise function   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "And this is a piecewise function. It's defined as essentially different lines. You see this right over here, even with all the decimals and the negative signs, this is essentially a line. It's defined by this line over this interval for x, this line over this interval of x, and this line over this interval of x. I want to see if we can graph it. I encourage you, especially if you have some graph paper, to see if you could graph this on your own first before I work through it. So let's think about this first interval. If when negative 10 is less than or equal to x, which is less than negative two, then our function is defined by negative 0.125x plus 4.75."}, {"video_title": "Graphing piecewise function   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "It's defined by this line over this interval for x, this line over this interval of x, and this line over this interval of x. I want to see if we can graph it. I encourage you, especially if you have some graph paper, to see if you could graph this on your own first before I work through it. So let's think about this first interval. If when negative 10 is less than or equal to x, which is less than negative two, then our function is defined by negative 0.125x plus 4.75. So this is going to be a line, a downward-sloping line. And the easiest way I can think about graphing it is let's just plot the endpoints here and then draw the line. So when x is equal to 10, so when, or sorry, when x is equal to negative 10, so we would have negative zero, actually let me write it this way."}, {"video_title": "Graphing piecewise function   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "If when negative 10 is less than or equal to x, which is less than negative two, then our function is defined by negative 0.125x plus 4.75. So this is going to be a line, a downward-sloping line. And the easiest way I can think about graphing it is let's just plot the endpoints here and then draw the line. So when x is equal to 10, so when, or sorry, when x is equal to negative 10, so we would have negative zero, actually let me write it this way. Let me do it over here where I do the, so we're gonna have negative 0.125 times negative 10 plus 4.75. That is going to be equal to, let's see, the negative times the negative is a positive, and then 10 times this is going to be, it's going to be 1.25 plus 4.75. That is going to be equal to six."}, {"video_title": "Graphing piecewise function   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "So when x is equal to 10, so when, or sorry, when x is equal to negative 10, so we would have negative zero, actually let me write it this way. Let me do it over here where I do the, so we're gonna have negative 0.125 times negative 10 plus 4.75. That is going to be equal to, let's see, the negative times the negative is a positive, and then 10 times this is going to be, it's going to be 1.25 plus 4.75. That is going to be equal to six. So we're going to have the point negative 10 comma six. And that point, and it includes, so x is defined there, it's less than or equal to. And then we go all the way to negative two."}, {"video_title": "Graphing piecewise function   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "That is going to be equal to six. So we're going to have the point negative 10 comma six. And that point, and it includes, so x is defined there, it's less than or equal to. And then we go all the way to negative two. So when x is equal to negative two, we have negative 0.125 times negative two plus 4.75 is equal to, see, negative times negative is positive. Two times this is going to be, is going to be positive 0.25 plus 4.75. It's going to be equal to positive five."}, {"video_title": "Graphing piecewise function   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "And then we go all the way to negative two. So when x is equal to negative two, we have negative 0.125 times negative two plus 4.75 is equal to, see, negative times negative is positive. Two times this is going to be, is going to be positive 0.25 plus 4.75. It's going to be equal to positive five. Now, we might be tempted, we might be tempted to just circle in this dot over here. But remember, this interval does not include negative two. It's up to it including, it's up to negative two, not including."}, {"video_title": "Graphing piecewise function   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "It's going to be equal to positive five. Now, we might be tempted, we might be tempted to just circle in this dot over here. But remember, this interval does not include negative two. It's up to it including, it's up to negative two, not including. So I'm gonna put a little open circle there, and then I'm gonna draw the line. And then I'm gonna draw, and I'm gonna draw the line. I am going to draw my best attempt, my best attempt at the line."}, {"video_title": "Graphing piecewise function   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "It's up to it including, it's up to negative two, not including. So I'm gonna put a little open circle there, and then I'm gonna draw the line. And then I'm gonna draw, and I'm gonna draw the line. I am going to draw my best attempt, my best attempt at the line. Now let's do the next interval. The next interval, this one's a lot more straightforward. We start at x equals negative two."}, {"video_title": "Graphing piecewise function   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "I am going to draw my best attempt, my best attempt at the line. Now let's do the next interval. The next interval, this one's a lot more straightforward. We start at x equals negative two. When x equals negative two, negative two plus seven is, negative two plus seven is five. So negative two, so negative two comma five. So it actually includes that point right over there."}, {"video_title": "Graphing piecewise function   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "We start at x equals negative two. When x equals negative two, negative two plus seven is, negative two plus seven is five. So negative two, so negative two comma five. So it actually includes that point right over there. So we're actually able to fill it in. And then when x is negative one, negative one plus seven is going to be positive six. Positive six, but we're not including x equals negative one, up to and including."}, {"video_title": "Graphing piecewise function   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "So it actually includes that point right over there. So we're actually able to fill it in. And then when x is negative one, negative one plus seven is going to be positive six. Positive six, but we're not including x equals negative one, up to and including. So it's going to be, it's going to be right over here. When x is negative one, we are approaching, or as x approaches negative one, we're approaching negative one plus seven is six. So that's that interval right over there."}, {"video_title": "Graphing piecewise function   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "Positive six, but we're not including x equals negative one, up to and including. So it's going to be, it's going to be right over here. When x is negative one, we are approaching, or as x approaches negative one, we're approaching negative one plus seven is six. So that's that interval right over there. And now let's look at this last interval. This last interval, when x is negative one, you're going to have, well this is just going to be positive 12 over 11, because we're multiplying it by negative one, plus 54 over 11, which is equal to 66 over 11, which is equal to positive six. So we're able to fill in that right over there."}, {"video_title": "Graphing piecewise function   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "So that's that interval right over there. And now let's look at this last interval. This last interval, when x is negative one, you're going to have, well this is just going to be positive 12 over 11, because we're multiplying it by negative one, plus 54 over 11, which is equal to 66 over 11, which is equal to positive six. So we're able to fill in that right over there. And then when x is equal to 10, you have negative 120 over 11. I just multiplied this times 10. 12 times 10 is 120, and we have the negative, plus 54 over 11."}, {"video_title": "Graphing piecewise function   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "So we're able to fill in that right over there. And then when x is equal to 10, you have negative 120 over 11. I just multiplied this times 10. 12 times 10 is 120, and we have the negative, plus 54 over 11. So this is the same thing. This is going to be, what is this? This is negative 66 over 11."}, {"video_title": "Graphing piecewise function   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "12 times 10 is 120, and we have the negative, plus 54 over 11. So this is the same thing. This is going to be, what is this? This is negative 66 over 11. Is that right? Let's see, if you, yeah, that is negative 66 over 11, which is equal to negative six. So when x is equal to 10, our function is equal to negative negative six."}, {"video_title": "Graphing piecewise function   Functions and their graphs   Algebra II   Khan Academy.mp3", "Sentence": "This is negative 66 over 11. Is that right? Let's see, if you, yeah, that is negative 66 over 11, which is equal to negative six. So when x is equal to 10, our function is equal to negative negative six. And so this one actually doesn't have any jumps in it. It could've, but we see. So there we have it."}, {"video_title": "How to find the average rate of change from a table   Functions   Algebra I   Khan Academy.mp3", "Sentence": "So this is x is equal to negative 5. When x is equal to negative 5, y of x is equal to 6. And when x is equal to negative 2, y of x is equal to 0. So to figure out the average rate of change of y of x with respect to x, this is going to be the change in y of x over the change of x of that interval. And the shorthand for change is this triangle symbol, delta. Delta y, I'll just write y, I could write delta y of x, it's delta y, change in y over our change in x. That's going to be our average rate of change over this interval."}, {"video_title": "How to find the average rate of change from a table   Functions   Algebra I   Khan Academy.mp3", "Sentence": "So to figure out the average rate of change of y of x with respect to x, this is going to be the change in y of x over the change of x of that interval. And the shorthand for change is this triangle symbol, delta. Delta y, I'll just write y, I could write delta y of x, it's delta y, change in y over our change in x. That's going to be our average rate of change over this interval. So how much did y change over this interval? So y went from a 6 to a 0. So let's say that we can kind of view this as our end point right over here."}, {"video_title": "How to find the average rate of change from a table   Functions   Algebra I   Khan Academy.mp3", "Sentence": "That's going to be our average rate of change over this interval. So how much did y change over this interval? So y went from a 6 to a 0. So let's say that we can kind of view this as our end point right over here. So this is our end, this is our start. And we could have done it the other way around, we would get a consistent result. But since this is higher up on the list, let's call this the start, and the x is a lower value, we'll call that our start, this is our end."}, {"video_title": "How to find the average rate of change from a table   Functions   Algebra I   Khan Academy.mp3", "Sentence": "So let's say that we can kind of view this as our end point right over here. So this is our end, this is our start. And we could have done it the other way around, we would get a consistent result. But since this is higher up on the list, let's call this the start, and the x is a lower value, we'll call that our start, this is our end. So we start at 6, we end at 0. So our change in y is going to be negative 6. We went down by 6 in the y direction."}, {"video_title": "How to find the average rate of change from a table   Functions   Algebra I   Khan Academy.mp3", "Sentence": "But since this is higher up on the list, let's call this the start, and the x is a lower value, we'll call that our start, this is our end. So we start at 6, we end at 0. So our change in y is going to be negative 6. We went down by 6 in the y direction. It's negative 6, you could say that's 0 minus 6. And our change in x, well we were at negative 5 and we go up to negative 2, we increased by 3. So we increased by 3."}, {"video_title": "How to find the average rate of change from a table   Functions   Algebra I   Khan Academy.mp3", "Sentence": "We went down by 6 in the y direction. It's negative 6, you could say that's 0 minus 6. And our change in x, well we were at negative 5 and we go up to negative 2, we increased by 3. So we increased by 3. So when we increased x by 3, we decreased y of x by 6. Or if we want to simplify this right over here, negative 6 over 3 is the same thing as negative 2. So our average rate of change of y of x over the interval from negative 5 to negative 2 is negative 2."}, {"video_title": "How to find the average rate of change from a table   Functions   Algebra I   Khan Academy.mp3", "Sentence": "So we increased by 3. So when we increased x by 3, we decreased y of x by 6. Or if we want to simplify this right over here, negative 6 over 3 is the same thing as negative 2. So our average rate of change of y of x over the interval from negative 5 to negative 2 is negative 2. Every time on average x increased 1, y went down by negative 2."}, {"video_title": "How to find the average rate of change from a table   Functions   Algebra I   Khan Academy.mp3", "Sentence": "So our average rate of change of y of x over the interval from negative 5 to negative 2 is negative 2. Every time on average x increased 1, y went down by negative 2."}]